Multivariable Calculus Background

Derivatives and Multivariable Models

• Important results from single-variable calculus (e.g. mean value theorem) hold for functions of multiple variables because they involve the value of a real-valued function at two points $x, \bar{x} \in \mathbb{R}^n$, and the line connecting these points can be parameterized using one variable (e.g. gradient, Hessian).

• There is no mean value theorem for continuously differentiable vector-valued functions (e.g. Jacobian).

Exercise 1

$$\begin{split}F(x) = \begin{bmatrix} x_1^2 + x_2^3 - x_3^4\\ x_1 x_2 x_3\\ 2 x_1 x_2 - 3 x_2 x_3 + x_1 x_3\\ \end{bmatrix}\end{split}$$

$$\begin{split}J(x) = \begin{bmatrix} 2 x_1 & 3 x_2^2 & - 4 x_3^3\\ x_2 x_3 & x_2 x_3 & x_1 x_2\\ 2 x_2 + x_3 & 2 x_1 - 3 x_3 & -3 x_2 + x_1\\ \end{bmatrix}\end{split}$$

Exercise 2

Let $f(x) = x_1^3 x_2^4 + \frac{x_2}{x_1}$.

$$\begin{split}\nabla f(x) = \begin{bmatrix} \frac{\partial f}{\partial x_1}(x)\\ \frac{\partial f}{\partial x_2}(x)\\ \end{bmatrix} = \begin{bmatrix} 3 x_1^2 x_2^4 - x_2 (x_1)^{-2}\\ 4 x_1^3 x_2^3 + (x_1)^{-1} \end{bmatrix}\end{split}$$

$$\begin{split}\nabla^2 f(x) = \begin{bmatrix} \frac{\partial^2 f}{\partial^2 x_1}(x) & \frac{\partial^2 f}{\partial x_1 \partial x_2}(x)\\ \frac{\partial^2 f}{\partial x_2 \partial x_1}(x) & \frac{\partial^2 f}{\partial^2 x_2}(x)\\ \end{bmatrix} = \begin{bmatrix} 6 x_1 x_2^4 + 2 x_2 (x_1)^{-3} & 12 x_1^2 x_2^3 - (x_1)^{-2}\\ 12 x_1^2 x_2^3 - (x_1)^{-2} & 12 x_1^3 x_2^2 \end{bmatrix}\end{split}$$

Exercise 3

Define $x(t) = x + tp$ and $g(t) = f(x(t)) = f(x + tp)$ where $t \in (0, 1)$.

For $0 \leq \alpha \leq 1$, applying the chain rule twice yields

$$\begin{split}\frac{d g(t)}{d t}[\alpha] &= \sum_{j = 1}^n \frac{\partial f(x(t))}{\partial x(t)_j}[x(\alpha)] \frac{d x(t)_j}{d t}[\alpha] & \quad & \text{definition of total differential}\\ &= \sum_{j = 1}^n \frac{\partial}{\partial x_j} f[x(\alpha)] p_j\\ &= \nabla f[x + \alpha p] \cdot p\end{split}$$

and

$$\begin{split}\frac{d^2 g}{d t^2}[\alpha] &= \frac{d}{dt} \left( \sum_{j = 1}^n \frac{\partial}{\partial x_j} f[x(\alpha)] p_j \right)\\ &= \sum_{i = 1}^n \sum_{j = 1}^n \frac{\partial^2}{\partial x_i \partial x_j} f[x(\alpha)] p_i p_j & \quad & \text{definition of total differential}\\ &= p^\top \nabla^2 f[x(\alpha)] p.\end{split}$$

Substituting $\alpha = 0$ yields $\frac{\partial^2 f}{\partial p^2} = p^\top \nabla^2 f(x) p$.

