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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!

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    $\begingroup$ One thing to note: $Av \neq 0$ does not imply that $\mathrm {Re}(Av) \neq 0$ if no more restrictions. $\endgroup$
    – xbh
    Jul 11, 2019 at 0:52
  • $\begingroup$ Use the fact that $v$ and $\overline v$ are linearly independent. $\endgroup$
    – amd
    Jul 11, 2019 at 0:54
  • $\begingroup$ If $A$ in invertible then every matrix $A_i$ in $A=A_1A_2A_3$ is invertible ( $ \det A=\det A_1 \det A_2 \det A_3$) $\endgroup$
    – Widawensen
    Jul 11, 2019 at 9:23

2 Answers 2

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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.

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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.

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  • $\begingroup$ Hmm... How do you know that $\operatorname{Re}v\ne0$? That doesn’t follow from $v\ne0$ alone. You’re assuming this when you assert that there is a $\gamma$ such that $\operatorname{Im}v = \gamma\operatorname{Re}v$. $\endgroup$
    – amd
    Jul 11, 2019 at 9:05
  • $\begingroup$ @amd Assume that $\operatorname{Re}v = 0$. Then $v= i \operatorname{Im}v$ so $$ia(\operatorname{Im} v) - b\operatorname{Im} v= (a+ib)v = Av = iA(\operatorname{Im}v)$$ so $b\operatorname{Im} v = 0$. We assumed $b \ne 0$ so $\operatorname{Im} v = 0$ which contradicts $v \ne 0$. $\endgroup$ Jul 11, 2019 at 9:32
  • $\begingroup$ @amd Ok, it isn't that obvious but we can WLOG assume that $\operatorname{Re} v \ne 0$, as a symmetric argument applies in the case $\operatorname{Im} v \ne 0$. Note that we must assume $b \ne 0$ since otherwise we have $\operatorname{Im} v = 0$ and thus the OP's statement doesn't hold. $\endgroup$ Jul 11, 2019 at 9:34
  • $\begingroup$ It’s much simpler than all of that: $\operatorname{Re}v\ne0$ and $\operatorname{Im}v\ne0$ are both immediate consequences of the linear independence of $v$ and $\overline v$. $\endgroup$
    – amd
    Jul 11, 2019 at 18:59
  • $\begingroup$ @amd True, but I wanted to avoid using linear independence of $v$ and $\overline{v}$ since you already covered that approach in your answer. $\endgroup$ Jul 11, 2019 at 19:09

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