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(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.

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  • $\begingroup$ Your matrix doesn't work, you need a negative sign in one of the entries. To find a matrix with the given properties, take your $A$ (after you fix it) and consider $P^{-1}AP$. If you don't make $P$ to simple, you can even do this by trial and error. I got $\begin{bmatrix} -8 & 4\\ -17& 8\end{bmatrix}$ on my first try. $\endgroup$
    – Git Gud
    Dec 2, 2014 at 16:10
  • $\begingroup$ Ah, yes. You're right. Thanks. I don't know why I thought that that worked. I need only to make either the 4 or the 1 negative in my original matrix to make it work. And, yes, your matrix works for the problem I asked. I guess I don't really understand how to implement the (P^1)(A)(P) aspect, which I imagine I should know. Where does the P come from? $\endgroup$ Dec 2, 2014 at 16:34
  • $\begingroup$ Similar matrices have the same eigenvalues, that's where $P$ comes from. $\endgroup$
    – Git Gud
    Dec 2, 2014 at 16:35

1 Answer 1

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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 $\chi(t)=t^2+4$.

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  • $\begingroup$ This was my original approach which I had abandoned, for some reason (I think I thought that I had too many unknowns, or something). Thank you! $\endgroup$ Dec 2, 2014 at 16:41
  • $\begingroup$ I think you should have $ad-bc=4$, so that then with $a=-d$ you have $-a^2-bc=4$, or $a^2+bc=-4$. $\endgroup$
    – Ian
    Dec 2, 2014 at 16:51
  • $\begingroup$ @Ian yes, thank you ! $\endgroup$ Dec 2, 2014 at 18:40

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