Applying the first fundamental theorem of calculus twice followed by the mean value theorem for functions of one variable yields

$$\begin{split}g(1) &= g(0) + \int_0^1 g'(t_1) dt_1\\ &= g(0) + \int_0^1 \left( g'(0) + \int_0^{t_1} g''(t_2) dt_2 \right) dt_1 & \quad & \text{since } g'(t_1) - g'(0) = \int_0^{t_1} g''(t_2) dt_2\\ &= g(0) + \int_0^1 g'(0) dt_1 + \int_0^1 dt_1 \int_0^{t_1} g''(t_2) dt_2\\ &= g(0) + g'(0) (1 - 0) + \int_0^1 g''(t_2) dt_2 \int_{t_2}^1 dt_1 & \quad & \text{changed order of integration in the double integral}\\ &= g(0) + g'(0) + \int_0^1 (1 - t_2) g''(t_2) dt_2\\ &= g(0) + g'(0) + g''(\xi) \int_0^1 (1 - t_2) dt_2 & \quad & \text{Mean Value Theorem with } \xi \in (0,1)\\ f(x + p) &= f(x) + \nabla f(x)^\top p + \frac{1}{2} p^\top \nabla^2 f[x(\xi)] p.\end{split}$$

Exercise 4

Invoking Lemma 4.2.1 for each component function in $\left\{ f_i \right\}_{i = 1}^n$ yields

$$\begin{split}\lim_{t \rightarrow 0} \frac{f_i(x + tp) - f_i(x)}{t} &= \nabla f_i(x)^\top p\\ \lim_{t \rightarrow 0} \frac{f_i(x + tp) - f_i(x)}{t} - \nabla f_i(x)^\top p &= 0\\ \left\vert \lim_{t \rightarrow 0} \frac{f_i(x + tp) - f_i(x)}{t} - \nabla f_i(x)^\top p \right\vert &= 0\\ \lim_{t \rightarrow 0} \left\vert \frac{f_i(x + tp) - f_i(x)}{t} - \nabla f_i(x)^\top p \right\vert &= 0 & \quad & \lvert \cdot \rvert \text{ is continuous.}\end{split}$$

Since $F$ is assumed to be continuously differentiable in an open set $D$ containing $x$, the consequent property $F'(x) = \nabla F(x)^\top = J(x)$ allows the previous statement to be written more compactly as

$$\begin{split}\lim_{t \rightarrow 0} \frac{F(x + tp) - F(x)}{t} &= \nabla F(x)^\top p\\ &= J(x) p\\ \lim_{t \rightarrow 0} \frac{F(x + tp) - F(x)}{t} - J(x) p &= \vec{0}\\ \left\Vert \lim_{t \rightarrow 0} \frac{F(x + tp) - F(x) - J(x)tp}{t} \right\Vert &= \left\Vert 0 \right\Vert\\ \lim_{t \rightarrow 0} \frac{\lVert F(x + tp) - F(x) - J(x)tp \rVert}{t} &= 0 & \quad & \lVert \cdot \rVert \text{ is continuous.}\end{split}$$

Thus $J(x) = \hat{J}(x)$.

Exercise 5

A useful fact to note is that if $I \subset \mathbb{R}$ is an interval and $f \colon I \mapsto \mathbb{R}$ is differentiable on $I$, then $f$ is Lipschitz on $I$ if and only if $f'$ is bounded on $I$.

Clearly if $f$ is Lipschitz on $I$, then $\lvert f'(x) \rvert \leq \gamma$ (4.1.11).

Suppose $\lvert f'(x) \rvert \leq M$ for all $x \in I$ and $M \in \mathbb{R}$. Let $x$ and $y$ be any two distinct elements of $I$. Applying the Mean Value Theorem gives

$$\begin{split}f'(w) &= \frac{f(x) - f(y)}{x - y}\\ \lvert f'(w) \rvert &= \left\vert \frac{f(x) - f(y)}{x - y} \right\vert\\ M \lvert x - y \rvert &\geq \left\vert f(x) - f(y) \right\vert\end{split}$$

where $w \in (x, y)$. Thus, $f$ is Lipschitz on $I$ with $\gamma = M$.

