---
title: Exam 3
tags: Linear algebra
GA: G-77TT93X4N1
---

# Exam 3

<!--

---

## 1

### ( a)
Let $V=\mathbb{R}^2$ with scalar field $\mathbb{R}$. We define
$$
T\left(\begin{bmatrix}x\\ y\end{bmatrix}\right) = \begin{bmatrix}-y\\ x\end{bmatrix}.
$$
If $(x, y)^T$ is an eigenvector we should bave
$$
\lambda \begin{bmatrix}x\\ y\end{bmatrix}=T\left(\begin{bmatrix}x\\ y\end{bmatrix}\right) = \begin{bmatrix}-y\\ x\end{bmatrix}.
$$
That is, $\lambda x = -y$ and $\lambda y = x$. Then
$$
x = \lambda y = \lambda (-\lambda x) = -\lambda^2 x,
$$
that gives $\lambda^2=-1$. Since our field is $\mathbb{R}$, there is no solution.

### ( b)

$$
T(p) = 2\langle p, P_0\rangle P_1(x) + \langle p, P_1\rangle P_2(x).
$$

### ( c)
$$
T(p) = 2\langle p, P_0\rangle P_0(x) + \langle p, P_1\rangle P_1(x).
$$
Then $T(P_0)=2P_0$, $T(P_1)=P_1$ and $T(P_2)=0$.

-->

---

## 2
At first, we have
$$
A = \begin{bmatrix}
1 & 2\\
2 & 1
\end{bmatrix}=\frac{1}{2}
\begin{bmatrix}
-1 & 1\\
1 & 1
\end{bmatrix}
\begin{bmatrix}
-1 & 0\\
0 & 3
\end{bmatrix}
\begin{bmatrix}
-1 & 1\\
1 & 1
\end{bmatrix}.
$$
So,
$$
\begin{align}
|A| &= \frac{1}{2}
\begin{bmatrix}
-1 & 1\\
1 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
0 & 3
\end{bmatrix}
\begin{bmatrix}
-1 & 1\\
1 & 1
\end{bmatrix}=\begin{bmatrix}
2 & 1\\
1 & 2
\end{bmatrix}= \left[|T|\right]_{\mu},\\
\sqrt{|A|} &= \frac{1}{2}
\begin{bmatrix}
-1 & 1\\
1 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
0 & \sqrt{3}
\end{bmatrix}
\begin{bmatrix}
-1 & 1\\
1 & 1
\end{bmatrix}= \left[\sqrt{|T|}\right]_{\mu},\\
\sqrt[4]{|A|} &= \frac{1}{2}
\begin{bmatrix}
-1 & 1\\
1 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
0 & \sqrt[4]{3}
\end{bmatrix}
\begin{bmatrix}
-1 & 1\\
1 & 1
\end{bmatrix} = \left[\sqrt[4]{|T|}\right]_{\mu}.
\end{align}
$$

---

## 3

[Chapter 6 extra note 1](https://hackmd.io/@teshenglin/2024LA2_ch6_1)

---

## 4

[Chapter 6 extra note 1](https://hackmd.io/@teshenglin/2024LA2_ch6_1)

---

## 5

Let $n\in\mathbb{N}$, given $p\in P_n$, we have $T^{n+1}(p)=\frac{d^{n+1}}{dx^{n+1}}p={\bf 0}$, where ${\bf 0}$ denotes the zero polynomial. So $T$ is nilpotent.

Let $n\in\mathbb{N}$, given $p=a_0 + a_1 x +\cdots a_n x^n$, if $p$ is an eigenvector corresponds to eigenvalue $\lambda$, then we should have
$$
T(p) = a_1 + \cdots +na_nx^{n-1} = \lambda(a_0 + a_1 x +\cdots a_n x^n).
$$
Equalizing the coefficients gives
$$
a_n = \cdots =a_1=0.
$$
Therefore, the only eigenvalue is $\lambda=0$ with eigenvectors $\text{span}\{1\}\setminus\{{\bf 0}\}$.

---

## 6.
Let $T\in\mathcal{L}(\mathbb{P}_2([-1, 1]))$ with $T(p) = \frac{d}{dx} p(x)$. Determine $T^{\dagger}(a+bx+cx^2)$, where $T^{\dagger}$ is the pseudo-inverse of $T$, and $a, b, c\in\mathbb{R}$.

