[solution] **Step 1: Compute $z/w$.** Multiply numerator and denominator by $\overline{w} = 1 + i$: \begin{align*} \frac{z}{w} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1 + 2i + i^2}{1 + 1} = \frac{2i}{2} = i. \end{align*} Geometrically: $z$ and $w$ both have modulus $\sqrt{2}$, so $|z/w| = 1$. The argument of $z$ is $\pi/4$ and the argument of $w$ is $-\pi/4$, so $\arg(z/w) = \pi/4 - (-\pi/4) = \pi/2$. The result $i$ has modulus $1$ and argument $\pi/2$, confirming the calculation. **Step 2: Compute $z^6$.** In polar form, $z = \sqrt{2}\,e^{i\pi/4}$, so by De Moivre's theorem: \begin{align*} z^6 = (\sqrt{2})^6 e^{i \cdot 6\pi/4} = 8 e^{i \cdot 3\pi/2} = 8(\cos(3\pi/2) + i\sin(3\pi/2)) = 8(0 - i) = -8i. \end{align*} Verification: $z^2 = (1+i)^2 = 2i$, $z^4 = (2i)^2 = -4$, $z^6 = z^4 \cdot z^2 = -4 \cdot 2i = -8i$. $\checkmark$ Geometrically, raising to the 6th power cubes the modulus ($(\sqrt{2})^6 = 8$) and multiplies the argument by 6 ($6 \cdot \pi/4 = 3\pi/2$), landing on the negative imaginary axis. [/solution]

[problem:Vector Identity by Index Notation | 2] Using index notation and the epsilon-delta identity, prove that for any vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d} \in \mathbb{R}^3$: \begin{align*} (\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{c} \times \mathbf{d}) = (\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{d}) - (\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \cdot \mathbf{c}). \end{align*} [/problem]

[solution] **Step 1: Express in index notation.** The left side is: \begin{align*} (\mathbf{a} \times \mathbf{b})_i (\mathbf{c} \times \mathbf{d})_i = (\varepsilon_{ijk} a_j b_k)(\varepsilon_{ipq} c_p d_q). \end{align*} **Step 2: Apply the epsilon-delta identity.** Contract on the repeated index $i$: \begin{align*} \varepsilon_{ijk}\varepsilon_{ipq} = \delta_{jp}\delta_{kq} - \delta_{jq}\delta_{kp}. \end{align*} **Step 3: Substitute and simplify.** \begin{align*} (\delta_{jp}\delta_{kq} - \delta_{jq}\delta_{kp}) a_j b_k c_p d_q &= a_j b_k c_j d_k - a_j b_k c_k d_j \\ &= (a_j c_j)(b_k d_k) - (a_j d_j)(b_k c_k) \\ &= (\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{d}) - (\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \cdot \mathbf{c}). \qquad \checkmark \end{align*} This identity, known as the **Binet-Cauchy identity** (or Lagrange's identity in $\mathbb{R}^3$), generalises the scalar triple product relation and, as a special case with $\mathbf{c} = \mathbf{a}$ and $\mathbf{d} = \mathbf{b}$, gives $|\mathbf{a} \times \mathbf{b}|^2 = |\mathbf{a}|^2|\mathbf{b}|^2 - (\mathbf{a} \cdot \mathbf{b})^2$, confirming that $|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta$. [/solution]

[problem:Subspace or Not | 1] Determine which of the following subsets of $\mathbb{R}^3$ are subspaces, with proof or counterexample: (a) $S_1 = \{(x, y, z) : x + 2y - z = 0\}$ (b) $S_2 = \{(x, y, z) : x^2 + y^2 + z^2 \leq 1\}$ (c) $S_3 = \{(x, y, z) : x = y^2\}$ (d) $S_4 = \{(x, y, z) : x = 2z\}$ [/problem]

