My question is regarding this answer.
The argument goes:
Given a matrix $A=\left( \begin{array}{cc} a&b \\ c&-a\\ \end{array} \right)$ with $a^2 +bc =-1$ we can assume, without loss of generality, that $A = \left( \begin{array}{cc} 0&-1 \\ 1&0\\ \end{array} \right)$.
I took this to mean $A$ is equivalent to $\left( \begin{array}{cc} 0&-1 \\ 1&0\\ \end{array} \right)$ after some change of basis, i.e. they are similar matrices. However, I'm having trouble finding the explicit change of basis.
I would like to know why we can assume $A = \left( \begin{array}{cc} 0&-1 \\ 1&0\\ \end{array} \right)$ and if this is intuitively obvious, why?
Low-tech answer: By explicit computation (using $a^2+bc=-1$), $A^2=-I$. Since $-1$ has no square in $\mathbb{R}$, $A$ has no eigenvalues over $\mathbb{R}$. So given any nonzero vector $v$, $Av$ is linearly independent from $v$. Fix any such $v$ and consider the basis $\{v,Av\}$. Since $A(Av)=-v$, the matrix of $A$ with respect to this basis is $\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}$.
High-tech answer: The trace of $A$ is $0$, and the determinant of $A$ is $-1$. The characteristic polynomial of $A$ is thus $$x^2-tr(A)x+\det(A)=x^2+1.$$ Since the characteristic polynomial is squarefree, there is only one possible rational canonical form (over $\mathbb{R}$) for matrices with this characteristic polynomial. Thus any two such matrices are similar.