We already know that one of its orthonormal basis is
$$
\beta = \left\{P_0(x)=\frac{1}{\sqrt{2}}, \quad
P_1(x)=\sqrt{\frac{3}{2}} x, \quad
P_2(x)=\sqrt{\frac{5}{8}} (3x^2-1)
\right\}.
$$

> > 1. Matrix representation
>
> The matrix representation of $T$ is
> $$
> A = [T]_{\beta} = \begin{bmatrix}
0 & \sqrt{3} & 0\\
0 & 0 & \sqrt{15} \\
0 & 0 & 0
\end{bmatrix},
> $$
> so
> $$
A^*A = \begin{bmatrix}
0 & 0 & 0\\
0 & 3 & 0 \\
0 & 0 & 15
\end{bmatrix}, \quad
|A| = \begin{bmatrix}
0 & 0 & 0\\
0 & \sqrt{3} & 0 \\
0 & 0 & \sqrt{15}
\end{bmatrix}.
$$
> That gives $\sigma_1=\sqrt{15}$, $\sigma_2=\sqrt{3}$ and ${\bf v}_1 = [0, 0, 1]^T$, ${\bf v}_2=[0, 1, 0]^T$.
>
> We can then calculate ${\bf w}$ as
> $$
> {\bf w}_1 = \frac{1}{\sqrt{15}}A{\bf v}_1 = [0, 1, 0]^T, \quad
> {\bf w}_2 = \frac{1}{\sqrt{3}}A{\bf v}_2 = [1, 0, 0]^T.
> $$
>
> As a result, we have the SVD of $A$:
> $$
> A = \begin{bmatrix}
{\bf w}_1 & {\bf w}_2
\end{bmatrix}\begin{bmatrix}
\sigma_1 & 0\\
0 & \sigma_2 \\
\end{bmatrix}\begin{bmatrix}
{\bf v}^T_1 \\
{\bf v}^T_2
\end{bmatrix} = \begin{bmatrix}
0 & 1\\
1 & 0\\
0 & 0
\end{bmatrix}\begin{bmatrix}
\sqrt{15} & 0\\
0 & \sqrt{3} \\
\end{bmatrix}\begin{bmatrix}
0 & 0 & 1\\
0 & 1 & 0
\end{bmatrix}.
> $$
> Regarding the pseudo-inverse, we have
> $$
> A^{\dagger} = \begin{bmatrix}
0 & 0\\
0 & 1\\
1 & 0
\end{bmatrix}\begin{bmatrix}
\frac{1}{\sqrt{15}} & 0\\
0 & \frac{1}{\sqrt{3}} \\
\end{bmatrix}\begin{bmatrix}
0 & 1 & 0\\
1 & 0 & 0
\end{bmatrix}.
> $$
>
> > $T$
>
> Similarly, the Schmidt decomposition of $T$ as
> $$
> T(p) = \sqrt{15}\langle p, P_2\rangle P_1(x)+\sqrt{3}\langle p, P_1\rangle P_0(x).
> $$
>
> and
> $$
> T^{\dagger}(p) =\frac{1}{\sqrt{15}}\langle p, P_1\rangle P_2(x)+\frac{1}{\sqrt{3}}\langle p, P_0\rangle P_1(x).
> $$
>
> Finally, we note that
> $$
> a + bx + cx^2 = \left(\sqrt{2}\,a+\frac{\sqrt{2}}{3}c\right)P_0(x) + \left(\sqrt{\frac{2}{3}}\,b\,\right)P_1(x) + \left(\frac{1}{3}\sqrt{\frac{8}{5}}\,c\,\right)P_2(x).
> $$
> So, let $p=a + bx + cx^2$ we already have
> $$
> \langle p, P_1\rangle = \sqrt{\frac{2}{3}}\,b, \quad
> \langle p, P_0\rangle = \sqrt{2}\,a+\frac{\sqrt{2}}{3}\,c.
> $$
> That gives
> $$
> \begin{align}
> T^{\dagger}(a + bx + cx^2) &=\frac{1}{\sqrt{15}}\left(\sqrt{\frac{2}{3}}\,b\right) P_2(x)+\frac{1}{\sqrt{3}}\left(\sqrt{2}\,a+\frac{\sqrt{2}}{3}\,c\right) P_1(x)\\
> &=\frac{b}{6}(3x^2-1) + \left(a+\frac{c}{3}\right)x.
> \end{align}
> $$
>

---

## 7

### (a)

Given ${\bf v}\in V$ with $\|{\bf v}\|\le 1$,
$$
\begin{align}
\|(T+S){\bf v}\| = \|T{\bf v}+S{\bf v}\| &\le \|T{\bf v}\|+\|S{\bf v}\| \\
&\le\max_{\|{\bf v}\|\le 1}\|T{\bf v}\|+\max_{\|{\bf v}\|\le 1}\|S{\bf v}\|\\
&=\sigma_1(T) + \sigma_1(S).
\end{align}
$$
So,
$$
\sigma_1(T+S) = \max_{\|{\bf v}\|\le 1}\|(T+S){\bf v}\|\le \sigma_1(T) + \sigma_1(S).
$$

### (b)

Given ${\bf v}_1\in V$ with $\|{\bf v}_1\|=1$ that is the eigenvector corresponds to the largest eigenvalue $\lambda_1$, i.e.,
$$
T{\bf v}_1 = \lambda_1{\bf v}_1.
$$
We have
$$
\lambda_1^2 = \langle T{\bf v}_1, T{\bf v}_1\rangle = \|T{\bf v}_1\|^2\le\max_{\|{\bf v}\|\le 1}\|T{\bf v}\|^2=\sigma_1^2.
$$
Since $\sigma_1\ge 0$, we obtain
$$
|\lambda_1|\le \sigma_1.
$$