Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given an $n \times n$ matrix $A$ with real entries such that $A^2=-I$. Prove that A has no real eigenvalues.

We can easily prove the following additional statements about $A$ by taking determinants of both sides of the given equality -

(a) $A$ is nonsingular;

(b)$n$ is even;


share|cite|improve this question
up vote 4 down vote accepted

Well, what does it mean for something to be a real eigenvalue of A? It means there's some vector we can multiply into A on the right and get back a scalar multiple of that vector.

But for any eigenvector, if we do this twice: $-v = (-I)v = A^2 v = A(A v) = A(\lambda v) = \lambda^2 v$.

You should be able to take it from here.

share|cite|improve this answer

Suppose $\lambda$ were a real eigenvalue of $A$; then there is a non-zero vector $\vec v \in \Bbb R^n$ such that $A\vec v = \lambda \vec v$. This implies that $A^2 \vec v = A(A\vec v) = A(\lambda \vec v) = \lambda (A \vec v) = \lambda^2 \vec v$; then $0 = (A^2 + I) \vec v = (\lambda^2 + 1) \vec v$; since $\vec v \ne 0$, we have $\lambda^2 + 1 = 0$, clearly impossible for real $\lambda$ ; thus $A$ has no real eigenvalues. This fact implies that $n$ is even, since otherwise the characteristic polynomial $p_A(\lambda) = \det(A - \lambda I)$ of $A$ always has at least one real root by virtue of the fact that $\deg p_A(\lambda) = n$ is odd. Furthermore, since $A^2 = -I$ and $n$ is even, we have $(\det A)^2 = \det A^2 = \det(-I) = 1$. Thus $\det A \ne 0$ so $A$ is nonsingular. Finally $A^2 + I = 0$ actually implies $\lambda^2 = -1$ for the complex eigenvalues of $A$, by a very slight extension of our previous argument, thus $\lambda = \pm i$; but $A$ real implies the $\lambda$ occur in complex conjugate pairs, so that $\det A$, being the product of $n / 2$ pairs $\lambda, \bar \lambda$ with $\lambda \bar \lambda = i(-i) = 1$, must itself be $1$. QED.

And that cover's all the OP's points, if I am not mistaken.

Hope this helps! Cheerio,

and as always,

Fiat Lux!!!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.