In this exercise, $f(x) = x \sin x^{-1}$, so $f'(x) = \sin(x^{-1}) - \frac{\cos x^{-1}}{x}$ and $I \in (-\infty, \infty)$. Since $f(x = 0) = 0$ and $x^{-1}$ becomes unbounded only when $x = 0$, $f'(x)$ is bounded by a finite value over $I$. Therefore, $f(x)$ is Lipschitz continuous.

However, $f$ is not differentiable at $x = 0$ because the limit does not exist:

$$\begin{split}f'(0) &= \lim_{\epsilon \rightarrow 0^+} \frac{f(0 + \epsilon) - f(0)}{\epsilon}\\ &= \lim_{\epsilon \rightarrow 0^+} \frac{\epsilon \sin(\epsilon^{-1})}{\epsilon}\\ &= \lim_{\epsilon \rightarrow 0^+} \sin \frac{1}{\epsilon}\\ &\neq \lim_{\epsilon \rightarrow 0^-} \sin \frac{1}{\epsilon}.\end{split}$$

Exercise 6

Let $g(t) = f(x + tp)$ where t in (0, 1) and $f \colon \mathbb{R}^n \mapsto \mathbb{R}$.

$$\begin{split}g(1) - g(0) &= \int_0^1 g'(t_1) dt_1\\ g(1) - g(0) - g'(0) &= \int_0^1 g'(t_1) - g'(0) dt_1\\ &= \int_0^1 \int_0^{t_1} g''(t_2) dt_2 dt_1 & \quad & g'(t_1) - g'(0) = \int_0^{t_1} g''(t_2) dt_2\\ g(1) - g(0) - g'(0) - \int_0^1 \int_0^{t_1} g''(0) dt_2 dt_1 &= \int_0^1 \int_0^{t_1} g''(t_2) dt_2 dt_1 - \int_0^1 \int_0^{t_1} g''(0) dt_2 dt_1\\ g(1) - g(0) - g'(0) - \frac{1}{2} g''(0) &= \int_0^1 \int_0^{t_1} g''(t_2) - g''(0) dt_2 dt_1\\ \left\Vert g(1) - g(0) - g'(0) - \frac{1}{2} g''(0) \right\Vert &= \left\Vert \int_0^1 \int_0^{t_1} g''(t_2) - g''(0) dt_2 dt_1 \right\Vert\\ \left\vert g(1) - g(0) - g'(0) - \frac{1}{2} g''(0) \right\vert &\leq \int_0^1 \int_0^{t_1} \left\Vert g''(t_2) - g''(0) \right\Vert dt_2 dt_1 & \quad & \text{double integral application of Lemma 4.1.10}\\ \left\vert f(x + p) - f(x) - \nabla f'(x)^\top p - \frac{1}{2} p^\top \nabla^2 f(x) p \right\vert &= \int_0^1 \int_0^{t_1} \left\Vert p^\top \left[ \nabla^2 f(x + t_2 p) - \nabla^2 f(x) \right] p \right\Vert dt_2 dt_1\\ &= \int_0^1 \int_0^{t_1} \left\Vert \sum_i \sigma_i p^\top u_i v_i^\top p \right\Vert dt_2 dt_1 & \quad & \text{apply Theorem 3.6.4 to Hessian difference}\\ &= \int_0^1 \int_0^{t_1} \left\Vert \sum_i \sigma_i \sum_j \sum_k p_j u_{ij} v_{ik} p_k \right\Vert dt_2 dt_1\\ &\leq \left\Vert p \right\Vert^2_\infty \int_0^1 \int_0^{t_1} \left\vert \sum_i \sigma_i \sum_j \sum_k u_{ij} v_{ik} \right\vert dt_2 dt_1\\ &\leq \left\Vert p \right\Vert^2_\infty \int_0^1 \int_0^{t_1} n \left\Vert \nabla^2 f(x + t_2 p) - \nabla^2 f(x) \right\Vert_\infty dt_2 dt_1\\ &\leq \left\Vert p \right\Vert^2 \int_0^1 \int_0^{t_1} n \left\Vert \nabla^2 f(x + t_2 p) - \nabla^2 f(x) \right\Vert dt_2 dt_1 & \quad & \text{(3.1.5)}\\ &\leq \left\Vert p \right\Vert^2 \int_0^1 \int_0^{t_1} \gamma \left\Vert t_2 p \right\Vert dt_2 dt_1 & \quad & \text{(4.1.11)}\\ &= \gamma \left\Vert p \right\Vert^3 \int_0^1 \int_0^{t_1} t_2 dt_2 dt_1\\ &= \frac{\gamma}{6} \left\Vert p \right\Vert^3\end{split}$$

