Chapter 4 Abstract Linear Spaces: Exercises - Linear Algebra with Python

Structure of each exercise: problem + hint (with a solution dropdown) → Python numerical verification (with execution output)

Some verification code blocks are shared by several exercises.

import numpy as np
np.set_printoptions(precision=4, suppress=True, linewidth=100)

# ── Visual hierarchy helpers ──────────────────────────────
def print_header(title):
    print("=" * 60)
    print(f"  {title}")
    print("=" * 60)

def print_step(step, desc):
    print(f"\n▶ Step {step}: {desc}")
    print("-" * 40)

def compare_print(label, actual, expected_desc):
    print(f"\n[{label}]")
    print(f"  Computed value: {actual}")
    print(f"  Expected value: {expected_desc}")

# ── Linear algebra utilities ─────────────────────────────
def rref(A, tol=1e-10):
    """Reduce A to RREF; return (RREF matrix, pivot column indices)"""
    M = A.astype(float).copy()
    r, pivot_cols = 0, []
    for c in range(M.shape[1]):
        idx = np.argmax(np.abs(M[r:, c])) + r
        if abs(M[idx, c]) < tol:
            continue
        M[[r, idx]] = M[[idx, r]]
        M[r] /= M[r, c]
        for i in range(len(M)):
            if i != r:
                M[i] -= M[i, c] * M[r]
        pivot_cols.append(c)
        r += 1
        if r == M.shape[0]:
            break
    return M, pivot_cols

def rank(A):
    _, piv = rref(np.array(A, dtype=float))
    return len(piv)

def null_basis(A, tol=1e-10):
    """Return a basis of the null space of A"""
    R, piv = rref(np.array(A, dtype=float))
    n = A.shape[1]
    free = [c for c in range(n) if c not in piv]
    basis = []
    for f in free:
        v = np.zeros(n)
        v[f] = 1.0
        for row, p in enumerate(piv):
            if row < R.shape[0]:
                v[p] = -R[row, f]
        basis.append(v)
    return basis

print("Helper functions loaded")

§4.1.1 Vector Spaces, Subspaces, and Spans¶

Vector Space Axioms and Subspaces¶

Exercise 1 (True or False: Vector Space Axioms)

(True or false) Decide whether each of the following statements is true. If it is true, briefly explain why; if it is false, give a counterexample or explain what is wrong.

In $\mathbb{R}^3$ , the sum $\mathbf{u} + \mathbf{v}$ of the vectors $\mathbf{u} = (1,3,-2)$ and $\mathbf{v} = (4,-3,-2)$ is still a vector in $\mathbb{R}^3$ .
In $\mathbb{R}^n$ , every vector $\mathbf{v}$ satisfies $0 \cdot \mathbf{v} = \mathbf{0}$ (where the 0 on the left is a scalar and the $\mathbf{0}$ on the right is the zero vector).
In $\mathbb{R}^4$ , $\mathbf{0} = (0,0,0,0)$ is the unique zero vector.
In $\mathbb{R}^n$ , for every vector $\mathbf{v}$ , its negative $-\mathbf{v}$ is unique and is not equal to $\mathbf{v}$ .
In $\mathbb{R}^n$ , if $\mathbf{u} + \mathbf{v} = \mathbf{u} + \mathbf{w}$ , then necessarily $\mathbf{v} = \mathbf{w}$ .
In $\mathbb{R}^4$ , if a scalar $c$ and a vector $\mathbf{v}$ satisfy $c\mathbf{v} = \mathbf{0}$ , then $c = 0$ .
In $\mathbb{R}^n$ , $(-1)\mathbf{v}$ is exactly the negative $-\mathbf{v}$ of $\mathbf{v}$ .
In $\mathbb{R}^n$ , for any vectors $\mathbf{u}, \mathbf{v}, \mathbf{w}$ , if $\mathbf{u} + \mathbf{w} = \mathbf{v} + \mathbf{w}$ , then $\mathbf{w}=\mathbf{0}$ .
In $\mathbb{R}^n$ , for any vectors $\mathbf{u}, \mathbf{v}, \mathbf{w}$ , if $\mathbf{v} + \mathbf{w} = \mathbf{0}$ , then $\mathbf{w} = -\mathbf{v}$ .

Hint. Distinguish carefully between “if and only if” and “if … then”; note that a single concrete counterexample is enough.

# Python verification: exer-ch4-tf-axioms
# Paired with the exercise label above; numerical examples do not replace a general proof.
import numpy as np

# (a) u + v is still in R^3
u = np.array([1, 3, -2])
v = np.array([4, -3, -2])
print("u + v =", u + v, "→ in R^3 ✓")

# (b) 0 * v = 0
v_gen = np.array([3.7, -2.1, 5.5])
print("0 * v =", 0 * v_gen, "→ the zero vector ✓")

# (d) Uniqueness of the negative: taking v=0 gives -v=0=v
zero = np.zeros(3)
print("-0 =", -zero, "= 0, so -v can equal v ✓ (when v=0)")

# (f) A counterexample for c*v=0
c = 2.0
v_zero = np.zeros(4)
print(f"c={c}, v=0: c*v = {c * v_zero} = 0, but c ≠ 0")

# (h) From u+w = v+w the conclusion is u=v, not w=0
u2 = np.array([1, 0])
v2 = np.array([1, 0])
w2 = np.array([3, 5])
print("u+w =", u2+w2, "= v+w =", v2+w2, "→ u=v, but w ≠ 0")

Exercise 2 (Checking Subspaces Axiom by Axiom)

Use the definition of a vector space (closure under addition and scalar multiplication, together with the eight axioms) to check, one by one, whether each of the following sets is a subspace of the given vector space. If it is not, state which requirement is violated.

In $\mathbb{R}^2$ , $W_0=\{ (m,n) \mid m, n \in \mathbb{Z} \}$ (the integer lattice points).
In $\mathbb{R}^2$ , $W_1=\{ (t,t) \mid t \in \mathbb{R} \}$ (the line $y=x$ ).
In $\mathbb{R}^2$ , $W_2=\{ (x,y) \mid xy=0,\; x, y \in \mathbb{R} \}$ ; first describe this set, then decide.
In $\mathbb{R}^3$ , $W_3=\{ (x,y,z) \mid x+y+z=1 \}$ (the plane $x+y+z=1$ ).
In $\mathbb{R}^3$ , $W_4=\{ (x,y,z) \mid x \geq 0, y \geq 0, z \geq 0 \}$ (the first octant).
In $\mathbb{R}^3$ , $W_5 = \{ (x,y,0) \mid x, y \in \mathbb{R} \}$ (the $xy$ -plane).

Hint. The three requirements for a subspace: it contains the zero vector, it is closed under addition, and it is closed under scalar multiplication.

# Python verification: exer-ch4-subspace-axioms
# Paired with the exercise label above; numerical examples do not replace a general proof.
# W0: integer lattice points — a counterexample to closure under scalar multiplication
v_int = np.array([1, 1])
c_frac = 0.5
print(f"W0: {c_frac} * {v_int} = {c_frac * v_int}  ← not an integer vector, so W0 is not a subspace")

# W2: union of the two coordinate axes — a counterexample to closure under addition
v1_ax = np.array([1, 0])
v2_ax = np.array([0, 1])
s = v1_ax + v2_ax
print(f"W2: {v1_ax} + {v2_ax} = {s}, product xy = {s[0]*s[1]} ≠ 0 → not in W2")

# W3: the plane x+y+z=1 — does not contain the zero vector
check_zero = 0 + 0 + 0
print(f"W3: 0+0+0={check_zero} ≠ 1 → the zero vector is not in W3, so W3 is not a subspace")

# W1: the line y=x — check closure
t1, t2, c = 2, 3, -1.5
v_w2_1 = np.array([t1, t1])
v_w2_2 = np.array([t2, t2])
print(f"W1 addition: {v_w2_1} + {v_w2_2} = {v_w2_1+v_w2_2}, still on y=x ✓")
print(f"W1 scalar: {c} * {v_w2_1} = {c * v_w2_1}, still on y=x ✓")

# W5: the xy-plane
w6_v = np.array([1.2, -3.4, 0])
w6_u = np.array([5.6, 0.7, 0])
c6 = 3.0
print(f"W5 addition: {w6_v} + {w6_u} = {w6_v+w6_u}, third component = 0 ✓")
print(f"W5 scalar: {c6} * {w6_v} = {c6*w6_v}, third component = 0 ✓")

Exercise 3 (Proof of the Three-Condition Subspace Theorem)

(The subspace test) Let $W$ be a nonempty subset of a vector space $V$ . Prove that

$W$ is a subspace of $V$ if and only if:

$\mathbf{0} \in W$ (the zero vector lies in $W$ )
for all $\mathbf{u}, \mathbf{v} \in W$ , we have $\mathbf{u} + \mathbf{v} \in W$ (closure under addition)
for every $\mathbf{v} \in W$ and every scalar $c$ , we have $c\mathbf{v} \in W$ (closure under scalar multiplication)

Hint. For the “if” direction, use conditions 1, 2, 3 to verify one by one the closure properties and the eight axioms required of a vector space; the “only if” direction follows directly from the definition of a subspace.

Exercise 4 (Closure under Linear Combinations as an Equivalent Condition)

Let $W$ be a nonempty subset of a vector space $V$ . Prove that $W$ is a subspace if and only if:

for all $\mathbf{u}, \mathbf{v} \in W$ and all scalars $c, d$ , we have $c\mathbf{u} + d\mathbf{v} \in W$ (closure under linear combinations).

Hint. Use the three-condition theorem of the previous exercise, choosing special values of $c, d$ .

Exercise 5 (Applying the Subspace Theorem)

Use the three-condition subspace theorem to decide whether each of the following sets is a subspace. If it is, prove it; if not, state which condition is violated and give a counterexample.

In $\mathbb{R}^3$ , $W_0 = \{ (x,y,z) \mid x \leq 0 \}$ .
In $\mathbb{R}^4$ , $W_1 = \{ (a,b,c,d) \mid a + 2b = c + 3d \}$ .
In $\mathbb{R}^3$ , $W_2 = \{ (x,y,z) \mid x^2 + y^2 + z^2 = 0 \}$ .
In $\mathbb{R}^3$ , $W_3 = \{ (x,y,z) \mid x-2y=0,\; 3y+2z=0\}$ .
In $\mathbb{R}^4$ , $W_4= \{ (a,b,c,d) \mid a = b \text{ or } c = d \}$ .
Let $\mathbf{u} = (1,0,1)$ and $\mathbf{v}= (0,1,1)$ ; in $\mathbb{R}^3$ , $W_5 = \{ a\mathbf{u} + b\mathbf{v} \mid a, b \in \mathbb{R} \}$ .
In $\mathbb{R}^n$ , $W_6 = \{ \mathbf{v} \in \mathbb{R}^n \mid A\mathbf{v} = \mathbf{0} \}$ , where $A$ is an $m \times n$ matrix.
In $\mathbb{R}^n$ , $W_7 = \{ \mathbf{v} \in \mathbb{R}^n \mid A\mathbf{v} = \mathbf{b} \}$ , where $A$ is an $m \times n$ matrix and $\mathbf{b} \neq \mathbf{0}$ .

Hint. The solution set of a homogeneous system of linear equations ( $W_6$ ) is an important example of a subspace; pay attention to the logic of “or” ( $W_4$ ).

# Python verification: exer-ch4-subspace-check
# Paired with the exercise label above; numerical examples do not replace a general proof.
# W1: a+2b = c+3d is closed under addition and scalar multiplication
def check_W1(v):
    a, b, c, d = v
    return np.isclose(a + 2*b, c + 3*d)

v1 = np.array([1.0, 2.0, 5.0, 0.0])  # 1+4=5, 5+0=5 ✓
v2 = np.array([3.0, 0.0, 0.0, 1.0])  # 3+0=3, 0+3=3 ✓
print(f"v1 in W1: {check_W1(v1)}, v2 in W1: {check_W1(v2)}")
print(f"v1+v2 in W1: {check_W1(v1+v2)}")
print(f"2.5*v1 in W1: {check_W1(2.5*v1)}")

# W4: a=b or c=d — a counterexample to closure under addition
u4 = np.array([1.0, 1.0, 0.0, 1.0])   # a=b ✓
v4 = np.array([0.0, 1.0, 1.0, 1.0])   # c=d ✓
s4 = u4 + v4
def check_W4(v):
    a, b, c, d = v
    return np.isclose(a, b) or np.isclose(c, d)
print(f"\nW4: u in W4={check_W4(u4)}, v in W4={check_W4(v4)}")
print(f"u+v={s4}, in W4={check_W4(s4)} ← counterexample, so W4 is not a subspace")

# W6: the kernel Av=0
A6 = np.array([[1,2,3],[0,1,2],[0,0,0]], dtype=float)
null_vecs = null_basis(A6)
print(f"\nW6 (ker A): basis vectors = {null_vecs}")
# Check that a linear combination stays in the kernel
if null_vecs:
    combo = 2*null_vecs[0]
    print(f"2*ker_vec in kernel: {np.allclose(A6 @ combo, 0)}")

Exercise 6 (Intersections and Unions of Subspaces)

Suppose that $W_0$ and $W_1$ are subspaces of a vector space $V$ . Prove that the intersection $W_0 \cap W_1$ is a subspace of $V$ . Is the union $W_0 \cup W_1$ a subspace as well? (Hint: think about the earlier examples.)

