(I apologize in advance for the formatting. This is my first post here. Advice is welcome in that regard.)
This is the second part of a two-part question. The first part asks the following:
"If 2i is an eigenvalue of a real 2 x 2 matrix A, find A^2."
For this part, I simply used the fact that the eigenvalues for a matrix {{a,c},{d,a}} are given by a +/- sqrt(cd). I then let a = 0, c = 1, and d = 4/c = 1, giving me the matrix {{0,1},{4,0}}. Sure enough, that has the eigenvalue 2i. I, then, just multiplied it by itself to get A^2. I assume, however, that I was supposed to use the "A^2" part of the problem as some sort of a "hint" for its solution, and that's why I'm getting caught up on the second part of the problem. Anyway, the "nonzero" part is the issue for me because the matrix I used for the first part clearly has zero entries, and I can't figure out how one is supposed to construct a matrix with the given properties without zeroes in the "a" positions (seeing as the eigenvalues are given by a +/- sqrt(cd)).
So, that's my problem.
Here is a simple calculation. Let $A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$. Then the characteristic polynomial of $A$ is $\chi(t)=t^2-t(a+d)+ad-bc$. We want that $\chi(t)=t^2+4$. Then $2i$ is a root. Now this means, we want to solve the equations $a+d=0$ and $ad-bc=4$. Setting $d=-a$ we can choose any real $a,b,c$ with $a^2+bc=-4$. For example, $a=1$, $b=1$ and $c=-5$. That is, $A=\begin{pmatrix} 1 & 1 \\ -5 & -1 \end{pmatrix}$.
Furthermore, in this case, by Cayley Hamilton we have $A^2=-4I_2$, because $\chi(t)=t^2+4$.