## Homework Problems

### #2

Let $A \in \mathbb{R}^{n \times n}$ be an arbitrary matrix, and let $A^S = (1/2)(A + A^T)$ be it’s symmetric counterpart. Show $x^TAx > 0$ for all $x \neq 0$ if and only if $A^S$ is positive definite.

• WTS $A^S$ is positive definite $\Leftrightarrow$ $x^T A x > 0$
• $\Rightarrow$ Suppose $A^S$ is positive definite
• Then for all $x \in \mathbb{R}^n, x^T A^S x > 0$ (defn of positive definite)
• Then $x^T \mathbf{(1/2) (A + A^T)} x > 0$ (rewrite $A^S$)
• Because $A^S \succ 0$, eigenvalues of $A^S$ are positive (property of positive definite)
• Then the eigenvalues of either $A$ or $A^T$ are positive
• Eigenvalues of $A$ are the eigenvalues of $A^T$ (property of eigenvalues)
• Then the eigenvalues of $A$ and $A^T$ must both be positive
• Because the eigenvalues of $A$ are positive, $A \succ 0$ (property of positive definite)
• Then $x^T A x > 0$ (definition of positive definite)
• Done $\checkmark$
• $\Leftarrow$ suppose $x^T A x > 0$
• Then $A$ is positive definite
• Then the eigenvalues of $A$ (and simmilarly $A^T$) are positive (property of positive definite)
• Then the eigenvalues of $(1/2) A$ and $(1/2) A^T$ are positive (property of eigenvalues)
• Then the eigenvalues of $(1/2)(A + A^T)$ are positive (property of eigenvalues)
• Then the eigenvalues of $\mathbf{A^S}$ are positive (rewrite $(1/2)(A + A^T)$)
• Then $A^S$ is positive definite (property of positive definite)
• Done $\checkmark$
• Then $A^S$ is positive definite $\Leftrightarrow$ $x^T A x > 0$
• Done $\checkmark$

### #3

Let $A \in \mathbb{R}^{n \times n}$ be a symmetric matrix. Assume that there exist a matrix $B \in \mathbb{m \times n}$ such that $A = B^T B$.

Show that $A$ is positive semidefinite.

• The eigenvalues $\lambda_k$ of $A = B^T B$ are non-negative, as $\lambda_k = \beta_k^2$ for each eigenvalue $\beta_k \in B$
• Non-negative eigenvalues implies $A \succeq 0$ (property of positive semidefinite)
• Done $\checkmark$

Show that if $B$ has full column-rank, then $A$ is positive definite.

• Suppose $B$ has full column-rank
• Then the columns of $B$ are linearly independent (defn of full column-rank)
• Then $A = B^T B$ is nonsingular (property of symmetric matrices)
• Then $A$ has nonzero eigenvalues (property of nonsingular matrices)
• Because $A$ is positive semi-definite, it has non-negative eigenvalues.
• Then the eigenvalues of $A$ must be strictly positive.
• Then $A$ is positive definite (property of positive definite)
• Done $\checkmark$

### #4

Let $A \in \mathbb{R}^{n \times n}$ be a symmetric matrix. If $\{ \lambda_i \}^{n}_{i=1}$ are the $n$ (real) eigenvalues of $A$, show that the Rayleigh Quotient satisfies $\frac{x^T A x}{\| x \|^2} \geq \min \{ \lambda_i \}$ for $x \in \mathbb{R}^n$ and $x \neq 0$

• Eigenvectors corresponding to distinct eigenvalues of A are orthogonal to each other (property of symmetric matrices)
• Write $x = \alpha_1 v_1 + \alpha_2 v_2 + \dots \alpha_n v_n$
• Write $A x = \sum_{i=1}^{n} \lambda_i \alpha_i v_i$
• Write $\| x \|^2 = x \cdot x = (\sum_{i=1}^{n} \alpha_i v_i) \cdot (\sum_{j=1}^{n} \alpha_j v_j) = \sum_{i=1}^{n} \alpha_i^2$
• Write $x^T A x = (\sum_{i=1}^{n} \alpha_i v_i) \cdot (\sum_{j=1}^{n} \alpha_j \lambda_j v_j) = \sum_{i=1}^{n} \alpha_{i}^{2} \lambda_i$
• Then
$\frac{x^T A x}{\| x \|^2} = \frac{\sum_{i=1}^{n} \alpha_i^2 \lambda_i}{\sum_{j=1}^{n} \alpha_{j}^{2}} = \sum_{i=1}^{n} \frac{\alpha_i^2}{\sum_{j=1}^n \alpha_j^2} \lambda_i$

Show that the quadratic form $x^T A x$ is coercive if and only if A is positive definite.

