Answers to Selected Exercises

This appendix gives full worked answers to a representative slice of the exercises — enough of each chapter's tiers (⭐ conceptual, ⭐⭐ hand, ⭐⭐⭐ proof / code, ⭐⭐⭐⭐ application) that you can calibrate your own solutions and see what a complete answer looks like. Many individual chapters' exercises.md files also carry their own "selected answers" block; this appendix complements those, not duplicates them.

How to use this key. Try the problem in full before you look here — the struggle is where the learning happens; reading a solution feels like understanding but isn't. When you do check, compare your reasoning, not just your final number. Two honest caveats specific to this subject: eigenvectors are determined only up to a nonzero scalar, so your representative may differ from ours by a scale factor or a sign and still be right; and singular/eigen-vectors carry sign (and, for repeated values, rotational) freedom, so for SVD problems compare the reconstruction $U\Sigma V^{\mathsf{T}}$ and the singular values, not the raw $U$ and $V$ entrywise. Every numeric answer below has been checked against numpy.

Chapter 2 — Vectors

Throughout, $\mathbf{u} = \begin{bmatrix} 3 \\ 1 \end{bmatrix}$, $\mathbf{v} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$, $\mathbf{w} = \begin{bmatrix} -2 \\ 4 \end{bmatrix}$.

2.1 (conceptual). Geometric: a vector is an arrow with a length and a direction, free to slide anywhere as long as those are preserved. Algebraic: a vector is an ordered list of numbers (its components). The bridge is a coordinate system: fix axes and an origin, and the arrow from the origin to a point becomes exactly the list of that point's coordinates.

2.3 (conceptual). Lay $\mathbf{u}$ then $\mathbf{v}$ tip-to-tail, and separately lay $\mathbf{v}$ then $\mathbf{u}$; both routes are the two sides of the same parallelogram and land on the same opposite corner, so $\mathbf{u}+\mathbf{v} = \mathbf{v}+\mathbf{u}$. In components, this is the commutativity of real-number addition: $u_i + v_i = v_i + u_i$ in each slot.

2.9 (hand). (a) $\mathbf{u}+\mathbf{v} = \begin{bmatrix} 4 \\ 3 \end{bmatrix}$; (b) $\mathbf{v}+\mathbf{u} = \begin{bmatrix} 4 \\ 3 \end{bmatrix}$ — identical, because addition is commutative. (c) $\mathbf{u}-\mathbf{v} = \begin{bmatrix} 2 \\ -1 \end{bmatrix}$; (d) $\mathbf{v}-\mathbf{u} = \begin{bmatrix} -2 \\ 1 \end{bmatrix}$ — these are negatives of each other, since $\mathbf{v}-\mathbf{u} = -(\mathbf{u}-\mathbf{v})$.

2.11 (hand). $2\mathbf{u} - \mathbf{v} + \tfrac12\mathbf{w} = \begin{bmatrix} 6 \\ 2 \end{bmatrix} - \begin{bmatrix} 1 \\ 2 \end{bmatrix} + \begin{bmatrix} -1 \\ 2 \end{bmatrix} = \begin{bmatrix} 4 \\ 2 \end{bmatrix}$.

2.13 (hand). $\lVert\mathbf{u}\rVert = \sqrt{3^2+1^2} = \sqrt{10}$, $\lVert\mathbf{v}\rVert = \sqrt{1^2+2^2} = \sqrt{5}$, $\lVert\mathbf{w}\rVert = \sqrt{(-2)^2+4^2} = \sqrt{20} = 2\sqrt{5}$. So $\mathbf{w}$ is longest ($\sqrt{20} \approx 4.47$).