Hint. For the union, try two different lines through the origin in $\mathbb{R}^2$ as a counterexample.

Spans¶

Exercise 7 (Membership in a Span and Its Geometric Description)

Answer the following questions.

(a) In $\mathbb{R}^2$ , let $\mathbf{v} =(2,3)$ .

Decide whether each of the following vectors lies in $\text{span}\{\mathbf{v}\}$ : $(4,6)$ , $(-4,-6)$ , $(1,1)$ .
Describe the geometric figure $\text{span}\{\mathbf{v}\}$ .

(b) In $\mathbb{R}^2$ , let $\mathbf{u} = (1,-2)$ and $\mathbf{v} =(-2,4)$ .

Decide whether the vectors $(1,0)$ , $(0,0)$ , $(3,-6)$ lie in $\text{span}\{\mathbf{u}, \mathbf{v}\}$ . If so, write each as a linear combination. Is the representation as a linear combination unique?
Describe the geometric figure $\text{span}\{\mathbf{u}, \mathbf{v}\}$ .

(c) In $\mathbb{R}^3$ , let $\mathbf{v} = (1,1,1)$ .

Decide whether the vectors $(2,2,2)$ and $(1,0,0)$ lie in $\text{span}\{\mathbf{v}, \mathbf{0}\}$ .
Describe the geometric figure $\text{span}\{\mathbf{v}, \mathbf{0}\}$ .

(d) In $\mathbb{R}^3$ , let $\mathbf{u} = (1,1,0)$ and $\mathbf{v} = (-1,1,1)$ .

Decide whether the vectors $(1,0,0)$ , $(0,1,0)$ , $(0,0,1)$ lie in $\text{span}\{\mathbf{u}, \mathbf{v}\}$ .
Describe the geometric figure $\text{span}\{\mathbf{u}, \mathbf{v}\}$ .

Hint. Deciding whether $\mathbf{w}\in\text{span}\{\mathbf{u},\mathbf{v}\}$ is equivalent to deciding whether the linear system $a\mathbf{u}+b\mathbf{v}=\mathbf{w}$ has a solution.

# Python verification: exer-ch4-span-basic
# Paired with the exercise label above; numerical examples do not replace a general proof.
# (a) span{(2,3)}
v = np.array([2, 3])
for w, name in [([4,6],'(4,6)'), ([-4,-6],'(-4,-6)'), ([1,1],'(1,1)')]:
    w = np.array(w, dtype=float)
    # solve c*v = w
    if not np.isclose(v[1]*w[0], v[0]*w[1]):
        print(f"{name} ∉ span{{v}}")
    else:
        c = w[0]/v[0] if not np.isclose(v[0], 0) else w[1]/v[1]
        print(f"{name} = {c:.4f} * v ∈ span{{v}}")

# (d) Vectors in span{u, v}
u7d = np.array([1.0, 1.0, 0.0])
v7d = np.array([-1.0, 1.0, 1.0])
A7d = np.column_stack([u7d, v7d])
for w, name in [([1,0,0],'(1,0,0)'), ([0,1,0],'(0,1,0)'), ([0,0,1],'(0,0,1)')]:
    w = np.array(w, dtype=float)
    try:
        x, res, _, _ = np.linalg.lstsq(A7d, w, rcond=None)
        in_span = np.isclose(A7d @ x, w).all()
        print(f"{name} ∈ span{{u,v}}: {in_span}")
    except:
        print(f"{name}: computation error")

Exercise 8 (Is a Vector in the Span?)

In $\mathbb{R}^3$ , let $\mathbf{u} = (1,2,-1)$ and $\mathbf{v} = (0,1,1)$ . Decide whether the vector $\mathbf{b} =(3,-1,1)$ lies in $\text{span}\{\mathbf{u}, \mathbf{v}\}$ .

Set $\mathbf{w} = x\mathbf{u} + y\mathbf{v}$ , write down the corresponding system of linear equations, and decide whether it has a solution.
Is $\mathbf{b}$ perpendicular to $\mathbf{u}$ and $\mathbf{v}$ ? For every $\mathbf{w} \in \text{span}\{\mathbf{u}, \mathbf{v}\}$ , must $\mathbf{w}$ be perpendicular to $\mathbf{b}$ ?

Hint. Use the linearity of the inner product: if $\mathbf{b}\perp\mathbf{u}$ and $\mathbf{b}\perp\mathbf{v}$ , then $\mathbf{b}$ is perpendicular to every vector in $\text{span}\{\mathbf{u},\mathbf{v}\}$ , and hence $\mathbf{b}∉\text{span}\{\mathbf{u},\mathbf{v}\}$ (unless $\mathbf{b}=\mathbf{0}$ ).

Exercise 9 (Is a Vector in a Span in

\mathbb{R}^4

?)

In $\mathbb{R}^4$ , let $\mathbf{u} = (1,1,1,0)$ and $\mathbf{v} = (0,1,-1,1)$ . Decide whether each of the following vectors lies in $\text{span}\{\mathbf{u}, \mathbf{v}\}$ :

$(2,5,-1,3)$
$(0,1,0,0)$

Hint. Row reduce the augmented matrix.

# Python verification: exer-ch4-span-check-r3, exer-ch4-span-r4
# Paired with the exercise label above; numerical examples do not replace a general proof.
# Q8: u=(1,2,-1), v=(0,1,1), b=(3,-1,1)
u8 = np.array([1.0, 2.0, -1.0])
v8 = np.array([0.0, 1.0, 1.0])
b8 = np.array([3.0, -1.0, 1.0])
A8 = np.column_stack([u8, v8])
x8, _, _, _ = np.linalg.lstsq(A8, b8, rcond=None)
print("R³ problem: b = A8 x8?", np.isclose(A8 @ x8, b8).all(), "→ b is not in span{u,v}")
print("  inner products: b·u =", np.dot(b8, u8), ", b·v =", np.dot(b8, v8), "(both zero: perpendicularity verified ✓)")

# Q9: u=(1,1,1,0), v=(0,1,-1,1)
u9 = np.array([1.0, 1.0, 1.0, 0.0])
v9 = np.array([0.0, 1.0, -1.0, 1.0])
A9 = np.column_stack([u9, v9])
for w, name in [([2,5,-1,3],'(2,5,-1,3)'), ([0,1,0,0],'(0,1,0,0)')]:
    w = np.array(w, dtype=float)
    x, _, _, _ = np.linalg.lstsq(A9, w, rcond=None)
    in_span = np.isclose(A9 @ x, w).all()
    print(f"R⁴ problem: {name} ∈ span{{u,v}}: {in_span}", end="")
    if in_span:
        print(f"  (coefficients: {x.round(4)})")
    else:
        print()

Exercise 10 (A Span Is a Subspace (Proof))

Let $\mathbf{v}_0, \mathbf{v}_1, \ldots, \mathbf{v}_{k-1}$ be vectors in a vector space $V$ . Prove that $W = \text{span}\{\mathbf{v}_0, \mathbf{v}_1, \ldots, \mathbf{v}_{k-1}\}$ is a subspace of $V$ .

Hint. Adding two linear combinations, or multiplying one by a scalar, gives another linear combination.

Exercise 11 (A Span Equal to

\mathbb{R}^3

)

In $\mathbb{R}^3$ , let $\mathbf{v}_0 = (1,1,0)$ , $\mathbf{v}_1 = (0,1,1)$ , and $\mathbf{v}_2 = (1,0,1)$ . Explain why $\text{span}\{\mathbf{v}_0, \mathbf{v}_1, \mathbf{v}_2\} = \mathbb{R}^3$ .

Hint. Compute the determinant $\det[\mathbf{v}_0|\mathbf{v}_1|\mathbf{v}_2]$ ; if it is nonzero, the three vectors are linearly independent and therefore span $\mathbb{R}^3$ .

# Python verification: exer-ch4-span-r3-full
# Paired with the exercise label above; numerical examples do not replace a general proof.
v0 = np.array([1.0, 1.0, 0.0])
v1 = np.array([0.0, 1.0, 1.0])
v2 = np.array([1.0, 0.0, 1.0])
M = np.column_stack([v0, v1, v2])
d = np.linalg.det(M)
print(f"det[v0, v1, v2] = {d:.4f}  ({'≠ 0, so they span R³ ✓' if abs(d) > 1e-10 else '= 0'})")
print(f"rank = {rank(M)}")

Exercise 12 (A Redundant Vector in a Span)

In $\mathbb{R}^3$ , let $\mathbf{v}$ be a nonzero vector. Explain why $\text{span}\{\mathbf{v}, 2\mathbf{v}\} = \text{span}\{\mathbf{v}\}$ , and interpret this geometrically.

Hint. $2\mathbf{v}$ is itself a scalar multiple of $\mathbf{v}$ , so it belongs to $\text{span}\{\mathbf{v}\}$ .

Exercise 13 (Spans of Sums and Subset Relations)

Let $\mathbf{u}$ and $\mathbf{v}$ be nonzero vectors in $\mathbb{R}^n$ .

Explain why $\text{span}\{\mathbf{u}, \mathbf{v}, \mathbf{u}+\mathbf{v}\} = \text{span}\{\mathbf{u}, \mathbf{v}\}$ .
Explain why $\text{span}\{\mathbf{u}\}$ is a subspace of $\text{span}\{\mathbf{u}, \mathbf{v}\}$ .

Hint. Use $\mathbf{u}+\mathbf{v}\in\text{span}\{\mathbf{u},\mathbf{v}\}$ ; a subspace is closed under linear combinations.

Exercise 14 (A Counterexample on Intersections of Spans of Three Vectors)

In $\mathbb{R}^3$ , $\mathbf{u}$ , $\mathbf{v}$ , $\mathbf{w}$ are nonzero vectors, no two of which are parallel. Let $W_0 = \text{span}\{\mathbf{u}, \mathbf{v}\}$ and $W_1 = \text{span}\{\mathbf{w}\}$ .

Explain why $W_0 \cap W_1$ need not be $\{\mathbf{0}\}$ , and give a concrete counterexample.
Looking at the counterexample in (a), explain why in this case $\text{span}\{\mathbf{u}, \mathbf{v}, \mathbf{w}\}\ne \mathbb{R}^3$ .

Hint. Let $\mathbf{w}$ lie in the plane $W_0$ (but not be parallel to $\mathbf{u}$ or $\mathbf{v}$ ).

Exercise 15 (Perpendicularity and Spans)

In $\mathbb{R}^3$ , $\mathbf{u}$ , $\mathbf{v}$ , $\mathbf{b}$ are nonzero vectors. Answer the following questions:

If $\mathbf{b}$ is perpendicular to $\mathbf{u}$ and $\mathbf{v}$ , must every nonzero vector $\mathbf{w} \in \text{span}\{\mathbf{u}, \mathbf{v}\}$ be perpendicular to $\mathbf{b}$ ?
Explain why, if $\mathbf{b} \perp \mathbf{u}$ and $\mathbf{b} \perp \mathbf{v}$ , then $\mathbf{b} ∉ \text{span}\{\mathbf{u}, \mathbf{v}\}$ .

Hint. Use the linearity of the inner product; if $\mathbf{b}\in\text{span}\{\mathbf{u},\mathbf{v}\}$ , compute $\mathbf{b}\cdot\mathbf{b}$ .

§4.1.2 Linear Independence, Bases, and Dimension¶

Linear Independence and Dependence¶

Exercise 16 (Testing for Linear Independence or Dependence)

Determine whether each of the following sets of vectors is linearly independent or linearly dependent. If a set is linearly dependent, give a linear combination with coefficients not all zero that equals $\mathbf{0}$ .

$(2,5)$
$(1,1)$ , $(-1,1)$
$(1,1)$ , $(-1,-1)$
$(1,2)$ , $(2,3)$ , $(3,1)$
$(1,3,5)$ , $(-2,-6,-10)$
$(1,3,5)$ , $(-2,6,-10)$ , $(0,1,0)$
$(1,3,5)$ , $(-2,6,-10)$ , $(1,0,0)$
$(1,3,5)$ , $(-2,6,-10)$ , $(1,0,0)$ , $(0,0,1)$
$(1,0,0,1)$ , $(0,1,1,0)$ , $(1,1,1,1)$
$(1,0,0,1)$ , $(0,1,1,0)$ , $(1,0,0,0)$
$(1,1,0,0)$ , $(0,1,1,0)$ , $(0,0,1,1)$ , $(1,0,0,1)$

Hint. Arrange the vectors in a matrix and row reduce; in $\mathbb{R}^n$ , more than $n$ vectors are always linearly dependent.