The additional factor of $n$ is rolled into $\gamma$. Since the definition of Lipschitz continuity only requires a norm to be defined over $\mathbb{R}^{m \times n}$, there does not seem to be any reason against using a matrix norm. The book focuses on $p$-norms where $p \geq 1$.

Exercise 7

$$\begin{split}F(v) - F(u) &= \int_u^v F'(z) dz & \quad & \text{(4.1.8)}\\ &= \int_0^1 J(u + t(v - u)) (v - u) dt & \quad & \text{change of variables where } z = u + t(v - u) \text{ and } dz = (v - u) dt\\ F(v) - F(u) - J(x) (v - u) &= \int_0^1 J(u + t(v - u)) (v - u) dt - J(x) (v - u)\\ \left\Vert F(v) - F(u) - J(x) (v - u) \right\Vert &= \left\Vert \int_0^1 \left[ J(u + t(v - u)) - J(x) \right] (v - u) dt \right\Vert\\ &\leq \int_0^1 \left\Vert \left[ J(u + t(v - u)) - J(x) \right] (v - u) \right\Vert dt & \quad & \text{(4.1.9)}\\ &\leq \int_0^1 \left\Vert J(u + t(v - u)) - J(x) \right\Vert \left\Vert v - u \right\Vert dt\\ &\leq \left\Vert v - u \right\Vert \int_0^1 \gamma \left\Vert u + t(v - u) - x \right\Vert dt & \quad & \text{(4.1.11)}\\ &= \gamma \left\Vert v - u \right\Vert \int_0^1 \left\Vert u + t(v - u) - \left[ tx + (1 - t)x \right] \right\Vert dt\\ &= \gamma \left\Vert v - u \right\Vert \int_0^1 \left\Vert tv - tx + (1 - t)u - (1 - t)x \right\Vert dt\\ &\leq \gamma \left\Vert v - u \right\Vert \int_0^1 t \left\Vert v - x \right\Vert + (1 - t) \left\Vert u - x \right\Vert dt\\ &= \gamma \left\Vert v - u \right\Vert \frac{\left\Vert v - x \right\Vert + \left\Vert u - x \right\Vert}{2}\end{split}$$

Exercise 8

$$\begin{split}F(x + p) - F(x) &= \int_0^1 J(x + tp) p dt\\ F(x + p) - F(x) - J(x) p &= \int_0^1 J(x + tp) p dt - J(x) p\\ \left\Vert F(x + p) - F(x) - J(x) p \right\Vert &= \left\Vert \int_0^1 \left[ J(x + tp) - J(x) \right] p dt \right\Vert\\ &\leq \int_0^1 \left\Vert \left[ J(x + tp) - J(x) \right] p \right\Vert dt & \quad & \text{(4.1.9)}\\ &\leq \int_0^1 \left\Vert J(x + tp) - J(x) \right\Vert \left\Vert p \right\Vert dt & \quad & \text{Theorem 3.1.3}\\ &\leq \left\Vert p \right\Vert \int_0^1 \gamma \left\Vert tp \right\Vert^\alpha dt & \quad & \text{Lipschitz Continuous is a subset of Holder Continuous}\\ &= \gamma \left\Vert p \right\Vert^{1 + \alpha} \int_0^1 t^\alpha dt\\ &= \left\Vert p \right\Vert^{1 + \alpha} \frac{\gamma}{1 + \alpha}\end{split}$$