2.17 (hand). Reading the second component first: $c_2 = 2$. Then $c_1 + c_2 = 5$ gives $c_1 = 3$. So $3\mathbf{e}_1 + 2\,(1,1) = (5,2)$. ✓

2.20 (proof). Claim: $\lVert c\mathbf{v}\rVert = |c|\,\lVert\mathbf{v}\rVert$. Proof: $\lVert c\mathbf{v}\rVert = \sqrt{\sum_i (c v_i)^2} = \sqrt{c^2 \sum_i v_i^2} = \sqrt{c^2}\,\sqrt{\sum_i v_i^2}$. The decisive step is $\sqrt{c^2} = |c|$, not $c$: the square root returns the nonnegative root, and a length must be nonnegative. If $c$ were negative, writing $c$ instead of $|c|$ would produce a negative "length." Hence $\lVert c\mathbf{v}\rVert = |c|\,\lVert\mathbf{v}\rVert$. $\blacksquare$

2.25 (application — navigation). (a) Ground velocity $= \begin{bmatrix} 4 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 3 \end{bmatrix} = \begin{bmatrix} 4 \\ 3 \end{bmatrix}$ km/h. (b) Ground speed $= \sqrt{4^2+3^2} = 5$ km/h. (c) Only the eastward component (4 km/h) closes the 1 km width, so the time is $1/4$ h $= 15$ min; in that time the northward drift is $3 \times \tfrac14 = 0.75$ km downstream.

Chapter 7 — Matrices as Functions

7.1 (conceptual). Multiplying a vector by a matrix transforms it linearly — it sends $\mathbf{x}$ to a new vector $A\mathbf{x}$ formed as the weighted sum of $A$'s columns, the weights being the components of $\mathbf{x}$. Geometrically, $A$ relocates every point of space (stretching, rotating, shearing, or projecting) while keeping the origin fixed and grid lines straight and evenly spaced.

7.2 (conceptual). (a) False — a matrix represents a linear map, but not every transformation is a matrix; only linear ones are. (Here read it as: every matrix gives a linear map — true.) (b) False — translation moves the origin, so it is not linear. (c) True — the $j$-th column is $A\mathbf{e}_j$, the image of the $j$-th basis vector. (d) False — $A\mathbf{x}$ is always a combination of the columns, so it lies in their span by definition.

7.8 (hand, weighted sum of columns). (a) $4\begin{bmatrix} 2 \\ 0 \end{bmatrix} + (-1)\begin{bmatrix} 0 \\ 3 \end{bmatrix} = \begin{bmatrix} 8 \\ -3 \end{bmatrix}$. (b) $3\begin{bmatrix} 1 \\ 0 \end{bmatrix} + 2\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 2 \end{bmatrix}$. (c) $5\begin{bmatrix} 0 \\ 1 \end{bmatrix} + 2\begin{bmatrix} -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -2 \\ 5 \end{bmatrix}$ (a $90°$ rotation of $(5,2)$).

7.14 (hand — superposition). Write $(3,-2) = 3\mathbf{e}_1 - 2\mathbf{e}_2$. By linearity, $T(3,-2) = 3T(\mathbf{e}_1) - 2T(\mathbf{e}_2) = 3(2,1) - 2(-1,3) = (6+2,\;3-6) = (8,-3)$. The matrix is $A = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$, and indeed $A(3,-2) = (8,-3)$. ✓

7.16 (proof). Claim: a linear $T$ satisfies $T(\mathbf{0}) = \mathbf{0}$ and $T(-\mathbf{v}) = -T(\mathbf{v})$. Proof: By homogeneity, $T(\mathbf{0}) = T(0\cdot\mathbf{v}) = 0\cdot T(\mathbf{v}) = \mathbf{0}$. Again by homogeneity with scalar $-1$, $T(-\mathbf{v}) = T((-1)\mathbf{v}) = (-1)T(\mathbf{v}) = -T(\mathbf{v})$. $\blacksquare$ (This is why a transformation that moves the origin, e.g. translation in 7.2b/7.3, cannot be linear.)

7.15 (hand — composition by tracking the basis). Scaling $S = \begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}$ then rotating $R = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ gives the combined matrix $RS = \begin{bmatrix} 0 & -1 \\ 2 & 0 \end{bmatrix}$. In the opposite order, $SR = \begin{bmatrix} 0 & -2 \\ 1 & 0 \end{bmatrix}$. They differ — order matters, because matrix multiplication is not commutative (a preview of Chapter 8).