[solution] **(a) $S_1$ is a subspace.** Let $\mathbf{u} = (u_1, u_2, u_3)$ and $\mathbf{v} = (v_1, v_2, v_3)$ satisfy $u_1 + 2u_2 - u_3 = 0$ and $v_1 + 2v_2 - v_3 = 0$. For any $\lambda, \mu \in \mathbb{R}$: \begin{align*} (\lambda u_1 + \mu v_1) + 2(\lambda u_2 + \mu v_2) - (\lambda u_3 + \mu v_3) = \lambda(u_1 + 2u_2 - u_3) + \mu(v_1 + 2v_2 - v_3) = 0. \end{align*} So $\lambda\mathbf{u} + \mu\mathbf{v} \in S_1$. Geometrically, $S_1$ is the plane through the origin with normal vector $(1, 2, -1)$. **(b) $S_2$ is not a subspace.** The vector $\mathbf{e}_1 = (1, 0, 0) \in S_2$ since $1 \leq 1$, but $2\mathbf{e}_1 = (2, 0, 0) \notin S_2$ since $4 > 1$. Closure under scalar multiplication fails. **(c) $S_3$ is not a subspace.** The point $(1, 1, 0)$ satisfies $1 = 1^2$, but $(1, -1, 0)$ also satisfies $1 = (-1)^2$. Their sum is $(2, 0, 0)$, which does not satisfy $2 = 0^2$. Alternatively, $\mathbf{0} = (0,0,0) \in S_3$, but $(1,1,0) \in S_3$ while $2(1,1,0) = (2,2,0) \notin S_3$ since $2 \neq 4$. **(d) $S_4$ is a subspace.** If $u_1 = 2u_3$ and $v_1 = 2v_3$, then $\lambda u_1 + \mu v_1 = 2(\lambda u_3 + \mu v_3)$, so $\lambda\mathbf{u} + \mu\mathbf{v} \in S_4$. This is a plane through the origin in $\mathbb{R}^3$. [/solution]

[problem:Dimension and Rank-Nullity | 2] Let $T: \mathbb{R}^4 \to \mathbb{R}^3$ be the linear map with matrix \begin{align*} A = \begin{pmatrix} 1 & 2 & 0 & -1 \\ 0 & 1 & 1 & 2 \\ 1 & 3 & 1 & 1 \end{pmatrix}. \end{align*} Find bases for $\ker(T)$ and $\operatorname{Im}(T)$, and verify the rank-nullity theorem. [/problem]

[solution] **Step 1: Row-reduce to find the kernel.** We solve $A\mathbf{x} = \mathbf{0}$: \begin{align*} \begin{pmatrix} 1 & 2 & 0 & -1 \\ 0 & 1 & 1 & 2 \\ 1 & 3 & 1 & 1 \end{pmatrix} \xrightarrow{R_3 \to R_3 - R_1} \begin{pmatrix} 1 & 2 & 0 & -1 \\ 0 & 1 & 1 & 2 \\ 0 & 1 & 1 & 2 \end{pmatrix} \xrightarrow{R_3 \to R_3 - R_2} \begin{pmatrix} 1 & 2 & 0 & -1 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{pmatrix}. \end{align*} There are two pivot columns (columns 1 and 2) and two free variables ($x_3 = s$, $x_4 = t$). Back-substituting: from $R_2$: $x_2 + x_3 + 2x_4 = 0$, so $x_2 = -s - 2t$. From $R_1$: $x_1 + 2x_2 - x_4 = 0$, so $x_1 = -2x_2 + x_4 = 2s + 4t + t = 2s + 5t$. So: \begin{align*} \mathbf{x} = s\begin{pmatrix} 2 \\ -1 \\ 1 \\ 0 \end{pmatrix} + t\begin{pmatrix} 5 \\ -2 \\ 0 \\ 1 \end{pmatrix}. \end{align*} A basis for $\ker(T)$ is $\left\{\begin{pmatrix} 2 \\ -1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 5 \\ -2 \\ 0 \\ 1 \end{pmatrix}\right\}$, so $\operatorname{nullity}(T) = 2$. **Step 2: Find a basis for the image.** The image is spanned by the columns of $A$. The pivot columns (columns 1 and 2) are linearly independent, and the non-pivot columns are linear combinations of them (as the row reduction reveals). A basis for $\operatorname{Im}(T)$ is $\left\{\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}\right\}$, so $\operatorname{rank}(T) = 2$. **Step 3: Verify rank-nullity.** $\operatorname{rank}(T) + \operatorname{nullity}(T) = 2 + 2 = 4 = \dim\mathbb{R}^4$. $\checkmark$ [/solution]

[problem:Diagonalisation and Matrix Powers | 2] Diagonalise the matrix $A = \begin{pmatrix} 1 & 1 \\ 4 & 1 \end{pmatrix}$ and hence compute $A^n$ for all $n \geq 1$. [/problem]

