Prove all $2\times2$ real matrices with eigenvalues $1$ and $-1$ can be represented as

Question

My exercise asks:

Prove that all $2\times2$ real matrices with eigenvalues $\lambda_1=1$ and $\lambda_2=-1$ can be represented as

\begin{equation} \begin{bmatrix} \cos\theta & a\sin\theta \\ \frac{1}{a}\sin\theta & -\cos\theta \end{bmatrix} \end{equation}

Starting like

\begin{equation} \begin{bmatrix} a & b \\ c & d \end{bmatrix} \end{equation}

Using the fact that $\operatorname{Tr}(A) = a+d =\lambda_1+\lambda_2 = 0 \Rightarrow a=-d$. And using the fact that $\det A = ad-bc=\lambda_1\lambda_2=-1 \Rightarrow-a^2-bc=-1$

\begin{equation} a^2+bc=1 \end{equation}

How can I complete the proof?

Answer 1

$a,b,c,d$ are fixed numbers and we have to show that they can be represented as $\cos\theta, k\sin\theta, \frac1k \sin\theta, -\cos\theta$

Let $\cos \theta=a \implies d = -a = -\cos \theta$
This can be done because there always exists a $\theta \in \mathbb{C}$ such that $\cos\theta = a$

Now, $$ a^2 + bc=1$$ $$ bc = 1-a^2 = \sin^2 \theta$$ You will always be able to choose a $k$ such that $b=k\sin\theta$
This is because $$ \sin^2\theta + \cos^2\theta=1$$ $$ \dfrac{b^2}{k^2} + a^2 = 1 $$ $$ k = \pm\dfrac{b}{\sqrt{1-a^2}} $$ This $k$ will also satisfy $c = \dfrac1k \sin\theta$ because $c = \dfrac{\sin^2\theta}{b}$

That completes the proof.

Answer 2

You cannot because it is wrong. For a counterexample, consider $\begin{pmatrix}2 & -3 \\ 1 & -2 \end{pmatrix}$. Since $\cos{\theta} \leq 1$, the above is a counter-example.