Exercise 9

$$\begin{split}F(x) &= \begin{pmatrix} 3x_1^2 - 2x_2\\ x_2^3 - \frac{1}{x_1} \end{pmatrix}\\ J(x) &= \begin{pmatrix} 6x_1 & -2\\ \frac{1}{x_1^2} & 3x_2^2 \end{pmatrix}\end{split}$$

When $x = \begin{pmatrix} 1 & 1 \end{pmatrix}^\top$,

$$\begin{split}J(x) = \begin{pmatrix} 6 & -2\\ 1 & 3 \end{pmatrix}.\end{split}$$

Using the forward difference approximation where $h = 0.01$ yields

$$\begin{split}F(x) = \begin{pmatrix} 1\\ 0 \end{pmatrix} \quad & \quad F(x + h e_1) = \begin{pmatrix} 1.0603\\ 0.0099 \end{pmatrix} \quad & \quad F(x + h e_2) = \begin{pmatrix} 0.98\\ 0.0303 \end{pmatrix}\end{split}$$

resulting in $A = \begin{pmatrix} 6.03 & -2\\ 0.99 & 3.03 \end{pmatrix}$.

The accuracy of the approximation using an $\mathcal{L}_1$-norm is $0.04$, which makes $\gamma \geq 8$.

Exercise 10

Define

$$\alpha = f(x + he_i) - f(x) - h \nabla f(x)_i - \frac{h^2}{2} \nabla^2 f(x)_{ii} + \mathcal{O}(h^2)$$

and

$$\beta = f(x - he_i) - f(x) + h \nabla f(x)_i - \frac{h^2}{2} \nabla^2 f(x)_{ii} + \mathcal{O}(h^2)$$

where $\left\vert \cdot \right\vert \leq \frac{\gamma h^3}{6}$ (4.1.13).

$$\begin{split}\alpha + \beta &= f(x + he_i) - 2f(x) + f(x - he_i) - h^2 \nabla^2 f(x)_{ii}\\ \left\vert \alpha + \beta \right\vert &= \left\vert f(x + he_i) - 2f(x) + f(x - he_i) - h^2 \nabla^2 f(x)_{ii} \right\vert\\ &\leq \frac{\gamma h^3}{3} \qquad \text{triangle inequality}\end{split}$$

Thus, $\left| A_{ii} - \nabla^2 f(x)_{ii} \right| \leq \frac{\gamma h}{3}$ where $A_{ii} = \frac{f(x + he_i) - 2f(x) + f(x - he_i)}{h^2}$.

Exercise 11

Substituting $p = he_i + he_j, p = he_i - he_j, p = -he_i + he_j,$ and $p = -he_i - he_j$ into Lemma 4.1.14 yields

$$\begin{split}\alpha &= f(x + he_i + he_j) - f(x) - h \nabla f(x)_i - h \nabla f(x)_j - \frac{h^2}{2} \left[ \nabla^2 f(x)_{ii} + 2 \nabla^2 f(x)_{ij} + \nabla^2 f(x)_{jj} \right] + \mathcal{O}(h^2)\\ \beta &= f(x + he_i - he_j) - f(x) - h \nabla f(x)_i + h \nabla f(x)_j - \frac{h^2}{2} \left[ \nabla^2 f(x)_{ii} - 2 \nabla^2 f(x)_{ij} + \nabla^2 f(x)_{jj} \right] + \mathcal{O}(h^2)\\ \eta &= f(x - he_i + he_j) - f(x) + h \nabla f(x)_i - h \nabla f(x)_j - \frac{h^2}{2} \left[ \nabla^2 f(x)_{ii} - 2 \nabla^2 f(x)_{ij} + \nabla^2 f(x)_{jj} \right] + \mathcal{O}(h^2)\\ \nu &= f(x - he_i - he_j) - f(x) + h \nabla f(x)_i + h \nabla f(x)_j - \frac{h^2}{2} \left[ \nabla^2 f(x)_{ii} + 2 \nabla^2 f(x)_{ij} + \nabla^2 f(x)_{jj} \right] + \mathcal{O}(h^2)\end{split}$$

