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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;

(c)$det(A)=1$.

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2 Answers 2

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.

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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!!!

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