Prove that the real and imaginary parts of an eigenvector are linearly independent. 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!
 A: 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 $b$.
A: Let one of the complex eigenvectors be $\vec{v} = \vec{u_1} + i\vec{u_2}$. 
Let us suppose, by way of contradiction, that $\exists k_1,k_2$ both non-zero such that $k_1\vec{u_1} + k_2\vec{u_2} = 0$ (i.e. $\vec{u_1}$  $\vec{u_2}$ are lin. dependent). 
Then, if $\vec{w}$ is the complex conjugate of $\vec{v}$, consider the following: 
$$k_1\vec{v} + ik_2\vec{w} = k_1\vec{u_1} + ik_1\vec{u_2} + ik_2\vec{u_1} + k_2\vec{u_2} = 0$$
Which  is a contradiction since eigenvectors corresponding to different eigenvalues are linearly independent.
A: More generally, if
$\lambda_1,\overline{\lambda}_1,\dotsc,\lambda_r,\overline{\lambda}_r$
are distinct complex eigenvalues of a real matrix,
with corresponding eigenvectors $a_k \pm i b_k$,
then $a_1,b_1,a_2,b_2,\dotsc,a_r,b_r$ are linearly independent
because they have same $(2r)$-dimensional span as the original list of eigenvectors.