# Python verification: exer-ch4-lindep-check
# Paired with the exercise label above; numerical examples do not replace a general proof.
groups = {
    "0: (2,5)": [[2,5]],
    "1: (1,1),(-1,1)": [[1,1],[-1,1]],
    "2: (1,1),(-1,-1)": [[1,1],[-1,-1]],
    "4: (1,3,5),(-2,-6,-10)": [[1,3,5],[-2,-6,-10]],
    "8: (1,0,0,1),(0,1,1,0),(1,1,1,1)": [[1,0,0,1],[0,1,1,0],[1,1,1,1]],
    "10: four vectors in R⁴": [[1,1,0,0],[0,1,1,0],[0,0,1,1],[1,0,0,1]],
}
for name, vecs in groups.items():
    M = np.array(vecs, dtype=float)
    r = rank(M)
    n_vecs = len(vecs)
    status = "linearly independent" if r == n_vecs else f"linearly dependent (rank={r})"
    print(f"  {name}: {status}")

Exercise 17 (An Equivalent Characterization of Linear Dependence (Proof))

Prove that a set of vectors $\{\mathbf{v}_0, \mathbf{v}_1, \ldots, \mathbf{v}_{k-1}\}$ is linearly dependent if and only if at least one of the vectors can be expressed as a linear combination of the others.

Hint. Start from the definition of linear dependence; assume $c_j\ne 0$ and solve for $\mathbf{v}_j$ .

Exercise 19 (Inheritance of Linear Independence)

Consider a set of vectors $\{\mathbf{v}_0, \mathbf{v}_1, \ldots, \mathbf{v}_{k-1}\}$ . Prove the following statements:

If $\{\mathbf{v}_0, \ldots, \mathbf{v}_{k-2}\}$ is linearly dependent, then $\{\mathbf{v}_0, \ldots, \mathbf{v}_{k-1}\}$ is necessarily linearly dependent.
If $\{\mathbf{v}_0, \ldots, \mathbf{v}_{k-1}\}$ is linearly independent, then $\{\mathbf{v}_0, \ldots, \mathbf{v}_{k-2}\}$ is necessarily linearly independent.

Hint. (1) Append $c_{k-1}=0$ to a dependence relation among the first $k-1$ vectors; (2) is the contrapositive of (1).

Exercise 20 (Determinants and Linear Independence)

In $\mathbb{R}^2$ , consider the vectors $\mathbf{v}_0=(3,1)$ and $\mathbf{v}_1=(-2,4)$ .

Rewrite the independence condition $x\mathbf{v}_0 + y\mathbf{v}_1=\mathbf{0}$ as a system of two linear equations in two unknowns.
Compute $\det(\mathbf{v}_0, \mathbf{v}_1)$ .
Use (1) and (2) to prove that $\mathbf{v}_0$ and $\mathbf{v}_1$ are linearly independent.
(Generalization) In $\mathbb{R}^2$ , two nonzero vectors $\mathbf{v}_0$ , $\mathbf{v}_1$ are linearly independent if and only if $\det(\mathbf{v}_0, \mathbf{v}_1)\ne 0$ . In $\mathbb{R}^3$ the analogous conclusion holds for three vectors.

Hint. A linear system has only the zero solution if and only if the determinant of its coefficient matrix is nonzero.

Bases and Dimension¶

Exercise 21 (Is It a Basis?)

Decide whether each of the following sets of vectors is a basis of the given space:

Are $(2,7)$ , $(3,-2)$ a basis of $\mathbb{R}^2$ ?
Are $(1,0,0)$ , $(0,2,0)$ a basis of $\mathbb{R}^3$ ?
Are $(1,0,0)$ , $(1,1,0)$ , $(1,1,1)$ a basis of $\mathbb{R}^3$ ?
Let $W = \{ (x,y,z)\mid x + y + z = 0 \}$ . Are $(1,-1,0)$ , $(0,-1,1)$ , $(1,0,-1)$ a basis of $W$ ?
Let $W = \text{span}\{ (1,0,0), (0,1,0), (1,1,0) \}$ . Are $(1,0,0)$ , $(0,0,1)$ , $(1,1,0)$ a basis of $W$ ?
Let $W = \text{span}\{ (1,0,1,0), (0,1,0,1) \}$ . Are $(1,0,1,0)$ , $(0,1,0,1)$ a basis of $W$ ? Are $(1,1,1,1)$ , $(-1,1,-1,1)$ also a basis of $W$ ?

Hint. Compute a determinant or row reduce, and check both linear independence and spanning of the target space.

# Python verification: exer-ch4-basis-check
# Paired with the exercise label above; numerical examples do not replace a general proof.
# (a) Are (2,7),(3,-2) a basis of R^2?
A_a = np.array([[2,3],[7,-2]], dtype=float)
print(f"(a) det = {np.linalg.det(A_a):.4f} → {'a basis ✓' if abs(np.linalg.det(A_a)) > 1e-10 else 'not a basis'}")

# (c) Are (1,0,0),(1,1,0),(1,1,1) a basis of R^3?
A_c = np.array([[1,1,1],[0,1,1],[0,0,1]], dtype=float)
print(f"(c) det = {np.linalg.det(A_c):.4f} → {'a basis ✓' if abs(np.linalg.det(A_c)) > 1e-10 else 'not a basis'}")

# (d) W = {x+y+z=0}: are the three vectors a basis?
vecs_d = np.array([[1,-1,0],[0,-1,1],[1,0,-1]], dtype=float)
r_d = rank(vecs_d)
print(f"(d) rank = {r_d} (dim W = 2); the three vectors are linearly dependent ({'yes' if r_d < 3 else 'no'}), so they are not a basis")

# (f) W = span{(1,0,1,0),(0,1,0,1)}: check both bases
B1 = np.array([[1,0,1,0],[0,1,0,1]], dtype=float)
B2 = np.array([[1,1,1,1],[-1,1,-1,1]], dtype=float)
print(f"(f) B1 rank = {rank(B1)}, B2 rank = {rank(B2)}")
# Check that B2 lies in W
print(f"  (1,1,1,1) = (1,0,1,0)+(0,1,0,1): {np.allclose(B2[0], B1[0]+B1[1])}")
print(f"  (-1,1,-1,1) = -(1,0,1,0)+(0,1,0,1): {np.allclose(B2[1], -B1[0]+B1[1])}")

Exercise 22 (Finding a Basis for a Subspace)

Find a basis for each of the following subspaces and state its dimension:

$W_0 = \{ (x,y,z) \mid x+y+z=0\}$
$W_1 = \{ (x,y,z) \mid x-2y=0,\; 3y+2z=0\}$
$W_2 = \{ (x,y,z,w) \mid x + 2y = z + 2w \}$
$W_3 = \{ (x,y,z,w) \mid x = z,\; y=w\}$
$W_4 = \text{span}\{ (1,1), (-1,-1) \}$
$W_5 = \text{span}\{ (1,2), (2,3), (3,1) \}$
$W_6 = \text{span}\{ (1,3,5), (-2, 6,-10), (0, 1, 0) \}$
$W_7 = \text{span}\{ (1,1,0,0), (0,1,1,0), (0,0,1,1), (1,0,0,1) \}$

Hint. Use row reduction (RREF) to find the free variables (for a homogeneous system) or the pivot columns (for a span).

# Python verification: exer-ch4-find-basis
# Paired with the exercise label above; numerical examples do not replace a general proof.
# (a) {x+y+z=0}
A_a = np.array([[1,1,1]], dtype=float)
null_a = null_basis(A_a)
print(f"(a) ker basis: {null_a}, dim={len(null_a)}")

# (b) {x-2y=0, 3y+2z=0}
A_b = np.array([[1,-2,0],[0,3,2]], dtype=float)
null_b = null_basis(A_b)
print(f"(b) ker basis: {null_b}, dim={len(null_b)}")

# (h) span{(1,1,0,0),(0,1,1,0),(0,0,1,1),(1,0,0,1)}
M_h = np.array([[1,1,0,0],[0,1,1,0],[0,0,1,1],[1,0,0,1]], dtype=float)
r_h = rank(M_h)
R_h, piv_h = rref(M_h)
print(f"(h) rank={r_h}, pivots in columns {piv_h}")
print(f"  RREF:\n{R_h}")

Sums and Direct Sums of Subspaces (Extension)¶

Exercise 23 (Sums of Subspaces)

Let $U$ and $W$ be subspaces of a finite-dimensional vector space $V$ . Define $U + W = \{\mathbf{u} + \mathbf{w} \mid \mathbf{u} \in U, \mathbf{w} \in W\}$ .

Prove that $U + W$ is a subspace of $V$ .
Let $U=\text{span}\{(1,0,0)\}$ and $W=\text{span}\{(0,0,1)\}$ . Describe $U+W$ and $U\cap W$ , and compute $\dim(U+W)$ .
Let $U=\text{span}\{(0,1,2)\}$ and $W=\text{span}\{(0,1,0),(0,0,1)\}$ . Describe $U+W$ and $U\cap W$ , and compute $\dim(U+W)$ .

Hint. For (2) and (3), first determine the dimension of $U\cap W$ , then use the dimension formula.

# Python verification: exer-ch4-subspace-sum
# Paired with the exercise label above; numerical examples do not replace a general proof.
# (b) U=span{(1,0,0)}, W=span{(0,0,1)}
U_b = np.array([[1.0,0.0,0.0]])
W_b = np.array([[0.0,0.0,1.0]])
UW_b = np.vstack([U_b, W_b])
print(f"(b) dim(U)={rank(U_b)}, dim(W)={rank(W_b)}, dim(U+W)={rank(UW_b)}")
print(f"  dimension formula check: {rank(U_b)}+{rank(W_b)}-dim(U∩W)={rank(UW_b)}")
# U∩W = {0} so dim(U∩W)=0

# (c) U=span{(0,1,2)}, W=span{(0,1,0),(0,0,1)}
U_c = np.array([[0.0,1.0,2.0]])
W_c = np.array([[0.0,1.0,0.0],[0.0,0.0,1.0]])
# (0,1,2) = 1*(0,1,0) + 2*(0,0,1) ∈ W
print(f"\n(c) (0,1,2) ∈ W: {np.isclose(np.linalg.lstsq(W_c.T, U_c.T, rcond=None)[1], 0).all()}")
check = U_c[0] - 1*W_c[0] - 2*W_c[1]
print(f"  (0,1,2) - (0,1,0) - 2*(0,0,1) = {check} (= 0 → U ⊂ W)")
UW_c = np.vstack([U_c, W_c])
print(f"  dim(U)={rank(U_c)}, dim(W)={rank(W_c)}, dim(U+W)={rank(UW_c)}")

Exercise 24 (The Dimension Formula)

Let $U$ and $W$ be subspaces of a finite-dimensional vector space $V$ . Prove that

\dim(U + W) = \dim(U) + \dim(W) - \dim(U \cap W)

(6)

(Hint: extend a basis of $U \cap W$ to a basis of $U$ and to a basis of $W$ , then combine them into a basis of $U+W$ .)

Exercise 25 (Direct Sums: Definition and Applications)

Let $U$ and $W$ be subspaces of a finite-dimensional vector space $V$ . If $U + W = V$ and $U \cap W = \{\mathbf{0}\}$ , then $V$ is called the direct sum of $U$ and $W$ , written $V = U \oplus W$ .

Prove that if $V = U \oplus W$ , then $\dim(V) = \dim(U) + \dim(W)$ .
In $\mathbb{R}^2$ , let $U = \text{span}\{(1,0)\}$ (the $x$ -axis) and $W = \text{span}\{(0,1)\}$ (the $y$ -axis). Verify that $\mathbb{R}^2 = U \oplus W$ .
In $\mathbb{R}^3$ , let $U = \text{span}\{(1,0,0),(0,1,0)\}$ (the $xy$ -plane) and $W = \text{span}\{(0,0,1)\}$ (the $z$ -axis). Verify that $\mathbb{R}^3 = U \oplus W$ .
In $\mathbb{R}^3$ , let $U = \{(x,y,z) \mid x+2y+3z=0\}$ and $W = \text{span}\{(1,2,3)\}$ . Verify that $\mathbb{R}^3 = U \oplus W$ .

Hint. For (4), use the dimension formula: $\dim U=2$ , $\dim W=1$ , $\dim(U\cap W)=?$

# Python verification: exer-ch4-direct-sum
# Paired with the exercise label above; numerical examples do not replace a general proof.
# (d) U = {x+2y+3z=0}, W = span{(1,2,3)}
A_U = np.array([[1,2,3]], dtype=float)
null_U = null_basis(A_U)
print(f"Basis of U (ker of [1,2,3]):")
for v in null_U:
    print(f"  {v}")

w_vec = np.array([1.0, 2.0, 3.0])
print(f"\nW = span{{(1,2,3)}}")
# Check that w ∉ U
print(f"(1,2,3) ∈ U? (1+4+9={1+4+9} ≠ 0): {'no ✓'}")

# dim(U)+dim(W) = 2+1 = 3 = dim(R^3)
UW = np.vstack([np.array(null_U), w_vec])
print(f"dim(U+W) = rank of combined = {rank(UW)} = dim(R³) ✓")
print(f"The direct sum holds: U ∩ W = {{0}}, U + W = R³ ✓")

Comprehensive True-or-False Questions¶

Exercise 26 (Comprehensive True or False)

Decide whether each of the following statements is true. If it is true, briefly explain why; if it is false, give a counterexample or explain what is wrong.