7.25 (application — robotics). The body-frame force is $R(30°)\mathbf{f}$ with $\mathbf{f}=(3,4)$, which is $\approx (0.598,\;4.964)$ N. Its magnitude is $\sqrt{0.598^2 + 4.964^2} = 5$ N — unchanged, because $\lVert\mathbf{f}\rVert = \sqrt{3^2+4^2}=5$. A rotation is an isometry: it reorients a vector without altering its length (Chapter 21).

7.26 (application — economics). (a) Each column answers "where does this month's employed (or unemployed) population go?"; the destinations must account for everyone, so each column sums to $1$ (conservation of people). (b) From $(900,100)$: after one month $A\mathbf{x} = (865,135)$; after two, $A(A\mathbf{x}) = (835.25,164.75)$. (c) The unemployed fraction is rising (from 10% toward higher). The long-run split is governed by the eigenvector for $\lambda=1$ (Chapter 23, 29).

Chapter 11 — The Determinant

11.1 (conceptual). The magnitude $|\det(A)|$ is the factor by which $A$ scales area/volume; the sign tells you whether $A$ preserves orientation ($+$) or flips it like a mirror ($-$); a vanishing determinant means $A$ collapses space to a lower dimension, so it is singular and non-invertible.

11.2 (hand). Area scales by $|\det(A)| = 5$. The unit square maps to area $5$; the triangle of area $6$ maps to area $5 \times 6 = 30$.

11.7 (hand). Using $ad-bc$: (a) $3\cdot2 - 1\cdot1 = 5$ — invertible, orientation preserved. (b) $1\cdot4 - 2\cdot3 = -2$ — invertible, orientation reversed. (c) $4\cdot3 - 6\cdot2 = 0$ — singular, not invertible (column 2 $= 1.5\times$ column 1, so the columns are dependent and the plane collapses to a line).

11.13 (hand). $\det = k^2 - 16$. Singular when $k^2 = 16$, i.e. $k = \pm 4$. At those values the two columns become proportional, so the transformation squashes the plane onto a single line.

11.14 (application — triangle area). Edge vectors from $(1,1)$: $\mathbf{a} = (4,2)-(1,1) = (3,1)$ and $\mathbf{b} = (2,5)-(1,1) = (1,4)$. Signed area $= \tfrac12 \det\begin{bmatrix} 3 & 1 \\ 1 & 4 \end{bmatrix} = \tfrac12(12 - 1) = \tfrac{11}{2} = 5.5$. The determinant is positive, so the vertices are listed counterclockwise.

11.17 (proof). Claim: $\det(A^{-1}) = 1/\det(A)$ for invertible $A$. Proof: From $AA^{-1} = I$ and multiplicativity, $\det(A)\det(A^{-1}) = \det(I) = 1$, so $\det(A^{-1}) = 1/\det(A)$. $\blacksquare$ This re-proves that a singular matrix has no inverse: if $\det(A) = 0$, the product $\det(A)\det(A^{-1})$ would be $0$, which can never equal $1$ — so no $A^{-1}$ can exist.

11.19 (proof). $\det(cA) = c^n\det(A)$. Scaling a single row by $c$ multiplies the determinant by $c$ (the row-scaling rule); $cA$ scales all $n$ rows, multiplying by $c$ once per row, i.e. by $c^n$. Geometrically, stretching every one of the $n$ dimensions by $c$ scales $n$-dimensional volume by $c^n$.

11.26 (application — market equilibrium). With $A = \begin{bmatrix} 4 & -1 \\ -2 & 3 \end{bmatrix}$, $\det(A) = 12 - 2 = 10 \neq 0$, so the equilibrium price vector is unique. Solving $A\mathbf{p} = (10,5)$ (Cramer's rule or np.linalg.solve) gives $\mathbf{p} = (3.5,\;4)$. Setting the $(1,1)$ entry to $a_{11} = 2/3$ makes $\det = 2/3 \cdot 3 - 2 = 0$; degeneracy means the two market equations are redundant, so prices are not pinned down (no unique equilibrium).

