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Given an $n\times n$ matrix A with real entries such that $A^2=-I$, prove (a) that $n$ is even and (b) that $A$ has no real eigenvalues. How do you do this? I have no idea where to start.

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For the first part, consider the determinant. For the second, consider what happens to an eigenvector with a real eigenvalue if you apply $A^2$ to it. – Tobias Kildetoft Aug 19 '13 at 18:45
If this is homework, you should tag it as such. If not, I commend you on your curiosity, and hope your self-study is both interesting and enjoyable. – hasnohat Aug 19 '13 at 18:49
Actually (b) implies (a), since every odd-dimensional algebraic equation has a real root. – njguliyev Aug 19 '13 at 18:52
The $1\times 1$ matrix $A \equiv \left({\rm i}\right)$ satisfies $A^{2} = \left(-1\right)$. $n = 1$ is odd. – Felix Marin Aug 20 '13 at 2:26
@FelixMarin The matrix is assumed to have real entries. – Potato Aug 20 '13 at 3:24

5 Answers 5

up vote 18 down vote accepted


Since the matrix has real entries, $\det A$ is real, so $\det A^2 = (\det A)^2$ is positive. But $\det -I = (-1)^n$, because we can just multiply down the main diagonal, so we must have that $n$ is even.


Suppose $A$ has a real eigenvalue $\lambda$ with corresponding eigenvector $v$. Then $$-v=-Iv=A^2v=A(Av)=A(\lambda v)=\lambda^2v.$$

This products a contradiction, because no real number squares to $-1$.

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I like your answer,+1. But $A$ real can still have negative detetminant, viz. $\begin{bmatrix } 1 & 0 \\ 0 & -1 \end{bmatrix}$. Perhaps you meant $\det(A)$ real instead. Then I think she'll fly a little better, Wilbur. Yours, Orville. Cheerio! ;) – Robert Lewis Aug 19 '13 at 19:21
@RobertLewis Thanks for the catch. That was indeed what I meant. – Potato Aug 19 '13 at 21:18
@RiaD Thanks. I fixed it. – Potato Aug 19 '13 at 21:38

HINT: $\det(-I)=\det(A^2)$. If $Ax=\lambda x$, then $-x=A^2x=\ldots\;$?

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Just the right amount of information to get the OP going (I hope). +1. – Tobias Kildetoft Aug 19 '13 at 18:52
The downvote would be much more useful if accompanied by an explanatory comment. – Brian M. Scott Aug 19 '13 at 19:10
Did you address that comment to me, or did the system decide to ping me automatically? (I upvoted). – Tobias Kildetoft Aug 19 '13 at 19:11
@Tobias: It must have been the system; I certainly didn’t do it. I think that the system does ping the author of the immediately preceding comment if the new one has no addressee. – Brian M. Scott Aug 19 '13 at 19:13
Ahh, thanks for clarifying. – Tobias Kildetoft Aug 19 '13 at 19:16

Take a look at Cayley Hamilton Theorem. An overkill but the two people above me already answered it! In particular, look at the minimal polynomial.

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Let's start with (b). Suppose $\lambda$ were a real eigenvalue of $A$. Then there would exist a nonzero vector $v \in R^n$ such that

$Av = \lambda v$.


$A^2v = \lambda^2 v$,


$(A^2 + I)v = (\lambda^2 + 1)v$;

but since $A^2 + I = 0$ this implies

$(\lambda^2 + 1)v =0$,

which since $v \ne 0$ implies

$\lambda^2 + 1 = 0$;

but no real $\lambda$ satisfies this equation. This contradiction shows $A$ has no real eigenvalues.

As for (a), if $n$ were odd, then the characteristic polynomial $\det(A - \lambda I)$ of $A$ would have odd degree $n$; but every polynomial with real coefficients and odd degree has at least one real root. But we have seen that $A$ can have no real eigenvalues. Thus $n$ must itself be even. QED.

Hope this helps. Cheers.

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To prove (a):

Consider the matrix $I_n$, the $n*n$ identity matrix. Clearly, $\det I_n = (-1)^n$. Now, consider $\det A$. As $A$ has only real entries, $\det A$ is real. Thus, $(\det A)^2$ is positive. Since $A^2 = I_n$, it is also the case that $(\det A)^2 = \det I_n = (-1)^n$, and as $(\det A)^2 > 0$, it must be the case that $\det I_n > 0$, so $n$ must be even.

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