Let A be a 2x2 matrix with complex eigenvalue x = a-bi and an associated eigenvector v.
Show that A(Rev) = aReV + bImv and A(Imv) = -bRev + amImv
I'm really stuck.
Your question does not many any sense. What is the meaning of "Rev" and "Imv". The real and imaginary parts of what? How can v have real and imaginary parts? Now, v is an element of R^2 and so you can think of "Rev" and "Imv" as the first and second components of v. However, even if you think that way then A(Rev) makes no sense. This is because A is a 2x2 matrix and Rev is just a number, it is not a matrix.
Actually the question does make sense, a concrete example clarifies things I think:
Let $\displaystyle A = \left(\begin{array}{cc}3 & -2 \\ 4 & -1\end{array}\right) $.
Then $\displaystyle \lambda = 1 \pm 2i$.
The eigenvector associated with $\displaystyle \lambda = 1 - 2i$ is $\displaystyle v = \left(\begin{array}{c} 1 \\ 1 + i \end{array}\right) = \left(\begin{array}{c} 1 \\ 1 \end{array}\right) + i \left(\begin{array}{c} 0 \\ 1 \end{array}\right)$.
Then $\displaystyle Re(v) = \left(\begin{array}{c} 1 \\ 1 \end{array}\right) $ and $\displaystyle Im(v) = \left(\begin{array}{c} 0 \\ 1 \end{array}\right) $.
It's easy to see that what's been asked to be shown works here.
Since my linear algebra is terrible (and I'm short on time) I'll leave it to more expert members to prove what's been asked to be shown.