1. ## Rotation matrix.

Explain what it means for a matrix $\displaystyle M$ to be orthogonal.
With respect to the standard basis in cartesian plane, the anticlockwise rotation through angle $\displaystyle \theta$ around the origin is represented by:

$\displaystyle {M}_{\theta }$ =

($\displaystyle \cos \theta$ $\displaystyle -\sin \theta$)
($\displaystyle \sin \theta$ $\displaystyle \cos \theta$)

Supposed to be 2x2 matrix sorry I don't know the tex for it?

Verify that $\displaystyle {M}_{\theta }$ is orthogonal and verify the identity $\displaystyle {M}_{\theta + \varphi} = {M}_{\theta}{M}_{\varphi}$

2. Originally Posted by TeaWithoutMussolini
Explain what it means for a matrix $\displaystyle M$ to be orthogonal.
With respect to the standard basis in cartesian plane, the anticlockwise rotation through angle $\displaystyle \theta$ around the origin is represented by:

$\displaystyle {M}_{\theta }$ =

($\displaystyle \cos \theta$ $\displaystyle -\sin \theta$)
($\displaystyle \sin \theta$ $\displaystyle \cos \theta$)

Supposed to be 2x2 matrix sorry I don't know the tex for it?

Verify that $\displaystyle {M}_{\theta }$ is orthogonal and verify the identity $\displaystyle {M}_{\theta + \varphi} = {M}_{\theta}{M}_{\varphi}$
I will find all the orthogonal matrices at $\displaystyle M_2(\mathbb{R})$.

Let $\displaystyle A=\begin{pmatrix}a & b\\ c & d\end{pmatrix}$ orthogonal matrix .

Hence,$\displaystyle AA^*=AA^T=I$.

In other words:

$\displaystyle A^T=A^{-1}=\frac{1}{|A|}\begin{pmatrix}d & -b\\ -c & a\end{pmatrix}.$

Now, from $\displaystyle AA^T=I$ we get:

$\displaystyle 1=|I|=|AA^T|=|A||A^T|=A^2$

Hence $\displaystyle |A|=\pm1$ .

Suppose that $\displaystyle |A|=1$, then:

$\displaystyle \begin{pmatrix}a & c\\ b & d\end{pmatrix}=\begin{pmatrix}d & -b\\ -c & a\end{pmatrix}$

So:$\displaystyle b=-c, a=d$.

$\displaystyle A^{+}=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}$

In the case in which $\displaystyle |A|=-1$ we will get:

$\displaystyle \begin{pmatrix}a & c\\ b & d\end{pmatrix}=\begin{pmatrix}-d & b\\ c & -a\end{pmatrix}$

So:$\displaystyle b=s, a=-d$

$\displaystyle A^{-}=\begin{pmatrix}a & b\\ b & -a\end{pmatrix}$

In both cases we can deduce from orthogonality that $\displaystyle a^2+b^2=1$.

$\displaystyle |a|\leq 1$ so exists angel $\displaystyle 0\leq \theta \leq 2\pi$ for which$\displaystyle a=cos\theta$ and then $\displaystyle b=sin\theta$.

Finally we can write he follwing:

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

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

3. For the first bit here is a link. Orthogonal matrix - Wikipedia, the free encyclopedia

The LaTex code is

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

For the last part verify the definitions that you have or the one's in the link and I would start with the left hand side of the equality. Use matrix multiplication and then use the sum identities for sine and cosine.

Good luck