Linear independence of real and imaginary parts of complex eigenvector 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!
 A: 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.
A: Assume $b \ne 0$, otherwise $\operatorname{Im} v = 0$ so your claim doesn't hold. We have
$$(a \operatorname{Re} v - b\operatorname{Im}v) + i(b\operatorname{Re}v + a \operatorname{Im} v) = (a+ib)v = Av = \operatorname{Re} Av + i\operatorname{Im}Av$$
so
$$\begin{cases}\operatorname{Re} Av = a \operatorname{Re} v - b\operatorname{Im}v \\ \operatorname{Im} Av = b \operatorname{Re} v +a\operatorname{Im}v \end{cases}$$
Now, assume that $\operatorname{Re} v$ and $\operatorname{Im} v$ are linearly dependent. If $\operatorname{Re} v = 0$, then $v = i\operatorname{Im} v$ so $$ia\operatorname{Im} v - b\operatorname{Im} v= (a+ib)v = Av = iA(\operatorname{Im}v)$$
It follows $b\operatorname{Im} v = 0$ so $\operatorname{Im} v = 0$ which contradicts $v \ne 0$.
Hence $\operatorname{Re} v \ne 0$ and $\exists \gamma \in \mathbb{R}$ such that $\operatorname{Im} v = \gamma \operatorname{Re} v$. Then also $\operatorname{Im} Av = A(\operatorname{Im} v) = \gamma (\operatorname{Re} v)= 
 \gamma \operatorname{Re} Av$ so
$$\begin{cases}\operatorname{Re} Av = a \operatorname{Re} v - b\gamma\operatorname{Re}v  = (a-b\gamma)\operatorname{Re}v\\ \gamma\operatorname{Re} Av = b \operatorname{Re} v +a\gamma\operatorname{Re}v =(b+a\gamma)\operatorname{Re}v\end{cases}$$
Since $v \ne 0$ we have $\operatorname{Re}v \ne 0$ so
$$(b+a\gamma)\operatorname{Re}v =  \gamma\operatorname{Re} Av = \gamma(a-b\gamma)\operatorname{Re}v \implies b+a\gamma = \gamma(a-b\gamma) = a\gamma - b\gamma^2$$
or $b(1+\gamma^2) = 0$. This implies $b = 0$ which is a contradiction.
We conclude that $\operatorname{Re} v$ and $\operatorname{Im} v$ are linearly independent.  
