Exam 3

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2

At first, we have

$$A=\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]=\frac{1}{2}\left[\begin{array}{cc}-1& 1\\ 1& 1\end{array}\right]\left[\begin{array}{cc}-1& 0\\ 0& 3\end{array}\right]\left[\begin{array}{cc}-1& 1\\ 1& 1\end{array}\right].$$
So,

$$\begin{array}{rl}|A|& =\frac{1}{2}\left[\begin{array}{cc}-1& 1\\ 1& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 3\end{array}\right]\left[\begin{array}{cc}-1& 1\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}2& 1\\ 1& 2\end{array}\right]={\left[|T|\right]}_{\mu },\\ \sqrt{|A|}& =\frac{1}{2}\left[\begin{array}{cc}-1& 1\\ 1& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& \sqrt{3}\end{array}\right]\left[\begin{array}{cc}-1& 1\\ 1& 1\end{array}\right]={\left[\sqrt{|T|}\right]}_{\mu },\\ \sqrt[4]{|A|}& =\frac{1}{2}\left[\begin{array}{cc}-1& 1\\ 1& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& \sqrt[4]{3}\end{array}\right]\left[\begin{array}{cc}-1& 1\\ 1& 1\end{array}\right]={\left[\sqrt[4]{|T|}\right]}_{\mu }.\end{array}$$

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3

Chapter 6 extra note 1

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4

Chapter 6 extra note 1

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5

Let

$n\in \mathbb{N}$, given

$p\in P_n$, we have

$T^{n+1}(p)=\frac{d^{n+1}}{d{x}^{n+1}}p=\mathbf{0}$, where

$\mathbf{0}$ 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

$$T(p)=a_1+\cdots +n{a}_n{x}^{n-1}=\lambda (a_0+a_{1}x+\cdots a_n{x}^n).$$
Equalizing the coefficients gives

$$a_n=\cdots =a_1=0.$$
Therefore, the only eigenvalue is

$\lambda =0$ with eigenvectors

$\text{span}\{1\}\setminus \{\mathbf{0}\}$.

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6.

Let

$T\in \mathcal{L}({\mathbb{P}}_2([-1,1]))$ with

$T(p)=\frac{d}{dx}p(x)$. Determine

$T^{†}(a+bx+c{x}^2)$, where

$T^{†}$ is the pseudo-inverse of

$T$, and

$a,b,c\in \mathbb{R}$.

We already know that one of its orthonormal basis is

$$\beta =\{P_0(x)=\frac{1}{\sqrt{2}},P_1(x)=\sqrt{\frac{3}{2}}x,P_2(x)=\sqrt{\frac{5}{8}}(3x^2-1)\}.$$

1. Matrix representation

The matrix representation of

$T$ is

$$A=[T{]}_{\beta }=\left[\begin{array}{ccc}0& \sqrt{3}& 0\\ 0& 0& \sqrt{15}\\ 0& 0& 0\end{array}\right],$$
so

$$A^{\ast }A=\left[\begin{array}{ccc}0& 0& 0\\ 0& 3& 0\\ 0& 0& 15\end{array}\right],|A|=\left[\begin{array}{ccc}0& 0& 0\\ 0& \sqrt{3}& 0\\ 0& 0& \sqrt{15}\end{array}\right].$$
That gives

${\sigma }_1=\sqrt{15}$,

${\sigma }_2=\sqrt{3}$ and

${\mathbf{v}}_1=[0,0,1{]}^T$,

${\mathbf{v}}_2=[0,1,0{]}^T$.

