I am told that for a matrix $A\in \mathbb{R}^{2,2}$ with complex eigenvalue $\lambda=a+ib$ and associated complex eigenvector $v \in \mathbb{C}^n$ then $A=PCP^{-1}$ where $P=\begin{bmatrix} \text{Re } v& \text{Im } v\end{bmatrix}$ and $C=\begin{bmatrix}a&-b\\b&a \end{bmatrix}$.
I know that $A$ must be invertible because it has 2 distinct eigenvalues; I know that $C$ must be invertible because $\text{det} C = a^2 + b^2 \ne 0$, but am not sure how we know that $P$ is invertible.
I know that $$Av = (a+ib)v = av + ibv $$
Further calculations show that $$\text{Re}(Av)=A(\text{Re}v)=a\text{Re}v + b \text{Im}v$$
Can I state that because $A$ is invertible, $Av \ne 0$ and thus $\text{Re}(Av) \ne 0$, which implies that $a\text{Re}v + b \text{Im}v = 0$ does not have a nontrivial solution and therefore $\text{Re}v, \text{Im}v$ must be linearly independent? Not sure if my reasoning is correct.
Any help greatly appreciated!
Best Answer
Observe that if $Av=\lambda v$, then $\overline{Av} = \overline{\lambda v} = \overline\lambda \overline v,$ but since $A$ is real, $\overline A=A$, so $\overline v$ is an eigenvector of $A$ with eigenvalue $\overline\lambda$. This implies that if $\lambda$ is complex, then $v$ and $\overline v$ are linearly independent. If we have $$c_1\Re(v)+c_2\Im(v) = \frac {c_1}2(v+\overline v) - i\frac{c_2}2(v-\overline v) = {c_1-ic_2\over2}v+{c_1+ic_2\over2}\overline v = 0,$$ then we must have $c_1=c_2=0$ because $v$ and $\overline v$ are linearly independent.