• A quadratic form $x^T A x$ is coercive if there exists a constant $c > 0$ such that $x^T A x \geq c \| x \|^2$ for all $x \in \mathbb{R}^n$
• WTS $A^S$ is positive definite $\Leftrightarrow$ $x^T A x$ is coercive
• $\Rightarrow$ Suppose $A^S$ is coercive
• Then there exists $c > 0$ such that $x^T A x \geq c \| x \|^2$ for all $x \in \mathbb{R}^n$
• Then $\frac{x^T A x}{\| x \|^2} \geq c$ for all $x \in \mathbb{R}^n$ (divide both sides by $\| x \|^2$)
• From before, we know $\frac{x^T A x}{\| x \|^2}$ is greater than or equal to the smallest eigenvalue of $A$.
• Then the smallest eigenvalue of $A$ must be positive
• Then all eigenvalues of $A$ must be positive
• Then $A$ is positive definite
• $\Leftarrow$ Suppose $A^S$ is positive definite
• Then the eigenvalues of $A$ are positive
• From before, we know $\frac{x^T A x}{\| x \|^2}$ is greater than or equal to the smallest eigenvalue of $A$.
• let $\lambda_k$ represent the smallest eigenvalue of $A$
• Then $x^T A x \geq \lambda_k \| x \|^2$ for all $x \in \mathbb{R}^n$ (multiply both sides by $\| x \|^2$)
• Because $A$ is positive definite, $\lambda_k$ is positive
• Then $A$ is coercive with positive $c = \lambda_k$

Show that if $a \in \mathbb{R}$ and $b \in \mathbb{R}^n$, then the quadratic function $f(x) = a + b^T x + \frac{1}{2} x^T A x$ is coercive if and only if $A$ is positive definite.

• A scalar function $f(x)$ is coercive if $\lim_{\|x\| \rightarrow \infty} \frac{f(x)}{\|x\|} = \infty$

### #5

To approximate a function $g$ over an interval $[0, 1]$ by a polynomial of degree n we minimize the integral $f(c) = \int_0^1 [g(x)-p_n(x)]^2 dx$ where $p_n(x) = c_0 + c_1 x + c_2 x^2 + \dots c_n x^n$. Find the equations satisfied by the optimal coefficients $\mathbf{c} = (c_0, c_1, c_2, \dots c_n)$ to minimize $f(x)$. Write your answer in terms of matrix $A \in \mathbb{R}^{n \times n}$, vector $b \in \mathbb{R}^n$, and scalar $\alpha$ given by

$A_ij = \int_0^1 x^{i+j} dx = \frac{1}{1+i+j}$, $b_i = \int_0^1 g(x)x^i dx$, $\alpha = \int_0^1 g^2(x) dx$

• WTS a system of equations to solve for $\mathbf{c}$ involving $A$, $b$ and $\alpha$
• The optimal $\mathbf{c}$ is such that $f(c) = f(c_0, c_1, \dots c_n) = \int_0^1 [g(x) - (c_0 + c_1 x + c_2 x^2 + \dots c_n) ]^2 dx$
• Then the optimal derivatives of the $n^{th}$ value of $f(c)$ with respect to $c_n$ is
• $\frac{\delta f}{\delta c_0} = 0 = \int_0^1 2 [g(x) - p_n(x)] (-1)$
• $\frac{\delta f}{\delta c_1} = 0 = \int_0^1 2 [g(x) - p_n(x)] (-x)$
• $\dots$
• $\frac{\delta f}{\delta c_n} = 0 = \int_0^1 2 [g(x) - p_n(x)] (-x)^n$
• Note that $\frac{\delta f}{\delta c_i} = \int_0^1 2 [g(x) - p_n(x)] (-x)^i = 0 \Rightarrow$
$b_i = \int_0^1 g(x) (x)^i dx = \int_0^1 p_n(x) (x)^i dx$ $\int_0^1 p_n(x) (x)^i dx = \int_0^1 [c_0 + c_1 x + c_2 x^2 + \dots + c_j x^j + \dots + c_n x^n ] x^i dx =$ $\int_0^1 c_0 x^i dx + \int_0^1 c_1 x^{i+1} dx + \int_0^1 c_2 x^{i+2} dx + \dots + \int_0^1 c_j x^{i+j} dx + \dots + \int_0^1 c_n x^{i+n} dx =$
• Then we have $\int_0^1 c_j x^{i+j} dx = A_{ij} c_j$
• Note that $\sum_{j=0}^n (\int_0^1 x^{i+j} dx) c_j = b_i$
• Then $\sum_{j=0}^n A_{i j} c_j = b_i$
• Then Ac = b
• done $\checkmark$