We can then calculate

$\mathbf{w}$ as

$${\mathbf{w}}_1=\frac{1}{\sqrt{15}}A{\mathbf{v}}_1=[0,1,0{]}^T,{\mathbf{w}}_2=\frac{1}{\sqrt{3}}A{\mathbf{v}}_2=[1,0,0{]}^T.$$

As a result, we have the SVD of

$A$:

$$A=\left[\begin{array}{cc}{\mathbf{w}}_1& {\mathbf{w}}_2\end{array}\right]\left[\begin{array}{cc}{\sigma }_1& 0\\ 0& {\sigma }_2\end{array}\right]\left[\begin{array}{c}{\mathbf{v}}_1^T\\ {\mathbf{v}}_2^T\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 1& 0\\ 0& 0\end{array}\right]\left[\begin{array}{cc}\sqrt{15}& 0\\ 0& \sqrt{3}\end{array}\right]\left[\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\end{array}\right].$$
Regarding the pseudo-inverse, we have

$$A^{†}=\left[\begin{array}{cc}0& 0\\ 0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}\frac{1}{\sqrt{15}}& 0\\ 0& \frac{1}{\sqrt{3}}\end{array}\right]\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\end{array}\right].$$

$T$

Similarly, the Schmidt decomposition of

$T$ as

$$T(p)=\sqrt{15}⟨p,P_2⟩P_1(x)+\sqrt{3}⟨p,P_1⟩P_0(x).$$

and

$$T^{†}(p)=\frac{1}{\sqrt{15}}⟨p,P_1⟩P_2(x)+\frac{1}{\sqrt{3}}⟨p,P_0⟩P_1(x).$$

Finally, we note that

$$a+bx+c{x}^2=(\sqrt{2}a+\frac{\sqrt{2}}{3}c)P_0(x)+\left(\sqrt{\frac{2}{3}}b\right)P_1(x)+\left(\frac{1}{3}\sqrt{\frac{8}{5}}c\right)P_2(x).$$
So, let

$p=a+bx+c{x}^2$ we already have

$$⟨p,P_1⟩=\sqrt{\frac{2}{3}}b,⟨p,P_0⟩=\sqrt{2}a+\frac{\sqrt{2}}{3}c.$$
That gives

$$\begin{array}{rl}{T}^{†}(a+bx+c{x}^2)& =\frac{1}{\sqrt{15}}\left(\sqrt{\frac{2}{3}}b\right)P_2(x)+\frac{1}{\sqrt{3}}(\sqrt{2}a+\frac{\sqrt{2}}{3}c)P_1(x)\\ & =\frac{b}{6}(3x^2-1)+(a+\frac{c}{3})x.\end{array}$$

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7

(a)

Given

$\mathbf{v}\in V$ with

$\Vert \mathbf{v}\Vert \le 1$,

$$\begin{array}{rl}\Vert (T+S)\mathbf{v}\Vert =\Vert T\mathbf{v}+S\mathbf{v}\Vert & \le \Vert T\mathbf{v}\Vert +\Vert S\mathbf{v}\Vert \\ & \le \underset{\Vert \mathbf{v}\Vert \le 1}{max}\Vert T\mathbf{v}\Vert +\underset{\Vert \mathbf{v}\Vert \le 1}{max}\Vert S\mathbf{v}\Vert \\ & ={\sigma }_1(T)+{\sigma }_1(S).\end{array}$$
So,

$${\sigma }_1(T+S)=\underset{\Vert \mathbf{v}\Vert \le 1}{max}\Vert (T+S)\mathbf{v}\Vert \le {\sigma }_1(T)+{\sigma }_1(S).$$

(b)

Given

${\mathbf{v}}_1\in V$ with

$\Vert {\mathbf{v}}_1\Vert =1$ that is the eigenvector corresponds to the largest eigenvalue

${\lambda }_1$, i.e.,

$$T{\mathbf{v}}_1={\lambda }_1{\mathbf{v}}_1.$$
We have

$${\lambda }_1^2=⟨T{\mathbf{v}}_1,T{\mathbf{v}}_1⟩=\Vert T{\mathbf{v}}_1{\Vert }^2\le \underset{\Vert \mathbf{v}\Vert \le 1}{max}\Vert T\mathbf{v}{\Vert }^2={\sigma }_1^2.$$
Since

${\sigma }_1\ge 0$, we obtain

$$|{\lambda }_1|\le {\sigma }_1.$$