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I have a real square matrix $A$. I am told to prove that there is $A$ such that $A^2+2A+5I=0$ if and only if $n$ is even. (If $A$ is 6x6, $n=6$)

I honestly have no clue how to start. Maybe I could turn this into a question with minimal polynomial and use $x^2+2x+5$. This polynomial doesn't have a root, and it is making me even more confused. Could someone help? Thank you.

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  • $\begingroup$ $x^2+1=0$ doesn't have a real root, yet it is satisfied by $\left[\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right]$. It is not really that weird that a polynomial with no real roots has a matrix "root". $\endgroup$
    – Arthur
    Nov 11 '18 at 22:59
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If $n$ is not even then $A$ has a real eigenvalue since that the characteristic polynomial of $A$ has an odd degree and hence has a real root.

Hence since the polynomial $x^2+2x+5$ has no real roots, it means that $n$ is even.

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    $\begingroup$ This is a very good hint yet to fully understand it the OP is going to have a very good understanding of roots of characteristic pol, miimal pol. and etc. to complete. +1 $\endgroup$
    – DonAntonio
    Nov 11 '18 at 23:01
  • $\begingroup$ @DonAntonio this is only half of an answer $\endgroup$
    – shalop
    Nov 12 '18 at 0:19
  • $\begingroup$ @Shalop That doesn't matter. We are allowed to post hints. $\endgroup$
    – Arthur
    Nov 12 '18 at 4:42
  • $\begingroup$ @Arthur sure, but at least say that it’s a hint $\endgroup$
    – shalop
    Nov 12 '18 at 12:15
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Take $\lambda \in \mathbb{C}$ such that $\lambda^2+2\lambda +5=0$ and pick $A \in Mat(2,\mathbb{C})$ such that $$A=\begin{pmatrix} \ Re(\lambda) & -Imm(\lambda) \\ Imm(\lambda) & Re(\lambda) \end{pmatrix} .$$

You can check this works for $n=2$ as the matrix multiplication is the same as complex multiplication and try to generalize this to $n=2k$ with generic $k$.

This proves that such a matrix exists if $n$ is even. To prove the inverse statement there is the answer upon mine

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  • $\begingroup$ I don't understand this. The matrix $\;A\;$ is given... $\endgroup$
    – DonAntonio
    Nov 11 '18 at 23:03
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    $\begingroup$ I understood there was and iff in his question,so one had to show also the existence of such a matrix in case $n$ is even,while the other implication had already been answered $\endgroup$ Nov 11 '18 at 23:07
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The condition implies that $(A+I)^2 = -4I_n$. What are the signs of determinants on both sides when $n$ is odd?

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