In $\mathbb{R}^2$ , if a set $W$ contains the zero vector and is closed under vector addition, then $W$ must be a subspace of $\mathbb{R}^2$ .
In $\mathbb{R}^3$ , the solution set of a homogeneous linear system $A\mathbf{x} = \mathbf{0}$ is always a subspace of $\mathbb{R}^3$ .
If a set of vectors $\{\mathbf{v}_0, \mathbf{v}_1, \mathbf{v}_2\}$ in $\mathbb{R}^4$ is linearly independent, then these three vectors can be extended to a basis of $\mathbb{R}^4$ .
In $\mathbb{R}^3$ , if $\mathbf{v}_0$ , $\mathbf{v}_1$ are linearly independent, then $\text{span}\{\mathbf{v}_0, \mathbf{v}_1\}$ is a plane through the origin.
If a subspace $W$ contains 3 linearly independent vectors, then $\dim(W) \geq 3$ .
If $W_0$ and $W_1$ are both subspaces of $\mathbb{R}^n$ , then $W_0 \cup W_1$ is also a subspace of $\mathbb{R}^n$ .
In $\mathbb{R}^3$ , if $\mathbf{v} \in \text{span}\{\mathbf{w}_0, \mathbf{w}_1\}$ , then $\text{span}\{\mathbf{v}, \mathbf{w}_0, \mathbf{w}_1\} = \text{span}\{\mathbf{w}_0, \mathbf{w}_1\}$ .
In the vector space $\mathbb{R}^n$ , if $\{\mathbf{v}_0, \ldots, \mathbf{v}_{k-1}\}$ is linearly independent and $k < n$ , then there must exist $\mathbf{w}$ such that $\{\mathbf{v}_0, \ldots, \mathbf{v}_{k-1}, \mathbf{w}\}$ is linearly independent.
If $\{\mathbf{u}_0, \mathbf{u}_1\}$ and $\{\mathbf{v}_0, \mathbf{v}_1\}$ are two different bases of $\mathbb{R}^2$ and $\mathbf{w}$ can be expressed as $\mathbf{u}_0+\mathbf{u}_1$ , then $\mathbf{w}$ cannot also be expressed as $\mathbf{v}_0+\mathbf{v}_1$ .

Hint. Note that (1) lacks closure under scalar multiplication; (6) is the question of intersections vs. unions of subspaces.

§4.2.1 Linear Mappings, Kernels, and Images¶

Exercise 27 (Is It a Linear Mapping?)

Decide whether each of the following mappings $T$ is linear. If it is, explain why; if not, state which condition is violated (or give a counterexample).

$T_0: \mathbb{R}^2 \to \mathbb{R}^2$ , $T_0(x,y) = (2x - y,\; x + 3y)$ .
$T_1: \mathbb{R}^2 \to \mathbb{R}^2$ , $T_1(x,y) = (x+1,\; y-1)$ .
$T_2: \mathbb{R}^3 \to \mathbb{R}^2$ , $T_2(x,y,z) = (x^2,\; y+z)$ .
$T_3: \mathbb{R}^3 \to \mathbb{R}^3$ , $T_3(x,y,z) = (y,\; z,\; 0)$ .
$T_4: \mathbb{R}^4 \to \mathbb{R}^2$ , $T_4(x,y,z,w) = (x,\; |w|)$ .
$T_5: \mathbb{R}^2 \to \mathbb{R}^4$ , $T_5(x,y) = (x,\; y,\; 0,\; 0)$ .
$T_6: \mathbb{R}^3 \to \mathbb{R}$ , $T_6(x,y,z) = xyz$ .

Hint. Check whether $T(c\mathbf{u}+d\mathbf{v})=cT(\mathbf{u})+dT(\mathbf{v})$ ; note that neither $|w|$ nor $x^2$ is linear.

# Python verification: exer-ch4-linmap-check
# Paired with the exercise label above; numerical examples do not replace a general proof.
import numpy as np

# T0: T0(x,y) = (2x-y, x+3y) — linear
def T0(v): return np.array([2*v[0]-v[1], v[0]+3*v[1]])
u, v, c = np.array([1.0,2.0]), np.array([-1.0,3.0]), 2.5
print("T0 linear:", np.allclose(T0(c*u + v), c*T0(u) + T0(v)), "✓")

# T1: T1(x,y) = (x+1, y-1) — not linear (T1(0) ≠ 0)
def T1(v): return np.array([v[0]+1, v[1]-1])
print("T1(0,0) =", T1(np.zeros(2)), "≠ (0,0) → not a linear mapping ✓")

# T2: T2(x,y,z) = (x^2, y+z) — not linear
def T2(v): return np.array([v[0]**2, v[1]+v[2]])
u2 = np.array([1.0,0.0,0.0])
print("T2(2u)=", T2(2*u2), ", 2*T2(u)=", 2*T2(u2), "→ not equal, so not linear ✓")

# T3: T3(x,y,z) = (y,z,0) — linear
A3 = np.array([[0,1,0],[0,0,1],[0,0,0]], dtype=float)
def T3(v): return A3 @ v
print("Matrix of T3:\n", A3)
print("T3 linear:", np.allclose(T3(c*u2 + np.array([0.5,1.5,2.5])),
                               c*T3(u2) + T3(np.array([0.5,1.5,2.5]))), "✓")

# T4: |w| — not linear
def T4(v): return np.array([v[0], abs(v[3])])
v4 = np.array([0,0,0,1.0])
print("T4(-v)=", T4(-v4), ", -T4(v)=", -T4(v4), "→ not equal ✓")

Exercise 28 (Kernel and Image of a Matrix Mapping)

For the linear mapping $T_A(\mathbf{x}) = A\mathbf{x}$ defined by each matrix $A$ below, find $\ker(T_A)$ , $\text{nullity}(T_A)$ , $\text{Im}(T_A)$ , and $\text{rank}(T_A)$ .

$A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$ , $T_A: \mathbb{R}^2 \to \mathbb{R}^2$ .
$A = \begin{pmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 0 \end{pmatrix}$ , $T_A: \mathbb{R}^3 \to \mathbb{R}^3$ .
$A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \end{pmatrix}$ , $T_A: \mathbb{R}^3 \to \mathbb{R}^2$ .

Hint. Row reduce the matrix: the free variables correspond to $\ker$ , and the pivot columns correspond to $\text{Im}$ .

# Python verification: exer-ch4-ker-im-matrix
# Paired with the exercise label above; numerical examples do not replace a general proof.
cases = [
    ("(1)", np.array([[1,2],[2,4]], dtype=float), 2),
    ("(2)", np.array([[1,0,2],[0,1,3],[0,0,0]], dtype=float), 3),
    ("(3)", np.array([[1,2,3],[2,4,6]], dtype=float), 3),
]
for label, A, n in cases:
    r = rank(A)
    nul = n - r
    print(f"Case {label}: rank={r}, nullity={nul}, rank+nullity={r+nul}={n} ✓")
    nb = null_basis(A)
    print(f"  ker basis: {nb}")
    # Im: use RREF to find pivot column indices, then take the corresponding columns of the original A
    R, piv = rref(A)
    print(f"  Im: span of columns {piv} of A = {A[:,piv].T.tolist()}")
    print()

Exercise 29 (Kernel and Image of a Projection)

The mapping $T: \mathbb{R}^2 \to \mathbb{R}^2$ is defined by

T(x,y) = \left(\frac{x+y}{2},\frac{x+y}{2}\right)

(7)

Explain why $T$ is a projection, and find $\ker(T)$ and $\text{Im}(T)$ .
Prove that $\mathbb{R}^2= \ker(T) \oplus \text{Im}(T)$ .
Explain why $\mathbf{v} \perp \mathbf{u}$ for any nonzero vectors $\mathbf{v} \in \ker(T)$ and $\mathbf{u} \in \text{Im}(T)$ .
Describe geometrically the preimage $T^{-1}(1,1)$ of $(1,1)$ . What is the geometric relation between $\ker(T)=T^{-1}(0,0)$ and $T^{-1}(1,1)$ ?

Hint. $T^2=T$ (the defining property of a projection), and $\ker(T)=\{(x,y)\mid x+y=0\}$ .

Exercise 30 (A Shift Mapping and Its Powers)

Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be defined by $T(x,y,z) = (y,z,0)$ .

Write down the matrix of $T$ .
Find $\ker(T)$ and $\text{Im}(T)$ , and explain why $\ker(T) \subset \text{Im}(T)$ .
Find $T^2(x,y,z)$ , and find $\ker(T^2)$ and $\text{Im}(T^2)$ .
Find $T^3(x,y,z)$ , and find $\ker(T^3)$ and $\text{Im}(T^3)$ .
In each of the three parts above, compute $\dim(\ker(\cdot))+\dim(\text{Im}(\cdot))$ . What do you observe?

Hint. $T$ is a “shift operator,” and $T^3=0$ (it is nilpotent).

# Python verification: exer-ch4-shift-power
# Paired with the exercise label above; numerical examples do not replace a general proof.
A = np.array([[0,1,0],[0,0,1],[0,0,0]], dtype=float)
A2 = A @ A
A3 = A @ A @ A

print("Matrix of T: A =\n", A)
print("\nT^2 =\n", A2)
print("T^3 =\n", A3, "← the zero matrix (nilpotent)")

for power, M in enumerate([A, A2, A3], 1):
    r = rank(M)
    nul = 3 - r
    print(f"\nT^{power}: rank={r}, nullity={nul}, sum={r+nul}=3 ✓")
    print(f"  ker basis: {null_basis(M)}")

§4.2.3–4.2.5 Change of Basis, Coordinates, and Matrix Representations¶

Exercise 31 (Coordinates of a Vector)

In $\mathbb{R}^4$ there is a vector $\mathbf{v}=(3,1,1,2)$ . Find its coordinates with respect to each of the following bases.

$\mathcal{B}_0$ : $(1,1,1,1)$ , $(1,1,-1,-1)$ , $(1,-1,1,-1)$ , $(-1,1,1,-1)$ .
$\mathcal{B}_1$ : $(1,0,0,0)$ , $(1,1,0,0)$ , $(1,1,1,0)$ , $(1,1,1,1)$ .

Hint. Solve the linear system $B\mathbf{c}=\mathbf{v}$ , where $B$ is the matrix whose columns are the basis vectors.

# Python verification: exer-ch4-coord-r4
# Paired with the exercise label above; numerical examples do not replace a general proof.
v = np.array([3.0, 1.0, 1.0, 2.0])

# Basis B0 (an orthogonal basis)
B0 = np.column_stack([(1, 1, 1, 1), (1, 1, -1, -1),
                      (1, -1, 1, -1), (-1, 1, 1, -1)]).astype(float)
coords0 = np.linalg.solve(B0, v)
print("Coordinates with respect to B0:", coords0.round(4))
print("Check: B0 @ coords0 =", (B0 @ coords0).round(4), "= v?", np.allclose(B0@coords0, v))

# Basis B1 (upper triangular)
B1 = np.column_stack([(1, 0, 0, 0), (1, 1, 0, 0),
                      (1, 1, 1, 0), (1, 1, 1, 1)]).astype(float)
coords1 = np.linalg.solve(B1, v)
print("\nCoordinates with respect to B1:", coords1.round(4))
print("Check: B1 @ coords1 =", (B1 @ coords1).round(4), "= v?", np.allclose(B1@coords1, v))

Exercise 32 (The Change-of-Basis Matrix)

In $\mathbb{R}^2$ , we are given the vector $\mathbf{v} = (-2,5)$ .

Find $[\mathbf{v}]_{\mathcal{B}_0}$ with respect to the basis $\mathcal{B}_0 = \{ (1,1), (1,-1) \}$ .
Find $[\mathbf{v}]_{\mathcal{B}_1}$ with respect to the basis $\mathcal{B}_1 = \{ (1,1), (2,0) \}$ .
Taking $\mathcal{B}_0$ as the old basis and $\mathcal{B}_1$ as the new basis, find the change-of-basis matrix $\mathbf{P}$ such that $[\mathbf{v}]_{\mathcal{B}_0} = \mathbf{P} [\mathbf{v}]_{\mathcal{B}_1}$ .
Using $B_0=\begin{pmatrix}1&1\\1&-1\end{pmatrix}$ and $B_1=\begin{pmatrix}1&2\\1&0\end{pmatrix}$ , verify that $\mathbf{P} = B_0^{-1} B_1$ .

Hint. $\mathbf{v}=B_0[\mathbf{v}]_{\mathcal{B}_0}=B_1[\mathbf{v}]_{\mathcal{B}_1}$ , and therefore $[\mathbf{v}]_{\mathcal{B}_0}=B_0^{-1}B_1[\mathbf{v}]_{\mathcal{B}_1}$ .