[solution] **Step 1: Find eigenvalues.** $\chi_A(t) = (1-t)^2 - 4 = t^2 - 2t - 3 = (t-3)(t+1)$. Eigenvalues: $\lambda_1 = 3$, $\lambda_2 = -1$. **Step 2: Find eigenvectors.** For $\lambda_1 = 3$: $(A - 3I)\mathbf{v} = \begin{pmatrix} -2 & 1 \\ 4 & -2 \end{pmatrix}\mathbf{v} = \mathbf{0}$, giving $2v_1 = v_2$. Take $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$. For $\lambda_2 = -1$: $(A + I)\mathbf{v} = \begin{pmatrix} 2 & 1 \\ 4 & 2 \end{pmatrix}\mathbf{v} = \mathbf{0}$, giving $2v_1 = -v_2$. Take $\mathbf{v}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$. **Step 3: Construct $P$ and $P^{-1}$.** Let $P = \begin{pmatrix} 1 & 1 \\ 2 & -2 \end{pmatrix}$. Then $\det P = -4$, so $P^{-1} = \frac{1}{-4}\begin{pmatrix} -2 & -1 \\ -2 & 1 \end{pmatrix} = \begin{pmatrix} 1/2 & 1/4 \\ 1/2 & -1/4 \end{pmatrix}$. Verification: $P^{-1}AP = \begin{pmatrix} 3 & 0 \\ 0 & -1 \end{pmatrix} = D$. $\checkmark$ **Step 4: Compute $A^n$.** Since $A = PDP^{-1}$, we have $A^n = PD^nP^{-1}$: \begin{align*} A^n &= \begin{pmatrix} 1 & 1 \\ 2 & -2 \end{pmatrix}\begin{pmatrix} 3^n & 0 \\ 0 & (-1)^n \end{pmatrix}\begin{pmatrix} 1/2 & 1/4 \\ 1/2 & -1/4 \end{pmatrix} \\ &= \begin{pmatrix} 3^n & (-1)^n \\ 2 \cdot 3^n & -2(-1)^n \end{pmatrix}\begin{pmatrix} 1/2 & 1/4 \\ 1/2 & -1/4 \end{pmatrix} \\ &= \begin{pmatrix} \frac{3^n + (-1)^n}{2} & \frac{3^n - (-1)^n}{4} \\ 3^n - (-1)^n & \frac{3^n + (-1)^n}{2} \end{pmatrix}. \end{align*} Verification for $n = 1$: $\frac{3 + (-1)}{2} = 1$, $\frac{3 - (-1)}{4} = 1$, $3 - (-1) = 4$, $\frac{3 + (-1)}{2} = 1$, giving $A^1 = \begin{pmatrix} 1 & 1 \\ 4 & 1 \end{pmatrix}$. $\checkmark$ [/solution]

[problem:Non-Diagonalisable Matrices and the Cayley-Hamilton Shortcut | 2] Show that $B = \begin{pmatrix} 3 & 1 \\ 0 & 3 \end{pmatrix}$ is not diagonalisable. Use the Cayley-Hamilton theorem to express $B^{-1}$ as a linear combination of $I$ and $B$. [/problem]

[solution] **Step 1: Non-diagonalisability.** The characteristic polynomial is $\chi_B(t) = (3-t)^2$, so $\lambda = 3$ has algebraic multiplicity $M_3 = 2$. The eigenspace is $\ker(B - 3I) = \ker\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \operatorname{span}\{(1,0)\}$, so $m_3 = 1 < 2 = M_3$. By the diagonalisation criterion, $B$ is not diagonalisable. **Step 2: Cayley-Hamilton.** The characteristic polynomial is $\chi_B(t) = t^2 - 6t + 9$. By Cayley-Hamilton: $B^2 - 6B + 9I = 0$, i.e., $B^2 = 6B - 9I$. **Step 3: Express $B^{-1}$.** From $B^2 - 6B + 9I = 0$, factor: $B(B - 6I) = -9I$, so $B \cdot \frac{1}{-9}(B - 6I) = I$. Therefore: \begin{align*} B^{-1} = \frac{1}{9}(6I - B) = \frac{1}{9}\begin{pmatrix} 3 & -1 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 1/3 & -1/9 \\ 0 & 1/3 \end{pmatrix}. \end{align*} Verification: $BB^{-1} = \begin{pmatrix} 3 & 1 \\ 0 & 3 \end{pmatrix}\begin{pmatrix} 1/3 & -1/9 \\ 0 & 1/3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. $\checkmark$ [/solution]

[problem:Spectral Theorem and Quadratic Form Classification | 3] Classify the quadratic form $Q(x_1, x_2, x_3) = 2x_1^2 + 2x_2^2 + 2x_3^2 + 2x_1 x_2 + 2x_1 x_3 + 2x_2 x_3$ as positive definite, negative definite, or indefinite. Find the principal axes and the form of $Q$ in the principal coordinate system. [/problem]