Chapter 18 — Dot Products, Norms, and Angles

Running pair: $\mathbf{u} = (1,2,3)$, $\mathbf{v} = (4,5,6)$, with $\mathbf{u}\cdot\mathbf{v} = 32$, $\lVert\mathbf{u}\rVert = \sqrt{14}$, $\lVert\mathbf{v}\rVert = \sqrt{77}$, angle $\approx 12.93°$.

18.4 (conceptual). Two vectors are orthogonal when $\mathbf{u}\cdot\mathbf{v} = 0$. Geometrically they meet at a right angle. The reason is the geometric formula $\mathbf{u}\cdot\mathbf{v} = \lVert\mathbf{u}\rVert\lVert\mathbf{v}\rVert\cos\theta$: at $\theta = 90°$, $\cos 90° = 0$, forcing the dot product to vanish (for nonzero vectors).

18.10 (hand). $\mathbf{a}\cdot\mathbf{b} = 2\cdot1 + (-1)\cdot4 + 3\cdot0 = 2 - 4 + 0 = -2$. Not orthogonal (the dot product is nonzero).

18.11 (hand). $\lVert(1,2,2)\rVert = \sqrt{1+4+4} = 3$; $\lVert(2,3,6)\rVert = \sqrt{4+9+36} = 7$. The unit vector along the second is $\tfrac17(2,3,6) = (\tfrac27,\tfrac37,\tfrac67)$.

18.12 (hand). $\cos\theta = \dfrac{(1,0)\cdot(1,\sqrt3)}{1 \cdot \lVert(1,\sqrt3)\rVert} = \dfrac{1}{\sqrt{1+3}} = \dfrac12$, so $\theta = 60°$.

18.13 (hand). (a) $3\cdot4 + 4\cdot(-3) = 0$ — orthogonal. (b) $1 - 2 + 1 = 0$ — orthogonal. (c) $0 + 0 + 0 + 0 = 0$ — orthogonal. All three pairs are orthogonal.

18.16 (hand — scalar projection). $\operatorname{comp}_{\mathbf{u}}\mathbf{v} = \dfrac{\mathbf{u}\cdot\mathbf{v}}{\lVert\mathbf{u}\rVert} = \dfrac{(3,4)\cdot(5,1)}{\lVert(3,4)\rVert} = \dfrac{15+4}{5} = \dfrac{19}{5} = 3.8$. The positive sign means $\mathbf{v}$ leans in the same general direction as $\mathbf{u}$ (the angle between them is acute).

18.18 (hand — three norms). For $\mathbf{v} = (2,-3,6)$: $\ell^1 = |2|+|-3|+|6| = 11$; $\ell^2 = \sqrt{4+9+36} = 7$; $\ell^\infty = \max(2,3,6) = 6$. Largest is $\ell^1$, smallest is $\ell^\infty$, and only the $\ell^2$ (Euclidean) norm comes from a dot product, $\lVert\mathbf{v}\rVert = \sqrt{\mathbf{v}\cdot\mathbf{v}}$.

18.21 (proof — Cauchy–Schwarz). Consider $f(t) = \lVert\mathbf{u} - t\mathbf{v}\rVert^2 = \lVert\mathbf{u}\rVert^2 - 2t(\mathbf{u}\cdot\mathbf{v}) + t^2\lVert\mathbf{v}\rVert^2 \ge 0$ for all $t$ (a norm-squared is never negative). As a quadratic in $t$ with a nonnegative value everywhere, its discriminant is $\le 0$: $4(\mathbf{u}\cdot\mathbf{v})^2 - 4\lVert\mathbf{u}\rVert^2\lVert\mathbf{v}\rVert^2 \le 0$, i.e. $(\mathbf{u}\cdot\mathbf{v})^2 \le \lVert\mathbf{u}\rVert^2\lVert\mathbf{v}\rVert^2$. Taking square roots gives $|\mathbf{u}\cdot\mathbf{v}| \le \lVert\mathbf{u}\rVert\lVert\mathbf{v}\rVert$. Equality holds exactly when $f(t) = 0$ for some $t$, i.e. $\mathbf{u} = t\mathbf{v}$ — the vectors are parallel. $\blacksquare$