# Python verification: exer-ch4-change-of-basis-r2
# Paired with the exercise label above; numerical examples do not replace a general proof.
v56 = np.array([-2.0, 5.0])

# Bases B0 = {(1,1),(1,-1)}, B1 = {(1,1),(2,0)}
B0 = np.column_stack([(1, 1), (1, -1)]).astype(float)
B1 = np.column_stack([(1, 1), (2, 0)]).astype(float)

coords_B0 = np.linalg.solve(B0, v56)
coords_B1 = np.linalg.solve(B1, v56)
print(f"[v]_B0 = {coords_B0.round(4)}")
print(f"[v]_B1 = {coords_B1.round(4)}")

# Change-of-basis matrix P = B0^{-1} B1
P = np.linalg.inv(B0) @ B1
print(f"\nChange-of-basis matrix P = B0^{{-1}} B1 =\n{P.round(4)}")

# Check [v]_B0 = P [v]_B1
print(f"P @ [v]_B1 = {(P @ coords_B1).round(4)}")
print(f"[v]_B0     = {coords_B0.round(4)}")
print(f"Equal? {np.allclose(P @ coords_B1, coords_B0)}")

Exercise 33 (Change of Basis and Matrix Representations)

The matrix of a linear mapping $T:\mathbb{R}^2 \to \mathbb{R}^2$ is $A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ , and we are given the new basis $\mathcal{B} = \{ (1,1), (-2,2) \}$ .

Let $\mathbf{v} = (1,0)$ . Compute $T(\mathbf{v})$ , $[\mathbf{v}]_{\mathcal{B}}$ , and $[T(\mathbf{v})]_{\mathcal{B}}$ .
Let $\mathbf{u} = (0,1)$ . Compute $T(\mathbf{u})$ , $[\mathbf{u}]_{\mathcal{B}}$ , and $[T(\mathbf{u})]_{\mathcal{B}}$ .
From the results above, compute $A_B$ (the matrix representation of $T$ with respect to $\mathcal{B}$ ).
Verify that $A_B = B^{-1} A B$ , where $B$ is the matrix whose columns are the basis vectors of $\mathcal{B}$ .
Can you explain the geometric meaning of choosing this basis? (Hint: $A$ is a reflection matrix.)

Hint. $A$ is the reflection across $y=x$ ; if we take as the new basis the direction along the mirror axis and the direction perpendicular to it, the matrix representation becomes especially simple.

# Python verification: exer-ch4-represent-matrix-change
# Paired with the exercise label above; numerical examples do not replace a general proof.
# T: R^2 → R^2, A = [[0,1],[1,0]] (reflection across y=x)
A58 = np.array([[0.0,1.0],[1.0,0.0]])
B58 = np.column_stack([(1, 1), (-2, 2)]).astype(float)

# A_B = B^{-1} A B
A_B = np.linalg.inv(B58) @ A58 @ B58
print("A_B = B^{-1} A B =\n", A_B.round(4))
print("\nNote: (1,1) is fixed by the reflection (eigenvalue 1); reflecting (-2,2) across y=x gives (2,-2)=-1*(-2,2)")
print("Hence A_B = diag(1,-1):", np.allclose(A_B, np.diag([1,-1])))

§4.2.6 The Rank-Nullity Theorem¶

Exercise 34 (Verifying the Rank-Nullity Theorem)

For each linear mapping $T:\mathbb{R}^n \to \mathbb{R}^m$ below, verify that $\text{rank}(T) + \text{nullity}(T) = n$ :

$T: \mathbb{R}^3 \to \mathbb{R}^2$ , $T(x,y,z)=(x+z,\;y+z)$ .
$T: \mathbb{R}^4 \to \mathbb{R}^4$ , with matrix $\text{diag}(0,1,4,9)$ .
$T: \mathbb{R}^3 \to \mathbb{R}^3$ , with matrix $D = \frac{1}{3}\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}$ .
$T:\mathbb{R}^2\to \mathbb{R}^4$ , $T(x,y) = (x-y,\;2x-2y,\;3x-3y,\;4x-4y)$ .

Hint. Row reduce to find the rank; $\text{nullity}=n-\text{rank}$ .

# Python verification: exer-ch4-rank-nullity
# Paired with the exercise label above; numerical examples do not replace a general proof.
cases_51 = [
    ("(1) T:R³→R²", np.array([[1,0,1],[0,1,1]], dtype=float), 3),
    ("(2) diag(0,1,4,9)", np.diag([0.0,1.0,4.0,9.0]), 4),
    ("(3) D (1/3 ones)", np.ones((3,3))/3, 3),
    ("(4) T:R²→R⁴", np.array([[1,-1],[2,-2],[3,-3],[4,-4]], dtype=float), 2),
]
for label, A, n in cases_51:
    r = rank(A)
    nul = n - r
    print(f"{label}: rank={r}, nullity={nul}, sum={r+nul}={n} {'✓' if r+nul==n else '✗'}")

§4.3.1 Concrete Vector Spaces: Complex Numbers and Quaternions¶

Exercise 35 (The Complex Numbers as a Real Vector Space)

Below we consider $\mathbb{C}$ as a vector space over the field $\mathbb{R}$ .

Explain why $\{1, i\}$ is a (real) basis of $\mathbb{C}$ , so that $\dim_\mathbb{R}(\mathbb{C})=2$ .
Explain why $T(z) = iz$ is a real-linear mapping.
Find the matrix representation of $T$ with respect to the basis $\{1, i\}$ .
Explain why the matrix in the previous part is exactly the rotation matrix for the rotation of $\mathbb{R}^2$ by $\frac{\pi}{2}$ .

Hint. $T(1)=i=(0,1)_{\{1,i\}}$ and $T(i)=-1=(-1,0)_{\{1,i\}}$ ; compare with the rotation matrix $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ .

Exercise 36 (Basic Operations with Quaternions)

In the quaternions $\mathbb{H}$ , let $q_0 = 1 + i + j + k$ and $q_1 = 3 - i + 2j - k$ .

Compute the conjugate $\bar{q}_0$ of $q_0$ , its norm $\|q_0\|$ , and the corresponding unit quaternion.
Compute the inverse $q_1^{-1}$ of $q_1$ (written in quaternion form).
Compute $q_0 + q_1$ .
Compute $q_0 q_1$ and $q_1 q_0$ (note that quaternion multiplication is not commutative).

Hint. $q^{-1}=\bar q/\|q\|^2$ ; the multiplication rules are $i^2=j^2=k^2=-1$ , $ij=k$ , $ji=-k$ , $jk=i$ , $kj=-i$ , $ki=j$ , $ik=-j$ .

# ============================================================
print_header("Basic Operations with Quaternions | exer-ch4-quaternion-basics")
# ============================================================

def q_mult(p, q):
    """Quaternion multiplication p*q, in the format (a,b,c,d) = a+bi+cj+dk"""
    a0,b0,c0,d0 = p; a1,b1,c1,d1 = q
    return np.array([a0*a1-b0*b1-c0*c1-d0*d1,
                     a0*b1+b0*a1+c0*d1-d0*c1,
                     a0*c1-b0*d1+c0*a1+d0*b1,
                     a0*d1+b0*c1-c0*b1+d0*a1])

def q_conj(p): return np.array([p[0], -p[1], -p[2], -p[3]])
def q_norm(p): return np.sqrt(np.sum(p**2))
def q_inv(p):  return q_conj(p) / np.dot(p, p)

q0 = np.array([1.0, 1.0, 1.0, 1.0])   # 1+i+j+k
q1 = np.array([3.0, -1.0, 2.0, -1.0])

print_step(1, "Conjugate, norm, and unit quaternion of q₀ = 1+i+j+k")
print("  q₀ =", q0)
print("  q₀* =", q_conj(q0))
compare_print("‖q₀‖", f"{q_norm(q0):.6f}", "√4 = 2")
print("  q̂₀ =", (q0 / q_norm(q0)).round(6))

print_step(2, "Check the inverse of q₁")
q1_inv = q_inv(q1)
print("  q₁⁻¹ =", q1_inv.round(6))
compare_print("q₁·q₁⁻¹ (should be (1,0,0,0))", q_mult(q1, q1_inv).round(6), "(1,0,0,0) ✓")

print_step(3, "q₀+q₁ and q₀·q₁ (checking noncommutativity)")
print("  q₀ + q₁ =", q0 + q1)
print("  q₀ · q₁ =", q_mult(q0, q1))
print("  q₁ · q₀ =", q_mult(q1, q0))
compare_print("q₀q₁ = q₁q₀? (should be False)",
    np.allclose(q_mult(q0, q1), q_mult(q1, q0)), "False → quaternion multiplication is not commutative ✓")

Exercise 37 (Quaternion Rotations)

Let $V=\{xi+yj+zk\in\mathbb{H}\mid x,y,z\in\mathbb{R}\}$ be the set of pure imaginary quaternions. Take $q=\cos\theta+\sin\theta\, k$ and define $T: V\to V$ by $T(\mathbf{w})=q\mathbf{w}\bar q$ .

Explain why $V$ is a three-dimensional real subspace of $\mathbb{H}$ .
Prove that $q\mathbf{w}\bar q\in V$ (that is, the values of $T$ lie in $V$ ).
Prove that $T$ is a real-linear mapping.
With $i,j,k$ as the basis, write down the matrix representation $M_\theta$ of $T$ .
Explain why $M_\theta$ is the matrix of the rotation by the angle $2\theta$ about the axis $k$ (that is, the $z$ -axis).

Hint. Compute $T(i)=q\,i\,\bar q$ , $T(j)=q\,j\,\bar q$ , and $T(k)=q\,k\,\bar q$ , expanding each one with the multiplication rules.

# Python verification: exer-ch4-quaternion-rotation
# Paired with the exercise label above; numerical examples do not replace a general proof.
# q = cos(θ) + sin(θ)k: check the rotation by 2θ about the k-axis
def q_mult(p, q):
    a1,b1,c1,d1 = p; a2,b2,c2,d2 = q
    return np.array([a1*a2-b1*b2-c1*c2-d1*d2,
                     a1*b2+b1*a2+c1*d2-d1*c2,
                     a1*c2-b1*d2+c1*a2+d1*b2,
                     a1*d2+b1*c2-c1*b2+d1*a2])
def q_conj(p): return np.array([p[0],-p[1],-p[2],-p[3]])

theta = np.pi / 6   # rotation by 2θ = π/3 = 60°

q = np.array([np.cos(theta), 0.0, 0.0, np.sin(theta)])
print(f"q = cos({theta:.4f}) + sin({theta:.4f})k = {q.round(4)}")

# pure imaginary quaternion (0,x,y,z)
w = np.array([0.0, 1.0, 0.0, 0.0])  # (the i direction)
result = q_mult(q_mult(q, w), q_conj(q))
print(f"T(i) = q*i*q̄ = {result.round(6)}")
print(f"Theoretical value (rotation by 2θ={2*theta:.4f} rad = {np.degrees(2*theta):.1f}°):")
print(f"  i → cos(2θ)i + sin(2θ)j = ({np.cos(2*theta):.4f}, {np.sin(2*theta):.4f})")

# Matrix representation M_θ
two_theta = 2 * theta
M_theta = np.array([
    [np.cos(two_theta), -np.sin(two_theta), 0],
    [np.sin(two_theta),  np.cos(two_theta), 0],
    [0, 0, 1]
])
print(f"\nRotation matrix M_θ (rotation about the z-axis by {np.degrees(two_theta):.1f}°):")
print(M_theta.round(4))

§4.3.2 Matrix Spaces and Their Subspaces¶

Exercise 38 (Matrix Spaces and Their Subspaces)

Let $M_{2 \times 2}$ denote the set of all $2\times 2$ real matrices.

Explain why $M_{2 \times 2}$ is a vector space and why the standard basis $\mathcal{E} = \{E_{00}, E_{01}, E_{10}, E_{11}\}$ is indeed a basis, so that $\dim M_{2\times 2}=4$ .
Prove that the following subsets of $M_{3\times 3}$ are subspaces, find a basis for each, and find its dimension:
- the space of diagonal matrices $W_0 = \{A \mid A_{ij}=0\text{ if }i\ne j\}$
- the space of symmetric matrices $W_1 = \{A \in M_{3\times 3} \mid A = A^{\top}\}$
- the space of skew-symmetric matrices $W_2 = \{A \in M_{3\times 3} \mid A = -A^{\top}\}$

Hint. A symmetric matrix has $n(n+1)/2$ free entries; a skew-symmetric matrix has diagonal entries 0 and $n(n-1)/2$ free entries.

Exercise 39 (Computing Conjugate Transposes)

For a square matrix $A \in M_{n \times n}(\mathbb{C})$ , define $A^* = \overline{A}^{\top}$ (the conjugate transpose). Compute $A^*$ for each of the following matrices:

$A=\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$
$A=\begin{pmatrix} 2 & i \\ -i & 0 \end{pmatrix}$
$A=\begin{pmatrix} 1-i & i & i \\ 0 & 3 & i \\ 0 & 0 & 4 \end{pmatrix}$

Hint. $A^*=(\bar A)^{\top}$ : take the conjugate and then the transpose (or transpose first and then take the conjugate; the order does not affect the result).

