# Show that $A^2+2A+5I=0$ has a solution if and only if $n$ is even.

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.

• $x^2+1=0$ 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". Nov 11 '18 at 22:59

## 3 Answers

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.

• 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 Nov 11 '18 at 23:01
• @DonAntonio this is only half of an answer Nov 12 '18 at 0:19
• @Shalop That doesn't matter. We are allowed to post hints. Nov 12 '18 at 4:42
• @Arthur sure, but at least say that it’s a hint Nov 12 '18 at 12:15

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

• I don't understand this. The matrix $\;A\;$ is given... Nov 11 '18 at 23:03
• 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 Nov 11 '18 at 23:07

The condition implies that $$(A+I)^2 = -4I_n$$. What are the signs of determinants on both sides when $$n$$ is odd?