18.24 (proof — Pythagorean theorem for vectors). Expand using $\lVert\mathbf{w}\rVert^2 = \mathbf{w}\cdot\mathbf{w}$ and bilinearity: $\lVert\mathbf{u}+\mathbf{v}\rVert^2 = \lVert\mathbf{u}\rVert^2 + 2(\mathbf{u}\cdot\mathbf{v}) + \lVert\mathbf{v}\rVert^2$. If $\mathbf{u}\perp\mathbf{v}$ then $\mathbf{u}\cdot\mathbf{v} = 0$ and the cross term vanishes, giving $\lVert\mathbf{u}+\mathbf{v}\rVert^2 = \lVert\mathbf{u}\rVert^2 + \lVert\mathbf{v}\rVert^2$. Conversely, if that identity holds, the cross term $2(\mathbf{u}\cdot\mathbf{v})$ must be $0$, so $\mathbf{u}\cdot\mathbf{v}=0$ and the vectors are orthogonal. The symmetry/bilinearity of the dot product is what makes the cross term collapse to a single $2(\mathbf{u}\cdot\mathbf{v})$. $\blacksquare$

Chapter 23 — Eigenvalues and Eigenvectors

The chapter's home matrix is $A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$, with eigenpairs $(3, (1,1))$ and $(1, (-1,1))$.

23.2 (conceptual). (a) False — eigenvectors must be nonzero by definition. (b) True — $\lambda = 0$ occurs exactly when $A$ is singular (its null space is the eigenspace $E_0$). (c) True — if $A\mathbf{v} = \lambda\mathbf{v}$ then $A(-\mathbf{v}) = \lambda(-\mathbf{v})$; eigenvectors come in whole lines. (d) False — a rotation (or any matrix with complex eigenvalues) has no real eigenvectors.

23.6 (conceptual). The classmate is wrong. An eigenvalue is the signed stretch factor along the eigen-direction; a negative eigenvalue like $-4$ simply means the eigenvector is flipped to point the opposite way and stretched by a factor of $4$. Eigenvalues can be negative, zero, or even complex.

23.8 (hand). (a) $A(3,3) = (2\cdot3+1\cdot3,\;1\cdot3+2\cdot3) = (9,9) = 3(3,3)$, so $(3,3)$ is an eigenvector with $\lambda = 3$. (b) $A(-2,2) = (-4+2,\;-2+4) = (-2,2) = 1\cdot(-2,2)$, so $\lambda = 1$. (c) $A(1,0) = (2,1)$, which is not a scalar multiple of $(1,0)$ — not an eigenvector.

23.11 (hand). For $A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}$: $\det(A-\lambda I) = (4-\lambda)(3-\lambda) - 2 = \lambda^2 - 7\lambda + 10 = (\lambda-5)(\lambda-2)$, so $\lambda = 5, 2$. For $\lambda=5$: $(A-5I)\mathbf{v} = \begin{bmatrix} -1 & 1 \\ 2 & -2 \end{bmatrix}\mathbf{v} = \mathbf{0}$ gives $\mathbf{v} = (1,1)$. For $\lambda=2$: $\begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix}\mathbf{v} = \mathbf{0}$ gives $\mathbf{v} = (1,-2)$. Check: $\lambda_1+\lambda_2 = 7 = \operatorname{tr}(A)$ and $\lambda_1\lambda_2 = 10 = \det(A)$. ✓

23.16 (hand — decoupling via the eigenbasis). Write $(4,2)$ in the eigenbasis: solving $c_1(1,1) + c_2(-1,1) = (4,2)$ gives $c_1 = 3$, $c_2 = -1$. Since $A$ scales each eigenvector by its eigenvalue, $A(4,2) = 3\cdot3(1,1) + (-1)\cdot1(-1,1) = (9,9) + (1,-1) = (10,8)$, and $A^4(4,2) = 3\cdot3^4(1,1) + (-1)\cdot1^4(-1,1) = 243(1,1) + (1,-1) = (244,242)$ — no matrix powers required.

