Prove that the real and imaginary parts of an eigenvector are linearly independent.

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 $b$.

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.