2 분 소요

# Questions

## Question 1

Rotations. Let $R(\theta)$ be the matrix given by

$R(\theta) = \begin{pmatrix}\cos{\theta}&-\sin{\theta}\\ \sin{\theta}&\cos{\theta}\end{pmatrix}.$

(a) Show that for any two numbers $\theta_{1}$, $\theta_{2}$ we have

$R(\theta_{1})R(\theta_{2}) = R(\theta_{1} + \theta_{2}).$

[You will have to use the addition formulas for sine and cosine.]

(b) Show that the matrix $R(\theta)$ has an inverse, and write down this inverse.

(c) Let $A = R(\theta)$. Show that

$A^{2} = \begin{pmatrix}\cos{2\theta}&-\sin{2\theta}\\ \sin{2\theta}&\cos{2\theta} \end{pmatrix}.$

(d) Determine $A^{n}$ for any positive integer $n$. Use induction.

## Question 2

For any vector $X$ in $R^{2}$ let $Y = R(\theta)X$ be its rotation by an angle $\theta$. Show that $\left\|Y\right\| = \left\|X\right\|.$

## Question 3

Let $A$ be a square matrix which is of the form

$\begin{pmatrix}a_{11}&*&\cdots\cdots&&*\\ 0&a_{22}&*&\cdots&*\\ \vdots&&\ddots&&\vdots\\ &&&&*\\ 0&\cdots\cdots&0&&a_{nn}\end{pmatrix}$

The notation means that all elements below the diagonal are equal to 0, and the elements above the diagonal are arbitrary. One may express this property by saying that

$a_{ij}=0 \text{ if } i>j.$

Such a matrix is called upper triangular. If $A$, $B$ are upper triangular matrices(of the same size) what can you say about the diagonal elements of $AB$?

(a)

$R(\theta_{1})R(\theta_{2}) = \begin{pmatrix}\cos{\theta_{1}}&-\sin{\theta_{1}}\\ \sin{\theta_{1}}&\cos{\theta_{1}}\end{pmatrix}\begin{pmatrix}\cos{\theta_{2}}&-\sin{\theta_{2}}\\ \sin{\theta_{2}}&\cos{\theta_{2}}\end{pmatrix}$ $=\begin{pmatrix}\cos{\theta_{1}}\cos{\theta_{2}}-\sin{\theta_{1}}\sin{\theta_{2}}&-\cos{\theta_{1}}\sin{\theta_{2}}-\sin{\theta_{1}}\cos{\theta_{2}}\\ \sin{\theta_{1}}\cos{\theta_{2}}+\cos{\theta_{1}}\sin{\theta_{2}}&-\sin{\theta_{1}}\sin{\theta_{2}}+\cos{\theta_{1}}\cos{\theta_{2}} \end{pmatrix}$ $=\begin{pmatrix}\cos{(\theta_{1}+\theta_{2})}&-\sin{(\theta_{1}+\theta_{2})}\\ \sin{(\theta_{1}+\theta_{2})}&\cos{(\theta_{1}+\theta_{2})} \end{pmatrix}=R(\theta_{1}+\theta_{2})$

(b)

The answer to the question is $R(-\theta)$

My answer to the question is $\begin{pmatrix}\cos{\theta}&\sin{\theta}\\ -\sin{\theta}&\cos{\theta} \end{pmatrix}$

$\because \begin{pmatrix}\cos{(-\theta)}&-\sin{(-\theta)}\\ \sin{(-\theta)}&\cos{(-\theta)}\end{pmatrix} = \begin{pmatrix}\cos{\theta}&\sin{\theta}\\ -\sin{\theta}&\cos{\theta} \end{pmatrix}$ $(\because \sin(-\theta) = -\sin(\theta),\text{ } \cos(-\theta) = \cos(\theta))$

(c)

$A = R(\theta)$ $A^2 = \begin{pmatrix}\cos{\theta}&-\sin{\theta}\\ \sin{\theta}&\cos{\theta} \end{pmatrix}\begin{pmatrix}\cos{\theta}&-\sin{\theta}\\ \sin{\theta}&\cos{\theta} \end{pmatrix}$ $=\begin{pmatrix}\cos^2{\theta}-\sin^2{\theta}&-2\sin{\theta}\cos{\theta}\\ 2\sin{\theta}\cos{\theta}&\cos^2{\theta}-\sin^2{\theta} \end{pmatrix} = \begin{pmatrix}\cos{2\theta}&-\sin{2\theta}\\ \sin{2\theta}&\cos{2\theta} \end{pmatrix}$ $(\because \sin{2\theta} = \sin{(\theta+\theta)} = \sin{\theta}\cos{\theta}+\cos{\theta} sin{\theta} = 2\sin{\theta}\cos{\theta})$ $(\because \cos{2\theta} = \cos{(\theta+\theta)} = \cos^2{\theta}-\sin^2{\theta})$

(d)

$A=\begin{pmatrix}\cos{\theta}&-\sin{\theta}\\ \sin{\theta}&\cos{\theta} \end{pmatrix}$ $A^{2}=\begin{pmatrix}\cos{2\theta}&-\sin{2\theta}\\ \sin{2\theta}&\cos{2\theta} \end{pmatrix}$ $\vdots$ $A^{k} = \begin{pmatrix}\cos{k\theta}&-\sin{k\theta} \\ \sin{k\theta}&\cos{k\theta} \end{pmatrix}?$