23.18 (proof). Claim: $\lambda = 0$ is an eigenvalue of $A$ iff $A$ is singular. Proof: $\lambda=0$ is an eigenvalue iff there is a nonzero $\mathbf{v}$ with $A\mathbf{v} = 0\cdot\mathbf{v} = \mathbf{0}$ — i.e. iff the null space $N(A)$ contains a nonzero vector — i.e. iff $A$ is singular ($\det(A)=0$). The eigenspace $E_0$ is exactly $N(A)$. $\blacksquare$

23.25 (application — Markov steady state). With state (subscribed, cancelled), the column-stochastic matrix is $P = \begin{bmatrix} 0.85 & 0.30 \\ 0.15 & 0.70 \end{bmatrix}$ (columns sum to $1$). Its eigenvalues are $1$ and $0.55$; since $|0.55| < 1$, the chain converges. The eigenvector for $\lambda = 1$, normalized to sum to $1$, is $(\tfrac{0.30}{0.45}, \tfrac{0.15}{0.45}) = (\tfrac23, \tfrac13)$. So in the long run $\tfrac23 \approx 66.7\%$ of the population is subscribed, independent of the starting mix.

23.26 (application — golden-ratio growth). The matrix $\begin{bmatrix} 1 & 4 \\ 0.25 & 0 \end{bmatrix}$ has characteristic equation $\det(A-\lambda I) = \lambda^2 - \lambda - 1 = 0$, whose roots are $\varphi = \tfrac{1+\sqrt5}{2} \approx 1.618$ and $\tfrac{1-\sqrt5}{2} \approx -0.618$. The dominant magnitude is $\varphi > 1$, so the population grows by a factor of about $1.618$ per season, and the stable juvenile-to-adult ratio is the dominant eigenvector, $\approx (6.47, 1)$. Power iteration confirms the per-season growth factor approaches $\varphi$.

Chapter 30 — The Singular Value Decomposition

Throughout, $A = U\Sigma V^{\mathsf{T}}$ with $\sigma_1 \ge \sigma_2 \ge \cdots \ge 0$. Compare reconstructions and singular values, not raw singular vectors (sign freedom).

30.1 (conceptual). Reading $A = U\Sigma V^{\mathsf{T}}$ right to left as it acts on $\mathbf{x}$: $V^{\mathsf{T}}$ rotates (reorients to the input principal axes), $\Sigma$ stretches each axis by its singular value, and $U$ rotates again (into the output axes). Rotate–stretch–rotate.

30.2 (conceptual). (a) True — every matrix has an SVD (the existence theorem of §30.5). (b) False — defective matrices are not diagonalizable. (c) False — singular values are always $\ge 0$. (d) True — a zero singular value signals rank deficiency.

30.8 (hand). For $A = \begin{bmatrix} 3 & 0 \\ 0 & -2 \end{bmatrix}$: $A^{\mathsf{T}}A = \begin{bmatrix} 9 & 0 \\ 0 & 4 \end{bmatrix}$, eigenvalues $9$ and $4$, so the singular values are $\sqrt9 = 3$ and $\sqrt4 = 2$ (positive — the singular value is $2$, not $-2$). The minus sign is absorbed into a singular vector: e.g. $\Sigma = \operatorname{diag}(3,2)$, $V = I$, $U = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ (any paired-sign variant is valid).

