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Give a $2 \times 2$ real matrix $A$ with eigenvalues $2+3i$, $2-3i$.

I would like hints only.

So far, I've been trying get somewhere with $\det[A-(2+3i)I] = 0$ and $\det[A-(2-3i)I] = 0$; which gives me a set of two equations:

1) $(x_1-2-3i)(y_2-2-3i)-x_2y_1 = 0$

2) $(x_1-2+3i)(y_2-2+3i)-x_2y_1 = 0$,

where $x_1$ is the first entry of $A$, $x_2$ is the $a_{1,2}$ entry of $A$, and so on.

I'm getting nowhere with this method, though...

Thanks,

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4 Answers 4

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Hint: $Tr(A) = A_{11} + A_{22} = 4$ (sum of eigenvalues)

Another Hint: $Det(A) = A_{11} * A_{22} - A_{12} * A_{21}= 13$ (product of eigenvalues)

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  • $\begingroup$ Hi @EmilianoSorbello - thanks for your hint. Can you offer another hint? As I mentioned to Rogelio Molina (below), I'm not sure how to proceed with just two equations in four unknowns... $\endgroup$
    – User001
    Jul 4, 2015 at 2:29
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    $\begingroup$ You don't need to have 4 equations for 4 variables because there's probably more than 1 matrix that applies. ANY real matrix with tr(A) = 4 and Det(A) = 13 WILL satisfy what you need, so you try to produce one with just that, or you can add more weird conditions yourself, and see what you get! like a = 1, or a = d, something like that $\endgroup$
    – Dleep
    Jul 4, 2015 at 2:32
  • $\begingroup$ ahhh....right.... :-) $\endgroup$
    – User001
    Jul 4, 2015 at 2:33
  • $\begingroup$ so I got my matrix and verified its eigenvalues were 2+3i and 2-3i with the quadratic formula. thanks so much! $\endgroup$
    – User001
    Jul 4, 2015 at 2:58
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Hint: start by finding the characteristic polynomial of $A$, and try working from there.

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  • $\begingroup$ Hi @Omnomnomnom, this was my first attempt, actually, before posting the question. Unfortunately, I got nowhere with this, and just got a couple of messy equations. So I am going to try using the trace and determinant of A to come up with something... thanks, as usual :-) $\endgroup$
    – User001
    Jul 4, 2015 at 2:32
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Hint: If $a$ and $b$ are real numbers, what are the eigenvalues of $\begin{bmatrix}a & -b \\ b & a\end{bmatrix}$?

Alternatively, notice that $\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}$ (the matrix which represents a rotation by $180^{\circ}$) has eigenvalues $\pm i$. Can you manipulate this matrix into one which has eigenvalues $2 \pm 3i$?

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Hint: What is the trace and determinant of the matrix?

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  • $\begingroup$ Hi @RogelioMolina, with yours and others hints, I get the equations ad - bc = 13, and a + d = 4. But I'm not sure where I can go from here. I have two equations in four unknowns... $\endgroup$
    – User001
    Jul 4, 2015 at 2:27
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    $\begingroup$ That is even better. In general if you have less equations than unkowns that means there are several (possibly an infinite number of) solutions. As you have now found you can take your pick. $\endgroup$ Jul 4, 2015 at 4:21
  • $\begingroup$ Yes - thanks so much @rogeliomolina! :-) $\endgroup$
    – User001
    Jul 4, 2015 at 7:11

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