Assume True:

$A^{k} = \begin{pmatrix}\cos{k\theta}&-\sin{k\theta} \\ \sin{k\theta}&\cos{k\theta} \end{pmatrix}$

Show True:

$A^{k+1} = \begin{pmatrix}\cos{(k+1)\theta}&-\sin{(k+1)\theta} \\ \sin{(k+1)\theta}&\cos{(k+1)\theta} \end{pmatrix}$

$Proof.$ $A^{k+1} = A^{k}A = AA^{k} = \begin{pmatrix}\cos{\theta}&-\sin{\theta}\\ \sin{\theta}&\cos{\theta} \end{pmatrix}\begin{pmatrix}\cos{k\theta}&-\sin{k\theta}\\ \sin{k\theta}&\cos{k\theta} \end{pmatrix}$ $=\begin{pmatrix}\cos{\theta}\cos{k\theta}-\sin{\theta}\sin{k\theta}&-\cos{\theta}\sin{k\theta}-\sin{\theta}\cos{k\theta}\\ \sin{\theta}\cos{k\theta}+\cos{\theta}\sin{k\theta}&-\sin{\theta}\sin{k\theta}+\cos{\theta}\cos{k\theta} \end{pmatrix}$ $=\begin{pmatrix}\cos{(k+1)\theta}&-\sin{(k+1)\theta} \\ \sin{(k+1)\theta}&\cos{(k+1)\theta} \end{pmatrix}$

$\because \begin{pmatrix}\cos{(k+1)\theta}&-\sin{(k+1)\theta} \\ \sin{(k+1)\theta}&\cos{(k+1)\theta} \end{pmatrix} = \begin{pmatrix}\cos{k\theta}\cos{\theta}-\sin{k\theta}\sin{\theta}&-(\sin{k\theta}\cos{\theta}+\cos{k\theta}\sin{\theta})\\ \sin{k\theta}\cos{\theta}+\cos{k\theta}\sin{\theta}&\cos{k\theta}\cos{\theta}-\sin{k\theta}\sin{\theta} \end{pmatrix}$

So $A^{k+1}$ is true.

$\therefore \text{The given statement is true for all k.}$

$Y = R(\theta)X = \begin{pmatrix}\cos{\theta}&-\sin{\theta}\\ \sin{\theta}&\cos{\theta} \end{pmatrix}\begin{pmatrix}x_{1}\\ x_{2}\end{pmatrix}$ $(\because \text{for any vector } X \text{ in } R^2.)$ $=\begin{pmatrix}\cos{\theta}&-\sin{\theta}\\ \sin{\theta}&\cos{\theta} \end{pmatrix}\begin{pmatrix}x1\\ x2\end{pmatrix}=\begin{pmatrix}{x_{1}}\cos{\theta}-{x_{2}}\sin{\theta}\\ {x_{1}}\sin{\theta}+{x_{2}}\cos{\theta} \end{pmatrix}$ $\therefore y_{1} = {x_{1}}\cos{\theta}-{x_{2}}\sin{\theta}, y_{2} = {x_{1}}\sin{\theta}+{x_{2}}\cos{\theta}$

$Y^{2} = {y_{1}}^2+{y_{2}}^2$ $={x_{1}}^2\cos^2{\theta}-2{x_{1}}{x_{2}}\cos{\theta}\sin{\theta}+{x_{2}}^2\sin^2{\theta}+{x_{1}}^2\sin^2{\theta}+2{x_{1}}{x_{2}}\sin{\theta}\cos{\theta}+{x_{2}}^2\cos^2{\theta}$ $={x_{1}}^2(\cos^2{\theta}+\sin^2{\theta})-2{x_{1}}{x_{2}}\cos{\theta}\sin{\theta}+2{x_{1}}{x_{2}}\sin{\theta}\cos{\theta}+{x_{2}}^2(\sin^2{\theta}+\cos^2{\theta})$ $={x_{1}}^2+{x_{2}}^2$ $(\because \sin^2{\theta}+\cos^2{\theta} = 1)$ $=X^{2}$

$\therefore \left\|Y\right\| = \left\|X\right\|.$

Let $B$ be the matrix given by

$B = \begin{pmatrix}b_{11}&*&\cdots\cdots&&*\\ 0&b_{22}&*&\cdots&*\\ \vdots&&\ddots&&\vdots\\ &&&&*\\ 0&\cdots\cdots&0&&b_{nn}\end{pmatrix}$

Also, The above-mentioned A was:

$\begin{pmatrix}a_{11}&*&\cdots\cdots&&*\\ 0&a_{22}&*&\cdots&*\\ \vdots&&\ddots&&\vdots\\ &&&&*\\ 0&\cdots\cdots&0&&a_{nn}\end{pmatrix}$

Thus,

$AB = \begin{pmatrix}a_{11}b_{11}&*&\cdots\cdots&&*\\ 0&a_{22}b_{22}&*&\cdots&*\\ \vdots&&\ddots&&\vdots\\ &&&&*\\ 0&\cdots\cdots&0&&a_{nn}b_{nn}\end{pmatrix}$

$\therefore \text{The diagonal elements of AB are } a_{11}b_{11}, \cdots, a_{nn}b_{nn}$

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