How to show linear independence of complex eigenvector solutions I'm having difficulty with a showing linear independence...
Let $t = a + ib, b \neq 0$ be an eigenvalue of real matrix $A$ with associated eigenvector $z = p + iq$
Then the two real solutions
$X_1(t) = (p\cos(bt) - q\sin(bt))e^{at}$
$X_2(t) = (p\sin(bt) + q\cos(bt))e^{at}$
are linearly independent solutions of $X' = AX$
How do I prove this?
 A: Two functions $x_1(t)$ and $x_2(t)$ are linearly independent if $ax_1(t) + bx_2(t) = 0$ implies that $a=b=0$. We will demonstrate this is satisfied below.

Suppose that $P$ and $Q$ (complex) are such that $PX_1(t) + QX_2(t) = 0$ for all $t$.
Thus $$(P\cdot p + Q \cdot q) \cos(bt) e^{at} + (Q\cdot p - P \cdot q) \sin(bt) e^{at} = 0$$
$\cos(bt)e^{at}$ and $\sin(bt)e^{at}$ are linearly independent, since $\cos$ is even and $\sin$ odd.
Thus $$P\cdot p + Q \cdot q = 0$$ and $$Q\cdot p - P \cdot q = 0$$ by linear independence. You could solve for $P$ and $Q$ to show that they must be zero. Otherwise you can use a bit of linear algebra.
This system of equations corresponds to the matrix equation $$\left( \begin{array}{cc} p& -q\\q & p\end{array}\right)\left(\begin{array}{c}P\\Q\end{array}\right) = \left(\begin{array}{c}0\\0\end{array}\right)$$
Since this matrix is invertible (determinant is nonzero), the only solution to this equation is $P=0$ and $Q=0$.

The only thing left to do, if you can't assume this already, is to show that $\cos(bt)e^{at}$ and $\sin(bt)e^{at}$ are linearly independent.
It the same sort of thing: Suppose that there are $S$ and $T$ for which $$S\cos(bt)e^{at} + T\sin(bt)e^{at} = e^{at}(S\cos(bt) + T\sin(bt))=0$$
for all $t$.
$e^{at}$ is never zero, which means that $S\cos(bt) + T\sin(bt)=0$ for all $t$. Note that this gives $S\cos(-bt)+T\sin(-bt) = S\cos(bt)-T\sin(bt)=0$ for all $t$.
Adding the two equations yields $2S\cos(bt) = 0$ for all $t$, and since $\cos(bt)$ is not identically zero, $S=0$. Plugging $S=0$ into the first equation this tells us that $T\sin(bt)=0$ for all $t$. Hence $T=0$.