# Python verification: exer-ch4-conj-transpose
# Paired with the exercise label above; numerical examples do not replace a general proof.
import numpy as np

cases_h = [
    ("(1)", np.array([[1,2],[3,4]], dtype=complex)),
    ("(2)", np.array([[2,1j],[-1j,0]], dtype=complex)),
    ("(3)", np.array([[1-1j,1j,1j],[0,3,1j],[0,0,4]], dtype=complex)),
]
for label, A in cases_h:
    A_star = A.conj().T
    is_hermite = np.allclose(A, A_star)
    print(f"{label} A* =\n{A_star}")
    print(f"  Hermitian? {is_hermite}\n")

Exercise 40 (Testing for Hermitian Matrices)

Let $A \in M_{n\times n}(\mathbb{C})$ . If $A = A^*$ , then $A$ is called a Hermitian matrix. Determine which of the following matrices are Hermitian:

A_0 = \begin{pmatrix} 3 & 2-i \\ 2+i & -1 \end{pmatrix}, \quad A_1 = \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix}, \quad A_2 = \begin{pmatrix} 1 & i \\ i & -1 \end{pmatrix}

(16)

Hint. The diagonal entries of a Hermitian matrix must be real; the $(i,j)$ entry is the conjugate of the $(j,i)$ entry.

# Python verification: exer-ch4-hermite-check
# Paired with the exercise label above; numerical examples do not replace a general proof.
cases_herm = [
    ("A0", np.array([[3,2-1j],[2+1j,-1]], dtype=complex)),
    ("A1", np.array([[1,2],[2,3]], dtype=complex)),
    ("A2", np.array([[1,1j],[1j,-1]], dtype=complex)),
]
for name, A in cases_herm:
    A_star = A.conj().T
    is_hermite = np.allclose(A, A_star)
    print(f"{name}* =\n{A_star}")
    print(f"  {name} Hermitian? {is_hermite}\n")
print("Expected: A0 and A1 are Hermitian, A2 is not")

Exercise 41 (Direct-Sum Decomposition into Symmetric and Skew-Symmetric Matrices)

Let $Sym_2$ and $Skew_2$ be the subspaces of $M_{2\times 2}$ consisting of the symmetric matrices and of the skew-symmetric matrices, respectively.

Prove that $M_{2\times 2} = Sym_2 \oplus Skew_2$ .
Write the matrix $\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ as the sum of a symmetric matrix and a skew-symmetric matrix.

Hint. $A=\frac{A+A^{\top}}{2}+\frac{A-A^{\top}}{2}$ ; the first term is symmetric and the second is skew-symmetric.

# Python verification: exer-ch4-sym-skew-direct-sum
# Paired with the exercise label above; numerical examples do not replace a general proof.
M64 = np.array([[1.0, 2.0],[0.0, 1.0]])
Sym = (M64 + M64.T) / 2
Skew = (M64 - M64.T) / 2
print("Original matrix M =\n", M64)
print("\nSymmetric part (A+A^T)/2 =\n", Sym)
print("Skew-symmetric part (A-A^T)/2 =\n", Skew)
print("\nSym + Skew = M?", np.allclose(Sym + Skew, M64))
print("Sym = Sym^T?", np.allclose(Sym, Sym.T))
print("Skew = -Skew^T?", np.allclose(Skew, -Skew.T))

Exercise 42 (Direct-Sum Decompositions of Matrix Spaces)

An important property of matrix spaces is that they decompose as direct sums of the symmetric and skew-symmetric (or Hermitian and skew-Hermitian) subspaces:

The real case. $M_n(\mathbb{R}) = Sym_n(\mathbb{R}) \oplus Skew_n(\mathbb{R})$

The complex case (over the field $\mathbb R$ ). Regarding complex matrices as a real vector space, $M_n(\mathbb C)=Herm_n(\mathbb C)\oplus\operatorname{Anti-Herm}_n(\mathbb C)$ .

Prove that every complex matrix $\mathbf A$ decomposes uniquely as

\mathbf A=\frac12(\mathbf A+\mathbf A^*)+\frac12(\mathbf A-\mathbf A^*),

(17)

where $\mathbf A^*$ denotes the conjugate transpose. For real matrices, simply write $\mathbf A^\top$ in place of $\mathbf A^*$ .

Hint. Proceed in three steps: both sets are real subspaces; their sum is all of $M_n(\mathbb C)$ ; their intersection contains only the zero matrix. Finally, cross-check with the dimension $\dim_{\mathbb R}M_n(\mathbb C)=2n^2$ .

Solution to Exercise 42

Step 1. Both are real subspaces of $M_n(\mathbb C)$ . The zero matrix satisfies $\mathbf 0^*=\mathbf 0=-\mathbf 0^*$ . If $\mathbf H_0,\mathbf H_1\in Herm_n(\mathbb C)$ and $c_0,c_1\in\mathbb R$ , then

(c_0\mathbf H_0+c_1\mathbf H_1)^*=\bar c_0\mathbf H_0^*+\bar c_1\mathbf H_1^*=c_0\mathbf H_0+c_1\mathbf H_1,

(18)

so $Herm_n(\mathbb C)$ is closed under real linear combinations; the same holds for $\operatorname{Anti-Herm}_n(\mathbb C)$ . Here we used the fact that the $c_k$ are real ( $\bar c_k=c_k$ ). If we take $c=i$ , then $(i\mathbf H)^*=-i\mathbf H$ , and $i\mathbf H$ becomes a skew-Hermitian matrix, so neither set is a complex subspace—which is exactly why the exercise specifies the field $\mathbb R$ .

Step 2 (existence). Let $\mathbf H=\frac12(\mathbf A+\mathbf A^*)$ and $\mathbf K=\frac12(\mathbf A-\mathbf A^*)$ . Since $(\mathbf A^*)^*=\mathbf A$ ,

\mathbf H^*=\tfrac12(\mathbf A^*+\mathbf A)=\mathbf H,\qquad \mathbf K^*=\tfrac12(\mathbf A^*-\mathbf A)=-\mathbf K,

(19)

and $\mathbf H+\mathbf K=\mathbf A$ . Hence $M_n(\mathbb C)=Herm_n(\mathbb C)+\operatorname{Anti-Herm}_n(\mathbb C)$ .

Step 3 (uniqueness). If $\mathbf X\in Herm_n(\mathbb C)\cap\operatorname{Anti-Herm}_n(\mathbb C)$ , then $\mathbf X=\mathbf X^*=-\mathbf X$ , which gives $\mathbf X=\mathbf 0$ . If $\mathbf A=\mathbf H_0+\mathbf K_0=\mathbf H_1+\mathbf K_1$ are two decompositions, then $\mathbf H_0-\mathbf H_1=\mathbf K_1-\mathbf K_0$ belongs to both subspaces and therefore equals $\mathbf 0$ , so the decomposition is unique. Together with Step 2,

M_n(\mathbb C)=Herm_n(\mathbb C)\oplus\operatorname{Anti-Herm}_n(\mathbb C).

(20)

Step 4 (dimension check). The diagonal entries of a Hermitian matrix are real ( $n$ real parameters), while each entry $a_{ij}$ with $i<j$ is an arbitrary complex number that determines $a_{ji}=\bar a_{ij}$ (2 real parameters each), so $\dim_{\mathbb R}Herm_n(\mathbb C)=n+2\cdot\frac{n(n-1)}2=n^2$ . Moreover $\operatorname{Anti-Herm}_n(\mathbb C)=i\,Herm_n(\mathbb C)$ , whose dimension is also $n^2$ . The two add up to $2n^2=\dim_{\mathbb R}M_n(\mathbb C)$ , consistent with the direct sum.

Step 5 (the real case). Replacing every $^*$ in Steps 1 through 3 by $^\top$ , the argument goes through word for word and gives $M_n(\mathbb R)=Sym_n(\mathbb R)\oplus Skew_n(\mathbb R)$ ; the dimension check is $\frac{n(n+1)}2+\frac{n(n-1)}2=n^2$ .

Step 6 (an example). Take $\mathbf A=\begin{pmatrix}1+2i&3\\i&4-i\end{pmatrix}$ , so that $\mathbf A^*=\begin{pmatrix}1-2i&-i\\3&4+i\end{pmatrix}$ and

\mathbf H=\begin{pmatrix}1&\frac{3-i}2\\\frac{3+i}2&4\end{pmatrix},\qquad \mathbf K=\begin{pmatrix}2i&\frac{3+i}2\\\frac{-3+i}2&-i\end{pmatrix}.

(21)

One checks directly that $\mathbf H^*=\mathbf H$ , $\mathbf K^*=-\mathbf K$ , and $\mathbf H+\mathbf K=\mathbf A$ .

$\underline{\text{Writing }\mathbf K=i\mathbf H'\text{ (where }\mathbf H'=\frac1{2i}(\mathbf A-\mathbf A^*)\text{ is Hermitian), }\mathbf A=\mathbf H+i\mathbf H'\text{ is exactly the matrix version of a complex number }z=a+ib\text{.}}$

# Python verification: exer-matrix-space-decomposition
# Paired with the exercise label above; numerical examples do not replace a general proof.
print_header("Direct-Sum Decomposition of Matrix Spaces | exer-matrix-space-decomposition")

print_step(1, "The hand-computed example: A = H + K")
A = np.array([[1+2j, 3],
              [1j,   4-1j]])
A_star = A.conj().T                      # conjugate transpose A*
H = (A + A_star) / 2
K = (A - A_star) / 2
print("H = (A + A*)/2 =\n", H)
print("K = (A - A*)/2 =\n", K)
print("H + K = A ?        ", np.allclose(H + K, A))
print("H* = H (Hermitian)?      ", np.allclose(H.conj().T, H))
print("K* = -K (skew-Hermitian)?", np.allclose(K.conj().T, -K))

print_step(2, "A random complex matrix: the decomposition holds for every A")
rng = np.random.default_rng(0)
n = 4
B = rng.standard_normal((n, n)) + 1j * rng.standard_normal((n, n))
HB, KB = (B + B.conj().T) / 2, (B - B.conj().T) / 2
print("H + K = B ?", np.allclose(HB + KB, B),
      "| H Hermitian?", np.allclose(HB.conj().T, HB),
      "| K skew-Hermitian?", np.allclose(KB.conj().T, -KB))

print_step(3, "Real dimensions: dim Herm + dim Anti-Herm = dim M_n(C) = 2n²")
def herm_basis(n):
    """A real basis of Herm_n(C): E_ii, E_ij+E_ji, i(E_ij-E_ji) (i<j)"""
    basis = []
    for i in range(n):
        E = np.zeros((n, n), dtype=complex); E[i, i] = 1
        basis.append(E)
    for i in range(n):
        for j in range(i + 1, n):
            S = np.zeros((n, n), dtype=complex); S[i, j] = S[j, i] = 1
            T = np.zeros((n, n), dtype=complex); T[i, j] = 1j; T[j, i] = -1j
            basis += [S, T]
    return basis

def realify(M):
    """Regard M_n(C) as the real vector space R^{2n²}: concatenate (Re, Im)"""
    return np.concatenate([M.real.ravel(), M.imag.ravel()])

HB_list = herm_basis(n)
KB_list = [1j * X for X in HB_list]      # Anti-Herm = i · Herm
rH = np.linalg.matrix_rank(np.array([realify(X) for X in HB_list]))
rK = np.linalg.matrix_rank(np.array([realify(X) for X in KB_list]))
rHK = np.linalg.matrix_rank(np.array([realify(X) for X in HB_list + KB_list]))
compare_print("dim_R Herm_n", rH, f"n² = {n**2}")
compare_print("dim_R Anti-Herm_n", rK, f"n² = {n**2}")
compare_print("dim_R (Herm + Anti-Herm)", rHK, f"2n² = {2*n**2} (= dim_R M_n(C), and the intersection is {{0}})")

print_step(4, "The real case: dim Sym + dim Skew = n²")
dim_sym, dim_skew = n * (n + 1) // 2, n * (n - 1) // 2
compare_print("dim Sym_n + dim Skew_n", dim_sym + dim_skew, f"n² = {n**2}")

§4.3.3 Function Spaces¶

Exercise 43 (Change of Basis in a Polynomial Space)

Let $P_2 = \{ ax^2+bx+c \mid a,b,c \in \mathbb{R} \}$ , with standard basis $\mathcal{E} = \{1, x, x^2\}$ .

Take the new basis $\mathcal{B} = \{1, x-1, (x-1)^2\}$ and explain why $\mathcal{B}$ is a basis of $P_2$ .
Compute the change-of-basis matrix $\mathbf{P}$ (the conversion from $\mathcal{B}$ to $\mathcal{E}$ ).
Express the polynomial $p(x)=1+x+x^2$ in terms of $\mathcal{B}$ (that is, expand it as a linear combination of the elements of $\mathcal{B}$ ).

Hint. $x=(x-1)+1$ and $x^2=(x-1)^2+2(x-1)+1$ .

Exercise 44 (The Differentiation Mapping)

Define the differentiation mapping $D: P_2 \to P_1$ by $D(p(x)) = p'(x)$ .

