I've been having issues to solve the following problem.
Prove that all 2 by 2 matrices (over $\mathbb{R}$) with eigenvalues $\lambda_1=1$ and $\lambda=-1$ can be represented as:
$$ \begin{bmatrix} \cos \theta & a \sin \theta \\ \frac{1}{a} \sin \theta & - \cos \theta \end{bmatrix} $$
or
$$ \begin{bmatrix} \cosh \theta & a \sinh \theta \\ -\frac{1}{a} \sinh \theta & - \cosh \theta \end{bmatrix} $$
Also, when a 2 by 2 matrices (over $\mathbb{R}$) with eigenvalues 1 and -1 are symmetric and orthogonal simultaneouly?
The thing is that I cannot wrap it out how to tackle it, is it related to change of basis?
Look, for example the matrix
$$ \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $$ it has eigenvalues 1 and -1, it is symmetric and orthogonal but I don't see clearly how to prove that it can be represented as the previous matrices (or basis). If it were concern with change of coordinates then I'll have to solve a system of equations, at least for the first vector as follows
\begin{align} 0= c_1 \cos \theta + c_2 a \sin \theta \\ 1= c_1 \frac{1}{a} \sin \theta - c_2 \cos \theta\\ \end{align}
But then, what's the meaning of $\mathbf{a}$? (beacuse I doesn't say anything about it). So is it concern with rotations?
Thanks a lot!
Just a sketch: We know that the characteristic polynomial of the matrix $A$reads $$\lambda^2-\operatorname{trace}(A)\lambda+\det(A),$$ on the other hand $$\lambda^2-(\lambda_1+\lambda_2)\lambda+\lambda_1\lambda_2.$$
Here the trace of the matrix $A$ vanishes and it's determinant is $-1$ so we may write $$A=\begin{pmatrix}c&x\\y&-c\end{pmatrix}.$$ Notice that $x=0$ is trivial; we assume $x\neq0$. I'll consider the case $-1< c<1$, the other case may be treated similarly.
Now if $-1\leq c\leq1$ there exist $\theta$ with $c=\cos(\theta)$. From $-1=\det(A)=-\cos^2(\theta)-xy$ and $\sin^2(\theta)+\cos^2(\theta)=1$ we conclude $xy=\sin^2(\theta)$ where $\sin(\theta)$ doesn't vanish (Why?). Hence we may let $x=a\sin(\theta)$ and $y=\sin(\theta)/a$.
For the remaining case use that $\cosh^2(\theta)-\sinh^2(\theta)=1$.