Linear Algebra: Matrices and Linear Equations - Multiplication of Matrices #2

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 \(AB\)?

Answers

Answer 1

(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 \(\begin{pmatrix}\cos{\theta}&\sin{\theta}\\ -\sin{\theta}&\cos{\theta} \end{pmatrix}\)

Both answers are correct.

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

(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.

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

Answer 2

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

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

\[\therefore \left\|Y\right\| = \left\|X\right\|.\]

Answer 3

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,

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

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