Verify that $D$ is a linear mapping.
Find the kernel $\ker(D)$ of $D$ .
Find the image $\text{Im}(D)$ of $D$ .
Verify that $3 = \dim(P_2) = \text{nullity}(D) + \text{rank}(D)$ .
Write down the matrix representation $M_D$ of $D$ with respect to the standard basis $\{1,x,x^2\}$ of $P_2$ and the standard basis $\{1,x\}$ of $P_1$ .

Hint. $D(1)=0$ , $D(x)=1$ , $D(x^2)=2x$ .

§4.3.4 Composite Spaces¶

Exercise 45 (Recommendation Models: How Do Interaction Features Increase Expressive Power?)

Consider a simplified recommendation model for teaching purposes, with the following setup:

User space $U=\mathbb R^2$ : the following 3 users are samples from this space, each represented by a preference vector

Xiaoming likes action movies: $u^{(0)} = (1, 0)$ (action preference = 1, romance preference = 0)
Xiaohong likes romance movies: $u^{(1)} = (0, 1)$
Xiaoli likes both: $u^{(2)} = (0.5, 0.5)$

Movie space $M=\mathbb R^2$ : the following 3 movies are samples from this space, each represented by a genre vector

Fast & Furious: a pure action movie, $m^{(0)} = (1, 0)$
Titanic: a pure romance movie, $m^{(1)} = (0, 1)$
Mr. & Mrs. Smith: an action romance, $m^{(2)} = (0.6, 0.4)$

For the current user and movie, we write $u=(u_0,u_1)$ and $m=(m_0,m_1)$ for their scalar components; the parenthesized superscripts serve only to label the samples. We now compare two modeling approaches:

Approach A: the direct-sum model $U \oplus M$

Feature representation: the user and movie features are simply placed side by side, with dimension $\dim(U \oplus M) = 2 + 2 = 4$
Xiaoming watching Fast & Furious: $(1, 0, 1, 0)$ ; Xiaohong watching Titanic: $(0, 1, 0, 1)$
Predicted rating: the linear model $r=w_0u_0+w_1u_1+w_2m_0+w_3m_1$

Approach B: the tensor-product model $U \otimes M$

Feature representation: all user-movie interactions are taken into account, with dimension $\dim(U \otimes M) = 2 \times 2 = 4$
Xiaoming watching Fast & Furious: $u^{(0)} \otimes m^{(0)} = (1, 0) \otimes (1, 0) = (1, 0, 0, 0)$
Xiaohong watching Titanic: $u^{(1)} \otimes m^{(1)} = (0, 1) \otimes (0, 1) = (0, 0, 0, 1)$
Predicted rating: $r=\sum_{i,j=0}^{1}W_{ij}u_i m_j$ , where $\mathbf W=(W_{ij})$ ; each $u_i m_j$ is a product of scalars.

Question 1 (computing tensor-product representations). Compute the tensor-product representations of the following user-movie combinations:

Xiaoli watching Mr. & Mrs. Smith: $u^{(2)} \otimes m^{(2)} = ?$
Xiaoming watching Titanic: $u^{(0)} \otimes m^{(1)} = ?$

Question 2 (understanding the difference between the models). Suppose the training data show that users who like action movies give action movies high ratings (the weight should be large) and romance movies low ratings (the weight should be small).

The direct-sum model can learn only “the average rating that people who like action movies give to all movies” and “the average rating that everyone gives to action movies”; it cannot directly express an interaction such as “people who like action movies rate action movies highly.”
In the tensor-product model, $W_{00}$ , $W_{01}$ , $W_{10}$ , $W_{11}$ are, respectively, the rating weights of action-preferring users for action movies, of action-preferring users for romance movies, of romance-preferring users for action movies, and of romance-preferring users for romance movies.

Explain why the tensor product captures the match between user preferences and movie genres better.

Question 3 (comparing actual predictions). Suppose the trained weights are:

direct-sum model: $w=(2,3,1,2)$ , rating $=2u_0+3u_1+m_0+2m_1$
tensor-product model: $\mathbf W=\begin{bmatrix}5&1\\1&4\end{bmatrix}$ , rating $=\sum_{i,j=0}^{1}W_{ij}u_i m_j$

Compute Xiaoli’s ( $u^{(2)} = (0.5, 0.5)$ ) predicted rating for Mr. & Mrs. Smith ( $m^{(2)} = (0.6, 0.4)$ ):

Prediction of the direct-sum model: $r_{\text{direct sum}} = ?$
Prediction of the tensor-product model: $r_{\text{tensor}} = ?$

Question 4 (explaining the difference). Based on your computations, explain:

What is the fundamental reason for the difference between the two models’ predictions?
In what situations does the tensor-product model clearly outperform the direct-sum model?
Summarize in one sentence: what is the core advantage of the tensor product over the direct sum?

Hints for thinking.

The direct sum simply “concatenates” the user features and the movie features, which amounts to saying “the user likes action movies” + “the movie is an action movie” → a high rating, but it does not express “the chemistry when a user who likes action movies meets an action movie.”
The tensor product expresses every possible “user preference × movie genre” combination, each with its own independent weight, which amounts to building a “preference-matching matrix.”

This example shows that interaction features can extend the expressive power of a linear rating model.

Solution to Exercise 45

Step 1 (Question 1). Using $(u_0,u_1)\otimes(m_0,m_1)=(u_0m_0,\,u_0m_1,\,u_1m_0,\,u_1m_1)$ :

$u^{(2)}\otimes m^{(2)}=(0.5\cdot0.6,\ 0.5\cdot0.4,\ 0.5\cdot0.6,\ 0.5\cdot0.4)=(0.3,\,0.2,\,0.3,\,0.2)$ .
$u^{(0)}\otimes m^{(1)}=(1\cdot0,\ 1\cdot1,\ 0\cdot0,\ 0\cdot1)=(0,\,1,\,0,\,0)$ .

Step 2 (Question 3).

r_{\text{direct sum}}=2(0.5)+3(0.5)+0.6+2(0.4)=1+1.5+0.6+0.8=3.9.

(22)

The tensor-product model can be written as $r=\mathbf u^\top\mathbf W\mathbf m=\langle\mathbf u|\mathbf W|\mathbf m\rangle$ . First compute $\mathbf W\mathbf m=(5\cdot0.6+1\cdot0.4,\ 1\cdot0.6+4\cdot0.4)=(3.4,\,2.2)$ , then take the inner product with $\mathbf u$ :

r_{\text{tensor}}=0.5\cdot3.4+0.5\cdot2.2=2.8.

(23)

Step 3 (Question 2). The direct-sum model $r=\mathbf a^\top\mathbf u+\mathbf b^\top\mathbf m$ ( $\mathbf a=(w_0,w_1)$ , $\mathbf b=(w_2,w_3)$ ) is additive: the user’s contribution does not depend on the movie, and the movie’s contribution does not depend on the user. Consequently, for every user, the difference between the ratings of the action movie and the romance movie is

r(\mathbf u,m^{(0)})-r(\mathbf u,m^{(1)})=w_2-w_3,

(24)

which does not depend on $\mathbf u$ . With the weights given in the problem, the difference is $1-2=-1$ for both Xiaoming and Xiaohong: the model cannot make “action fans prefer action movies” and “romance fans prefer romance movies” hold at the same time.

The tensor-product model $r=\mathbf u^\top\mathbf W\mathbf m$ is linear in $\mathbf u$ and in $\mathbf m$ separately, that is, it is bilinear; the rating difference becomes

\mathbf u^\top\mathbf W\big(m^{(0)}-m^{(1)}\big),

(25)

which varies with the user. For Xiaoming the difference is $W_{00}-W_{01}=5-1=4$ , and for Xiaohong it is $W_{10}-W_{11}=1-4=-3$ ; the opposite signs express exactly “the match between preference and genre.” From the point of view of the feature space, $r=\operatorname{vec}(\mathbf W)\cdot(\mathbf u\otimes\mathbf m)$ is still a linear model; only the features have been replaced by the products $u_im_j$ .

Step 4 (Question 4).

The fundamental reason: the features of the direct-sum model are only the $u_i$ and $m_j$ themselves, and the rating is the sum of the contributions of the two sides; the features of the tensor-product model are the $u_im_j$ , and the rating contains the multiplicative interaction of the two sides. In this example $r_{\text{tensor}}=5(0.30)+1(0.20)+1(0.30)+4(0.20)=1.5+0.2+0.3+0.8$ : Xiaoli and Mr. & Mrs. Smith are both of mixed type, the cross terms $u_0m_1+u_1m_0=0.5$ make up half of the product features, and the off-diagonal entries of $\mathbf W$ are small ( $W_{01}=W_{10}=1$ ), which means that the tensor-product model considers the “mismatched preference and genre” part to be worth very few points; so its rating is lower than the direct-sum model’s 3.9. The direct-sum model looks only at the preferences and the genres of the two sides separately and does not deduct points for a mismatch.
When users’ tastes differ widely and ratings depend mainly on “whether the movie suits one’s taste” (different groups give opposite ratings to the same genre), the tensor-product model clearly outperforms the direct-sum model. The price is the number of parameters: with $U=\mathbb R^p$ and $M=\mathbb R^q$ , the direct sum needs only $p+q$ weights while the tensor product needs $pq$ , which takes more data to estimate. In this example $p=q=2$ and both happen to be 4, but that is only a coincidence.
In one sentence: $\underline{\text{The tensor product turns products of components from the two spaces into features, letting a linear model express interactions; the direct sum can only add up the contributions of the two sides.}}$

# Python verification: exer-netflix-tensor-vs-direct
# Paired with the exercise label above; numerical examples do not replace a general proof.
print_header("Recommendation Models: Direct Sum vs. Tensor Product | exer-netflix-tensor-vs-direct")

users  = {"Xiaoming": np.array([1.0, 0.0]), "Xiaohong": np.array([0.0, 1.0]), "Xiaoli": np.array([0.5, 0.5])}
movies = {"Fast & Furious": np.array([1.0, 0.0]), "Titanic": np.array([0.0, 1.0]), "Mr. & Mrs. Smith": np.array([0.6, 0.4])}

print_step(1, "Question 1: tensor-product representations u ⊗ m (np.kron, flattened in C order)")
print("  Xiaoli ⊗ Mr. & Mrs. Smith =", np.kron(users["Xiaoli"], movies["Mr. & Mrs. Smith"]))
print("  Xiaoming ⊗ Titanic   =", np.kron(users["Xiaoming"], movies["Titanic"]))

w = np.array([2.0, 3.0, 1.0, 2.0])          # weights of the direct-sum model
W = np.array([[5.0, 1.0],
              [1.0, 4.0]])                    # weight matrix of the tensor-product model

def r_direct(u, m):
    return w @ np.concatenate([u, m])         # features (u, m) ∈ U ⊕ M

def r_tensor(u, m):
    return u @ W @ m                          # u^T W m

print_step(2, "Question 3: Xiaoli's predicted rating for Mr. & Mrs. Smith")
u, m = users["Xiaoli"], movies["Mr. & Mrs. Smith"]
compare_print("Direct-sum model", r_direct(u, m), "2(0.5)+3(0.5)+0.6+2(0.4) = 3.9")
compare_print("Tensor-product model", r_tensor(u, m), "u^T W m = 2.8")
print("  The tensor-product model is also a linear model on U⊗M: vec(W)·(u⊗m) =", W.ravel() @ np.kron(u, m))

print_step(3, "Questions 2 and 4: one user's rating difference between an action movie and a romance movie")
for name in ["Xiaoming", "Xiaohong"]:
    u = users[name]
    d_dir = r_direct(u, movies["Fast & Furious"]) - r_direct(u, movies["Titanic"])
    d_ten = r_tensor(u, movies["Fast & Furious"]) - r_tensor(u, movies["Titanic"])
    print(f"  {name}: direct-sum difference = {d_dir:+.1f}, tensor-product difference = {d_ten:+.1f}")
print("→ For every user the direct-sum model gives the difference w_2 - w_3 = -1, independent of the user;")
print("  the tensor-product difference changes sign with the user's preference, expressing the match of preference × genre.")

print_step(4, "Number of parameters: dim(U⊕M) = p+q vs dim(U⊗M) = pq")
for p, q in [(2, 2), (10, 20), (100, 1000)]:
    print(f"  p={p:>4}, q={q:>5}: direct sum {p+q:>6} weights, tensor product {p*q:>7} weights")

Exercise 46 (Compressing One Day of Casino Roulette Data)

Background. A casino runs American roulette and needs to design a data system that records each day’s operations.

Game setup.

Roulette outcomes: 38 possibilities (0, 00, 1-36)
Bet types: about 150 (combinations such as straight-up numbers, colors, odd/even, ranges, columns, and dozens)
Daily volume: thousands of spins, tens of thousands of bets, and a varying number of players
Recording requirement: store all of a day’s game data in a fixed format

System constraints.

