# 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!

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

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.

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.

• 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 $\operatorname{Im}v = \gamma\operatorname{Re}v$.
– amd
Jul 11, 2019 at 9:05
• @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 $v \ne 0$. Jul 11, 2019 at 9:32
• @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. Jul 11, 2019 at 9:34
• 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 $\overline v$.
– amd
Jul 11, 2019 at 18:59
• @amd True, but I wanted to avoid using linear independence of $v$ and $\overline{v}$ since you already covered that approach in your answer. Jul 11, 2019 at 19:09