By the triangle inequality,

$$\begin{split}\left| \alpha - \beta - \eta + \nu \right| &\leq \left| \alpha \right| + \left| \beta \right| + \left| \eta \right| + \left| \nu \right|\\ &\leq \frac{\gamma h^3}{6} \left[ \left\Vert e_i + e_j \right\Vert^3 + \left\Vert e_i - e_j \right\Vert^3 + \left\Vert -e_i + e_j \right\Vert^3 + \left\Vert -e_i - e_j \right\Vert^3 \right]\\ &\leq \frac{16}{3} \gamma h^3.\end{split}$$

Solving for $\nabla^2 f(x)_{ij}$ yields

$$\begin{split}\left| \alpha - \beta - \eta + \nu \right| &= \left| f(x + he_i + he_j) - f(x + he_i - he_j) - f(x - he_i + he_j) + f(x - he_i - he_j) - 4h^2 \nabla^2 f(x)_{ij} \right|\\ &\leq \frac{16}{3} \gamma h^3.\end{split}$$

Hence $\left| A_{ij} - \nabla^2 f(x)_{ij} \right| \leq \frac{4}{3} \gamma h$ where

$$A_{ij} = \frac{f(x + he_i + he_j) - f(x + he_i - he_j) - f(x - he_i + he_j) + f(x - he_i - he_j)}{4h^2}.$$

The number of function evaluations are still the same (unless $f(x)$ is already computed), but this new approximation reduces the upper bound by $20\%$.

Exercise 12

Define $A, M \in \mathbb{R}^{n \times n}$ where $M = M^\top$ and $\hat{A} = \frac{A + A^\top}{2}$.

$$\begin{split}\left\Vert M - \hat{A} \right\Vert_F^2 &= \text{tr}\left( \left( M - \hat{A} \right)^\top \left( M - \hat{A} \right) \right)\\ &= \text{tr}\left(M^\top M\right) - \text{tr}\left(M^\top \hat{A}\right) - \text{tr}\left(\hat{A}^\top M\right) + \text{tr}\left(\hat{A}^\top \hat{A}\right) & \quad & \text{trace operator is a linear mapping}\\ &= \text{tr}\left(M^\top M\right) - 2\text{tr}\left(M^\top \hat{A}\right) + \text{tr}\left(\hat{A}^\top \hat{A}\right) & \quad & \text{trace equivalence under transpose}\\ &= \text{tr}\left(M^\top M\right) - \text{tr}\left(M^\top A\right) - \text{tr}\left(M^\top A^\top\right) + \text{tr}\left(\hat{A}^\top \hat{A}\right)\\ &= \text{tr}\left(M^\top M\right) - \text{tr}\left(M^\top A\right) - \text{tr}\left(A^\top M\right) + \text{tr}\left(\hat{A}^\top \hat{A}\right) & \quad & \text{trace is invariant under cyclic permutations}\\ &= \text{tr}\left(M^\top M\right) - 2 \text{tr}\left(M^\top A\right) + \frac{1}{4} \left\Vert A + A^\top \right\Vert_F^2\\ &\leq \text{tr}\left(M^\top M\right) - 2 \text{tr}\left(M^\top A\right) + \left\Vert A \right\Vert_F^2 & \quad & \text{triangle inequality applied to matrix norm}\\ &\leq \text{tr}\left(M^\top M\right) - 2 \text{tr}\left(M^\top A\right) + \text{tr}\left(A^\top A\right)\\ &= \left\Vert M - A \right\Vert_F^2\\ \left\Vert M - \hat{A} \right\Vert_F &\leq \left\Vert M - A \right\Vert_F\end{split}$$

Exercise 13

Proof by contradiction; assume that $x$ is a local minimizer of $f$ (i.e. $\nabla^2 f(x)$ is not indefinite) and $\nabla^2 f(x)$ is negative definite (i.e. $\forall p \in \mathbb{R}^n, p^\top \nabla^2 f(x) p < 0$).

