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

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 ?

• I assume $n$ is even? – user7530 Sep 25 '13 at 15:28
• $\det(A^2)=\det(A)^2=\det(-I)=-1$, so maybe your question's wrong – Shuchang Sep 25 '13 at 15:30
• @Shuchang: $\det(-I)=(-1)^n.$ Hence user7530's comment. – Cameron Buie Sep 25 '13 at 15:32
• @user7530: Why did u remove the post? – mrs Sep 25 '13 at 15:33
• @CameronBuie Oh, what was I thinking. Thanks – Shuchang Sep 25 '13 at 15:39

## 1 Answer

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.

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