30.10 (hand). For the shear $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$: $A^{\mathsf{T}}A = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}$, eigenvalues $\tfrac{3\pm\sqrt5}{2}$, so $\sigma_1 = \varphi \approx 1.618$ and $\sigma_2 = 1/\varphi \approx 0.618$. The eigenvalues of $A$ are both $1$ — they report no stretching at all, whereas the singular values correctly measure that the shear stretches by up to $\varphi$. Singular values measure stretch; eigenvalues measure invariant directions, and the two coincide only for symmetric matrices.

30.11 (hand — norms from singular values). With $\sigma = (5,3,1)$: (a) operator norm $\lVert A\rVert_2 = \sigma_1 = 5$; (b) Frobenius norm $\lVert A\rVert_F = \sqrt{5^2+3^2+1^2} = \sqrt{35} \approx 5.916$; (c) condition number $\kappa = \sigma_1/\sigma_3 = 5/1 = 5$. For a $3\times3$ matrix, $|\det A| = \sigma_1\sigma_2\sigma_3 = 5\cdot3\cdot1 = 15$.

30.13 (hand — rank-one). For $A = \mathbf{u}\mathbf{v}^{\mathsf{T}}$ with $\mathbf{u}=(1,2,2)$, $\mathbf{v}=(3,4)$: the single nonzero singular value is $\sigma_1 = \lVert\mathbf{u}\rVert\,\lVert\mathbf{v}\rVert = 3 \cdot 5 = 15$. A rank-one outer product has exactly one nonzero singular value (its rank is $1$), with left singular vector $\mathbf{u}/\lVert\mathbf{u}\rVert$ and right singular vector $\mathbf{v}/\lVert\mathbf{v}\rVert$.

30.15 (proof — eigenvalues of $A^{\mathsf{T}}A$ are nonnegative). Let $A^{\mathsf{T}}A\,\mathbf{v} = \lambda\mathbf{v}$ with $\mathbf{v}$ a unit eigenvector. Then $\lambda = \lambda(\mathbf{v}\cdot\mathbf{v}) = \mathbf{v}^{\mathsf{T}}(A^{\mathsf{T}}A)\mathbf{v} = (A\mathbf{v})^{\mathsf{T}}(A\mathbf{v}) = \lVert A\mathbf{v}\rVert^2 \ge 0$. So every eigenvalue of $A^{\mathsf{T}}A$ is nonnegative — which is exactly why the singular values $\sigma_i = \sqrt{\lambda_i}$ are real and well defined. $\blacksquare$

30.17 (proof — $|\det A| = \prod\sigma_i$). Take determinants of $A = U\Sigma V^{\mathsf{T}}$: $\det A = \det U \cdot \det\Sigma \cdot \det V^{\mathsf{T}}$. For orthogonal matrices $|\det U| = |\det V| = 1$ (Chapter 21), and $\det\Sigma = \prod_i \sigma_i$. Taking absolute values, $|\det A| = \prod_i \sigma_i$. $\blacksquare$ (Geometric reading: the rotations don't change volume, so the volume-scaling factor is the product of the axis stretches.)

30.19 (code — verified numbers). For $A = \begin{bmatrix} 4 & 11 & 14 \\ 8 & 7 & -2 \end{bmatrix}$, np.linalg.svd(A) gives singular values $\sigma_1 \approx 18.974$ and $\sigma_2 \approx 9.487$. Reconstructing with only the top singular value, $A_1 = \sigma_1\mathbf{u}_1\mathbf{v}_1^{\mathsf{T}}$, the Frobenius error $\lVert A - A_1\rVert_F \approx 9.487$ — exactly $\sigma_2$. This is the Eckart–Young theorem in action: dropping the smaller singular value costs precisely its magnitude in Frobenius error (Chapter 31).

A final word

If your answer differs from the key, do not immediately assume you are wrong — re-derive, and find the first step where the two paths diverge. For eigen- and SVD problems, remember the scaling and sign freedoms above before concluding there is an error. And for every coding problem, the real check is the one the chapters insist on: compute it both by hand and with numpy, and make sure the two agree.