Say we have a 2 by 2 matrix $A$ with real entries and $A$ has a complex eigenvector $V = a+bi$ with corresponding complex eigenvalue $\lambda$. How do I prove that the vectors $\mathrm{Re}(V) = a$ and $\mathrm{Im}(V) = b$ are linearly independent? This is a common fact that is used to produce real solutions for a system of differential equations with complex eigenvalues and vectors.
Assumptions and Facts:
- We know that $\overline{V} = a-bi$ is also an eigenvector with eigenvalue $\overline{\lambda}$.
- We obviously have that $V$ and $\overline{V}$ are linearly independent. (For one thing, eigenvectors of distinct eingenvalues must be LI.)
I have started a few ways by trying to find a contradiction, assuming that there is a nonzero $k \in \mathbb{R}$ such $\mathrm{Re}(V)$ = $k\mathrm{Im}(V)$. I tried using the definitions of $\mathrm{Re}(V)$ and $\mathrm{Im}(V)$ in terms of $V$ and $\overline{V}$ but that didn't get me anywhere.
I tried starting with the fact that $V$ and $\overline{V}$ are LI to show this directly implies that $V$ and $\overline{V}$ line must be LI. I also couldn't work this through. Please help me out!
Best Answer
I think I may have found the proof, let me know if this is solid:
Using the knowledge that $V = a+ib$ and $\overline{V} = a-ib$ are LI, we know that for any $k_1, k_2 \in \mathbb{C}$: $$ k_1 (a+ib) + k_2 (a-ib) = 0 \implies k_1 = k_2 = 0$$ Thus $$ (k_1+k_2)a + (k_1-k_2)ib = 0 \implies k_1 = k_2 = 0$$ So for any two numbers $c_1$ and $c_2$, assume: $$ c_1 a + c_2 b = 0 $$ Then set $$k_1 = \frac{c_1+c_2 i}{2} \; ; \; k_2 = \frac{c_1-c_2 i}{2} $$ so the statement becomes $$ (k_1+k_2)a + (k_1-k_2)ib = 0 $$ which implies $ k_1 = k_2 = 0$ which in turn implies $ c_1 = c_2 = 0$.
So we have linear independence of $a$ and $b$.