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?
Best Answer
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.