Given $A = \begin{bmatrix} 0 & 1 \\ -4 & 0 \end{bmatrix}$ find a matrix C of the form $ C = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$ and a matrix P such that $A = PCP^{-1}$. Then factor the C matrix as a product of a scaling matrix of the form $rI_2$ and a rotation matrix of the form $\begin{bmatrix} \cos(\varphi) & -\sin(\varphi) \\ \sin(\varphi) & \cos(\varphi) \end{bmatrix}$.
I can get everything correct except for the factoring C part. I get the angle being $-\pi / 2$, but I get the wrong C unless I use $r = -2,$ but I thought that isn't allowed.
*edit: The theorem in my textbook:
Let A be a 2×2 matrix with a complex eigenvalue $\lambda = a – bi \ (b\neq0)$ and an associated eigenvector $\vec{v}$ in $\mathbb{C^2}$. Then
$$ A = PCP^{-1}, \ \text{where} \ P = \begin{bmatrix} Re(\vec{v}) & Im(\vec{v}) \end{bmatrix} and C = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$$
Best Answer
Write $R$ for the rotation matrix, so that $C=rIR$. Taking determinants, $$\det(C)=r^2$$ because $\det(R)=1$. But $\det(C)=\det(A)=4$, so we can take $r=\pm2$. Depending which $C$ you have chosen (there are two possibilities) you will then get $\phi=\pm\pi/2$.