Diagonalization with Complex Eigenvalues when no Numbers are Given

Question

I have run into a confusing problem in my linear algebra class:

Let $A$ be a real $2 \times 2$ matrix with a complex eigenvalue $\lambda = a-bi (b\ne 0)$ and an associated eigenvector $V$ in $\mathbb{C}^2$.

Show that $A(\Re V)= a\Re V + b\Im V$ and $A(\Im v) = -b\Re V + a\Im V$.

I suspect that I should solve it using this result:

However, I don't know how to invert a P matrix that is defined as $P = [\Re V, \Im V]$, nor how to multiply this $2\times 2$ matrix A by $[\Re V, \Im V]$ when the values aren't given. It would be much appreciated if I could get some direction on this problem and/or a solution. Thank you!

Answer 1

It is in fact very straightforward; you don't need the equation

$A = PCP^{-1}, \; \text{etc}; \tag 0$

instead, we may work directly from the definition

$V = \Re V + i\Im V; \tag 1$

we thus have, since $V$ is an eigenvector for eigenvalue $a - bi$,

$A\Re V + iA\Im V = A(\Re V + i\Im V) = AV$ $= (a - bi)(\Re V + i \Im V) = (a\Re V + b \Im V) + i(a \Im V - b\Re V); \tag 2$

equating real and imaginary parts, using the fact that $A$, $\Re V$ and $\Im V$ are real:

$A\Re V = a\Re V + b\Im V, \tag 3$

$A\Im V = a\Im V - b\Re V, \tag 4$

which was to be proved.

Answer 2

For the eigenvector $V$ we have: $$ AV=A(\Re V+i\Im V)= (a-ib)(\Re V+i\Im V) = $$ $$ = a\Re V+ia\Im V -ib \Re V-ib \cdot i\Im V$$

reordering you find the result