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I have stumbled upon this question in my textbook while preparing for a test in Linear Algebra.

I couldn't solve this question at all.

The question is:

Let A be a matrix so that $$A^2 = -I$$ Prove that A has no real eignevalues.

A friend who managed to solve this, gave me a hint: to use the definition of the eigenvalue:


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2 Answers 2

up vote 3 down vote accepted

Note that any eigenvalue $\lambda$ of $A$ satisfies

$Av = \lambda v \tag{1}$

for some vector $v \ne 0$, so that

$A^2 v = A(Av) = A(\lambda v) = \lambda(Av) = \lambda(\lambda v) = \lambda^2 v, \tag{2}$


$0 = (A^2 + I)v = A^2v + v = (\lambda ^2 + 1)v, \tag{3}$

and since $v \ne 0$ this implies

$\lambda^2 + 1 = 0; \tag{4}$

but equation (4) has no real solutions, and thus the desired conclusion immediately follows. QED.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

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@Amzoti: thanks, and glad you're still around! –  Robert Lewis Mar 8 '14 at 17:12

If $Av=\lambda v$ then $A^2v=\lambda^2v$ but on the other side $-Iv=-v$ so we get $\lambda^2=-1$ which has no real solution.

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