No relational database: multi-table structures and variable-length records cannot be used
Records must be formatted: each record must be either a single-person record or a multi-person statistical summary with the individual ids removed; no record may cover several people (for example, the betting record of 20 people)
Multiple records are allowed: a day may have several records
Key business information must be kept: especially the flow of funds and risk indicators, though different forms of data can retain different information
Support for later analysis: the records must support revenue analysis, risk assessment, and anomaly detection

Questions.

a) The essential challenge of compressing information:

Going from tens of thousands of transactions to a single record, what information must inevitably be sacrificed?
Which information is most critical for analyzing the casino’s operations and must be kept?

b) Comparing three ways of thinking in terms of mathematical structures: analyze the strengths and weaknesses of the following three ways of organizing the data:

Cartesian-product thinking (expanding record by record): if you insist on the Cartesian product, how do you deal with the excessive volume of data? (Hint: consider sampling, or recording only special transactions.)
Direct-sum thinking (separate statistics for each dimension): compute statistics for each dimension (time period, bet type, outcome) separately and then concatenate them. Think about it: why does keeping only statistical averages lose the key information about cash flow? For example, does knowing that “the average win is $100” let you work out the casino’s actual profit or loss?
Tensor-product thinking (keeping the joint distribution): construct an “outcome space ⊗ bet-distribution space” matrix that records, for each outcome that occurs, how the amounts are distributed over the bet types. Think about it: how does this preserve the association between outcomes and bets?

c) Concrete design and computation: design a concrete record format for each method, and compute:

Cartesian-product sampling: recording only the 1000 largest transactions
Direct-sum statistics vector: the concatenation of the statistics for each dimension
Tensor-product matrix: storing a 38×150 sparse matrix

Compare their storage requirements, the types of information they can recover, and the queries and analyses they support.

d) Practical scenarios: for each of the following business needs, which method is most suitable?

Computing “the casino’s net revenue when a red number comes up”
Detecting “abnormal betting patterns when a particular outcome occurs”
Estimating “the maximum possible loss in a single day” (risk management)
Tracking “the wins and losses of individual VIP players” (customer management)

Guiding questions.

The Cartesian product keeps the details but cannot truly compress
The direct sum severs the associations between dimensions, like knowing “how many people bet on red” and “how many times red came up” without knowing “how much the people who bet on red won when red came up”
The tensor product keeps the interaction between the two spaces, which is exactly what matters for business analysis
Consider: in information theory, the loss of which information is unacceptable?

Solution to Exercise 46

This is an open-ended design problem; what follows is one reference design. Notation: outcomes are numbered $o=0,\dots,37$ ( $\mathbf e_o\in\mathbb R^{38}$ is the standard basis), and bet types are numbered $b=0,\dots,149$ ; spin $t$ produces outcome $o_t$ , and the total stakes of the various bet types on that spin form $\mathbf s_t\in\mathbb R^{150}$ . The payout odds $k_b$ of bet type $b$ are a fixed rule; define the payoff table

G_{ob}=\begin{cases}-k_b,&b\text{ wins when the outcome is }o\\ +1,&b\text{ loses when the outcome is }o\end{cases}

(26)

that is, the revenue the casino earns per unit stake. The payoff table is determined by the rules of the game and need not be stored every day.

Step 1 (Question a). What must inevitably be sacrificed is personal identity (the constraints require removing individual ids), chronological order, and the record-by-record correspondence telling “which bet belongs to which spin.” What must be kept is the cash flow: the casino’s net revenue is

\text{net revenue}=\sum_t\sum_b G_{o_t b}\,(\mathbf s_t)_b,

(27)

which depends at the same time on “which outcome came up” and “what was bet at the time,” that is, on the joint information of the two; no record that keeps only the separate marginal statistics can compute it.

Step 2 (Question b-1, the Cartesian product). Each transaction is an element of the Cartesian product “spin × outcome × bet type × amount × player,” and storing the transactions one by one is just the raw data, with no compression. The way to deal with the volume is to keep only special transactions (for example, the 1000 largest, or all transactions of VIPs), each as a single-person record. The advantage is that individuals and details are kept; the drawback is that after sampling the totals are no longer exact, and the day’s net revenue cannot be computed.

Step 3 (Question b-2, the direct sum). Compute statistics for each dimension separately and concatenate them: the outcome counts $\mathbf n\in\mathbb R^{38}$ , the total amount for each bet type $\mathbf T\in\mathbb R^{150}$ , and the total for each time period $\in\mathbb R^{24}$ , giving a vector in $\mathbb R^{38}\oplus\mathbb R^{150}\oplus\mathbb R^{24}$ , whose dimensions add up to 212. The problem is that marginal statistics cannot determine the joint distribution. The smallest counterexample has only two outcomes, “red” and “black,” and two bets paying 1 to 1, “bet on red” and “bet on black”:

\mathbf X=\begin{pmatrix}100&0\\0&100\end{pmatrix},\qquad \mathbf Y=\begin{pmatrix}0&100\\100&0\end{pmatrix}

(28)

( $\text{row}_o$ is the outcome and $\text{col}_b$ the bet type). The two have exactly the same row sums and column sums, but in $\mathbf X$ every bet hits and the casino’s net revenue is -200, while in $\mathbf Y$ every bet misses and the net revenue is +200. Likewise, “the average win is $100” is only an average over the winners; without the number of winners and the total stakes of the losers, the profit or loss cannot be deduced.

Step 4 (Question b-3, the tensor product). Define

\mathbf S=\sum_t \mathbf e_{o_t}\otimes\mathbf s_t\in\mathbb R^{38}\otimes\mathbb R^{150},

(29)

that is, $S_{ob}$ = “the total stake on bet type $b$ over the spins on which $o$ came up.” Each spin contributes a rank-one tensor, and the whole day’s data is their sum; the record is a multi-person statistic with no individual ids, which satisfies the constraints. The net revenue becomes the Frobenius inner product of the payoff table with $\mathbf S$ :

\text{net revenue}=\sum_{o,b}G_{ob}S_{ob}=\langle\mathbf G,\mathbf S\rangle_F=\operatorname{tr}(\mathbf G^\top\mathbf S).

(30)

Hence $\mathbf S$ preserves the cash flow exactly, and it also preserves “how the bets are distributed when outcome $o$ occurs” ( $\text{row}_o(\mathbf S)$ ). Adding the outcome counts $\mathbf n$ gives per-spin averages.

Step 5 (Question c).

Method	Record format	Storage (number of values)	Recoverable information	Supported analyses
Cartesian-product sampling	1000 records × (spin number, outcome, type, amount, player id)	5000	Full details of large transactions, individuals	Tracing individual cases, VIPs; cannot compute totals
Direct-sum statistics	$\mathbf n\oplus\mathbf T\oplus$ time periods	$38+150+24=212$	The marginals of each dimension	Popular outcomes, popular bet types; cannot compute profit or loss
Tensor product	$\mathbf S$ ( $38\times150$ ) + $\mathbf n$	$5700+38=5738$	The joint distribution of outcome × bet type	Exact net revenue, conditional analysis; no personal information

On “sparse”: $\mathbf S$ is sparse only if a day has few spins, so that many $(o,b)$ combinations never occur; with thousands of spins, nearly every outcome occurs dozens of times and every spin has bets of every type, so $\mathbf S$ is close to dense. Even so, 5738 numbers are far fewer than the tens of thousands of raw transactions.

Step 6 (Question d).

Net revenue when red comes up: $\sum_{o\in\text{red}}\sum_b G_{ob}S_{ob}$ ; it suffices to take the rows of $\mathbf G$ and $\mathbf S$ for the red outcomes, so the tensor product is most suitable.
Abnormal betting when a particular outcome occurs: compare $\text{row}_o(\mathbf S)/n_o$ with the rows for the other outcomes; for example, an unusually large amount bet on 17 on the spins where 17 came up may indicate bets placed after the result was known. The tensor product is suited to detection, and the large transactions in the Cartesian-product sample can then be used to trace individual cases.
The maximum possible loss in a single day: this requires the exposure of each spin, $\max_o\big(-\sum_b G_{ob}(\mathbf s_t)_b\big)$ , which in turn requires the per-spin $\mathbf s_t$ ; but $\mathbf S$ sums over different spins and can no longer be taken apart. Risk assessment should therefore also store the de-identified per-spin betting vectors $\mathbf s_t$ (multiple records, each a multi-person statistic for one spin, which satisfies the constraints); this is a Cartesian-product-style expansion spin by spin. The whole day’s $\mathbf S$ can give only the profit or loss that actually occurred.
Wins and losses of individual VIPs: these require player ids, which only Cartesian-product-style single-person records can provide.

$\underline{\text{The direct sum adds dimensions and keeps only marginals; the tensor product multiplies dimensions and keeps the joint distribution of the two spaces, and cash flow is precisely a linear function of the joint distribution.}}$

# Python verification: exer-roulette-tensor
# Paired with the exercise label above; numerical examples do not replace a general proof.
# For readability, the bet types are simplified to 45: 38 straight-up bets (paying 35 to 1), red/black and odd/even (paying 1 to 1), and three dozen bets (paying 2 to 1).
print_header("Compressing One Day of Casino Roulette Data | exer-roulette-tensor")

# ── Outcome encoding: index 0 → "0", index 1 → "00", index k+1 → number k (k = 1..36) ──
n_out = 38
number = np.array([None, None] + list(range(1, 37)), dtype=object)
RED = {1, 3, 5, 7, 9, 12, 14, 16, 18, 19, 21, 23, 25, 27, 30, 32, 34, 36}

def wins(o, b):
    """Whether bet type b wins when the outcome is o"""
    if b < 38:
        return o == b                                   # straight-up bet
    k = number[o]
    if k is None:
        return False                                    # 0 and 00 make every outside bet lose
    return [k in RED, k not in RED, k % 2 == 1, k % 2 == 0,
            1 <= k <= 12, 13 <= k <= 24, 25 <= k <= 36][b - 38]

odds = np.array([35] * 38 + [1, 1, 1, 1, 2, 2, 2])     # payout per unit stake on a win
n_bet = len(odds)
# G[o, b]: the casino's revenue per unit stake on bet type b when the outcome is o (losing bet +1, winning bet -odds)
G = np.array([[-odds[b] if wins(o, b) else 1 for b in range(n_bet)]
              for o in range(n_out)], dtype=float)

print_step(1, "Simulating one day: individual transactions (raw data in Cartesian-product form)")
rng = np.random.default_rng(2024)
n_spin = 3000
outcome = rng.integers(0, n_out, size=n_spin)
rows = []                                               # (spin, outcome, bet type, amount)
for t in range(n_spin):
    for _ in range(rng.poisson(6)):
        rows.append((t, outcome[t], rng.integers(0, n_bet), rng.choice([10, 25, 50, 100, 500])))
rows = np.array(rows)
print(f"  spins {n_spin}, bets {len(rows)}, raw data {rows.size} numbers")
net_raw = sum(amt * G[o, b] for _, o, b, amt in rows)
print(f"  casino net revenue computed from the individual records = {net_raw:,.0f}")

print_step(2, "Tensor-product record: S = Σ_t e_{o_t} ⊗ s_t (a 38 × 45 matrix)")
S = np.zeros((n_out, n_bet))
np.add.at(S, (rows[:, 1], rows[:, 2]), rows[:, 3])
N = np.bincount(outcome, minlength=n_out)              # number of occurrences of each outcome
compare_print("net revenue = ⟨G, S⟩_F = Σ G[o,b] S[o,b]", f"{np.sum(G * S):,.0f}", f"{net_raw:,.0f} (agrees with the individual records)")
red_rows = [o for o in range(n_out) if number[o] in RED]
print(f"  net revenue when red comes up = {np.sum(G[red_rows] * S[red_rows]):,.0f}")
print(f"  fraction of nonzero entries = {np.count_nonzero(S) / S.size:.1%} (with many spins, S is in fact nearly dense)")

print_step(3, "Direct-sum record: outcome counts ⊕ amount per bet type (38 + 45 dimensions)")
d = np.concatenate([N, S.sum(axis=0)])
print(f"  length of the direct-sum vector = {d.size}")

print_step(4, "Counterexample: same marginals, different net revenue (2 outcomes × 2 bets, paying 1 to 1)")
G2 = np.array([[-1.0, 1.0],      # red comes up: bets on red win, bets on black lose
               [1.0, -1.0]])     # black comes up: bets on red lose, bets on black win
X = np.array([[100.0, 0.0], [0.0, 100.0]])   # scenario in which the bets hit
Y = np.array([[0.0, 100.0], [100.0, 0.0]])   # scenario in which the bets miss
for name, M in [("X", X), ("Y", Y)]:
    print(f"  {name}: outcome marginals {M.sum(axis=1)}, bet marginals {M.sum(axis=0)}, "
          f"casino net revenue {np.sum(G2 * M):+.0f}")
print("→ The direct sum stores only the two sets of marginals and cannot tell X from Y; the tensor product S keeps the joint distribution and can.")

print_step(5, "Comparison of storage (counted as the number of stored values)")
print(f"  Cartesian-product sampling (largest 1000 records × 5 fields, including player id): {1000 * 5}")
print(f"  direct-sum statistics vector: {d.size}")
print(f"  tensor-product matrix + outcome counts: {S.size + N.size}")

✓ All exercises completed.