If $A^2=-I$, Prove that $\det{A}=1$

Question

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 ?

$\det(A^2)=\det(A)^2=\det(-I)=-1$, so maybe your question's wrong — Shuchang, Sep 25 '13 at 15:30
note that det(A) is the product of all roots of characteristic polynomia(n even). now, what are the possible roots? does real coefficients help? — Mehdi, Sep 25 '13 at 15:40

user7530 · Accepted Answer · 2013-09-25T17:03:02.393

9

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.

edited Sep 25 '13 at 17:03

answered Sep 25 '13 at 15:28

user7530

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2

Got it . The eigenvalues are $i$ and $-i$ and the determinant is $1$.Thanks – thehe Sep 25 '13 at 15:41

If $A^2=-I$, Prove that $\det{A}=1$

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