Since the Hessian is a real symmetric square matrix, its eigendecomposition exists as

$$\nabla^2 f(x) = Q \Lambda Q^\top = \sum_{i = 1}^n \lambda_i q_i q_i^\top.$$

Pick $p = c q_i$ where $q_i$ could be any of the eigenvectors that has a corresponding negative eigenvalue and $c \in \mathbb{R} \setminus 0$ such that $x + p \in D$.

By the continuity of $\nabla^2 f$, there exists an open convex subset $D$ such that for every $0 \leq t \leq 1, x + tp \in D$. Applying Lemma $4.1.5$ for some $z \in [x, x + p]$ yields

$$\begin{split}f(x + p) &= f(x) + \nabla f(x)^\top p + \frac{1}{2} p^\top \nabla^2 f(x) p\\ &= f(x) + \frac{1}{2} p^\top \nabla^2 f(x) p\\ &= f(x) + \frac{1}{2} c^2 \lambda_i\\ f(x + p) &< f(x)\end{split}$$

which is a contradiction because $x$ is a local minimizer of $f$. Thus, $\nabla^2 f(x)$ is positive semidefinite.

This proof can be generalized to functions that are Lipschitz continuous. If $\nabla^2 f \in \text{Lip}_\gamma(D)$ at $x$, then there exists an open convex set $D \subset \mathbb{R}^n$ with $x \in D$, which matches the assumptions of Theorem 4.3.2.

Exercise 14

If $x_*$ is a local minimizer of $f$, Theorem 4.3.2 states that $\nabla^2 f(x_*)$ must be positive semidefinite. Since $\nabla^2 f(x_*)$ is nonsingular, none of its eigenvalues are zero, which means each of them must be positive in this case. This is equivalent to being positive definite.

If $\nabla^2 f(x_*)$ is positive definite, Theorem 4.3.2 states that $x_*$ is a local minimizer of $f$.

Exercise 15

If $f$ has a unique minimizer $x_*$, then $\nabla f(x_*) = 0 = Ax_* + b$. Since $Ax_* = -b$ has a unique solution, it is equivalent to saying $A$ has full rank and is therefore invertible. According to Theorem 4.3.2, $\nabla^2 f(x) = A$ must be positive semidefinite. Hence when $x_*$ is the unique minimizer, $A$ must be positive definite in order to yield $x_* = -A^{-1} b$.

If $A$ is positive definite, Theorem 4.3.2 states that $x$ is a local minimizer when $\nabla f(x) = Ax + b = 0$. Setting $x = -A^{-1} b$ satisfies that condition. The fundamental theorem of Linear Algebra states that $x$ is a unique solution because the nullspace of $A$ is zero.

Exercise 16

$$\begin{split}\DeclareMathOperator*{\argmin}{arg\,min} \argmin_{x \in \mathbb{R}^n} f(x) &= \argmin_{x \in \mathbb{R}^n} \left\Vert Ax - b \right\Vert_2\\ &= \argmin_{x \in \mathbb{R}^n} \left\Vert Ax - b \right\Vert_2^2\\ &= \argmin_{x \in \mathbb{R}^n} x^\top A^\top Ax - 2 b^\top Ax + b^\top b\end{split}$$

The necessary conditions for $x_*$ to be a local minimizer of $f$ are $\nabla f(x_*) = 0$ and $\nabla^2 f(x_*)$ is positive semidefinite. The sufficient condition is that $\nabla^2 f(x_*)$ must be positive definite. Solving for the gradient and the Hessian yields

$$\nabla f(x) = 2 x^\top A^\top A - 2 b^\top A \quad \text{and} \quad \nabla^2 f(x) = 2 A^\top A.$$

Solving for $x_*$ indeed satisfies $A^\top A x_* = A^\top b$; according to Theorem 4.3.2, the Hessian is positive semidefinite, so Theorem 3.6.5 is needed to compute the pseudo-inverse of $A$.