Given an $n \times n$ matrix $A$ with real entries such that $A^2 = -I$, prove that $\det(A) = 1$.
This question is multi-part, but I happen to be stuck on this one. The previous parts showed: $A$ is nonsingular, $n$ is even, and $A$ has no real eigenvalues.
I know that $\det(A)^2 = 1$ since $A$ has real entries and $n$ is even, but am not sure how to show that $\det(A)$, which can be either $1$ or $-1$, is not $-1$. Does anyone know how to continue from here?
On
One way is to prove by induction that $A$ is similar to the matrix
$$\begin{pmatrix} \begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix} & \dots & 0\\ \vdots & \ddots & \vdots\\ 0 & \dots & \begin{pmatrix} 0 & -1\\ 1 & 0\end{pmatrix} \end{pmatrix}$$
Then the answer is immediate.
On
Let $A$ be an $n\times n$ real matrix. If $\lambda_1,...,\lambda_n$ are the eigenvalues, up to multiplicity. Then since $\lambda_j^2$ are also eigenvalues of $A^2$, $\lambda_j=\pm i$. then the characteristics equation can be written in the form $$\chi_A(t)=\prod_{j=1}^n(t-\lambda_j).$$ Since $A$ is real matrix, the coefficients of $\chi_A$ are real. Hence the eigenvalues are in conjugate pairs i.e. the algebraic multiplicity of $i$ and $-i$ are equal. $$\det(A)=\prod_{j=1}^n\lambda_j=i^{\frac{n}{2}}(-i)^{\frac{n}{2}}=1.$$
I hope this helps
Since $A$ has no real eigenvalues yet is a real matrix, all its eigenvalues must come in complex conjugate pairs $a\pm bi$ with $b\ne0$. But $(a+bi)(a-bi)=a^2+b^2>0$, so the product of all eigenvalues of $A$ – its determinant – must be positive. Hence $\det A=1$.