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If $A^2=-I$ , where $A$ is a square matrix of order $n$ and which contains real entries only and $I$ is identity matrix. Then how can we prove that $\det(A)=1$?.

I could prove that $n$ should be an even integer. But could not proceed to prove that $\det(A)$ can take only $1$, finding out few matrices which satisfies such properies (of small order) also verifies the given statement that the determinant is only $1$ and not $-1$.

Can anyone help with a hint ?

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    $\begingroup$ I assume $n$ is even? $\endgroup$
    – user7530
    Sep 25, 2013 at 15:28
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    $\begingroup$ $\det(A^2)=\det(A)^2=\det(-I)=-1$, so maybe your question's wrong $\endgroup$
    – Shuchang
    Sep 25, 2013 at 15:30
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    $\begingroup$ @Shuchang: $\det(-I)=(-1)^n.$ Hence user7530's comment. $\endgroup$ Sep 25, 2013 at 15:32
  • $\begingroup$ @user7530: Why did u remove the post? $\endgroup$
    – Mikasa
    Sep 25, 2013 at 15:33
  • $\begingroup$ @CameronBuie Oh, what was I thinking. Thanks $\endgroup$
    – Shuchang
    Sep 25, 2013 at 15:39

1 Answer 1

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Let $\lambda$ be an eigenvalue of $A$, with eigenvector $v$.

Then $$v^HA^2v = v^H(-I)v.$$ What can you conclude about $\lambda$?

Now use the fact that the eigenvalues of a real matrix must come in complex conjugate pairs, and that the determinant is the product of the eigenvalues.

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    $\begingroup$ Got it . The eigenvalues are $i$ and $-i$ and the determinant is $1$.Thanks $\endgroup$
    – thehe
    Sep 25, 2013 at 15:41

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