[solution] **Step 1: Write the symmetric matrix.** The general quadratic form $\sum_{i,j} A_{ij} x_i x_j$ with $A_{ij} = A_{ji}$ satisfies: the diagonal entries are the coefficients of $x_i^2$, and the off-diagonal entries $A_{ij}$ (with $i \neq j$) are *half* the coefficient of $x_i x_j$. So: \begin{align*} A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix}. \end{align*} **Step 2: Find eigenvalues.** Observe that $A = I + J$ where $J = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$. The matrix $J$ has rank 1 with eigenvalue $3$ (eigenvector $(1,1,1)^T/\sqrt{3}$) and eigenvalue $0$ with multiplicity 2 (any vector orthogonal to $(1,1,1)^T$). Therefore $A = I + J$ has eigenvalues $1 + 3 = 4$ (with eigenvector $(1,1,1)^T$) and $1 + 0 = 1$ (with multiplicity 2). **Step 3: Find eigenvectors for $\lambda = 1$.** We need the null space of $A - I = J = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$. This is the plane $x_1 + x_2 + x_3 = 0$. Two orthogonal vectors in this plane: $\mathbf{u}_2 = \frac{1}{\sqrt{2}}(1, -1, 0)^T$ and $\mathbf{u}_3 = \frac{1}{\sqrt{6}}(1, 1, -2)^T$ (found by Gram-Schmidt or inspection). **Step 4: Construct $P$ and diagonalise.** With $\mathbf{u}_1 = \frac{1}{\sqrt{3}}(1,1,1)^T$: \begin{align*} P = \begin{pmatrix} 1/\sqrt{3} & 1/\sqrt{2} & 1/\sqrt{6} \\ 1/\sqrt{3} & -1/\sqrt{2} & 1/\sqrt{6} \\ 1/\sqrt{3} & 0 & -2/\sqrt{6} \end{pmatrix}, \qquad P^TAP = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}. \end{align*} **Step 5: Classification.** In principal coordinates $\mathbf{y} = P^T\mathbf{x}$: $Q = 4y_1^2 + y_2^2 + y_3^2$. All eigenvalues are positive ($4, 1, 1$), so $Q$ is **positive definite**: $Q(\mathbf{x}) > 0$ for all $\mathbf{x} \neq \mathbf{0}$. The principal axes are the columns of $P$: the direction $(1,1,1)/\sqrt{3}$ (along which the form stretches by factor 4) and the plane perpendicular to it (along which it stretches by factor 1). The level surface $Q = 1$ is an ellipsoid, elongated along the direction orthogonal to $(1,1,1)$. [/solution]

[problem:Failure of Diagonalisation and Jordan Blocks | 3] Let $A = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$. Show that $A$ is not diagonalisable. Compute $A^2$, $A^3$, and $(I - A)^{-1}$ directly, and explain why the pattern terminates. [/problem]

[solution] **Step 1: Non-diagonalisability.** The characteristic polynomial is $\chi_A(t) = (-t)^3 = -t^3$ (since $A$ is upper triangular, the eigenvalues are the diagonal entries). The only eigenvalue is $\lambda = 0$ with $M_0 = 3$. The eigenspace is $\ker(A) = \operatorname{span}\{(1,0,0)^T\}$, so $m_0 = 1 < 3 = M_0$. By the diagonalisation criterion, $A$ is not diagonalisable. **Step 2: Powers of $A$.** \begin{align*} A^2 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \qquad A^3 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = 0. \end{align*} The matrix $A$ is **nilpotent**: $A^3 = 0$. Each power of $A$ shifts the superdiagonal up: $A$ moves entries one step, $A^2$ moves them two steps, and $A^3$ tries to move them three steps — but there are only three rows, so everything falls off. This is the hallmark of a single Jordan block for eigenvalue $0$. **Step 3: Compute $(I - A)^{-1}$.** Since $A$ is nilpotent with $A^3 = 0$, the geometric series $\sum_{k=0}^{\infty} A^k$ terminates: \begin{align*} (I - A)^{-1} = I + A + A^2 = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}. \end{align*} Verification: $(I - A)(I + A + A^2) = I + A + A^2 - A - A^2 - A^3 = I - A^3 = I$. $\checkmark$ The series terminates precisely because $A$ is nilpotent. For the analogous scalar identity $\frac{1}{1-x} = 1 + x + x^2 + \cdots$, the series only converges for $|x| < 1$. The matrix version converges (in fact terminates) whenever the spectral radius of $A$ is less than $1$ — and for nilpotent matrices, all eigenvalues are $0$, so the spectral radius is $0 < 1$. [/solution]