On the eigenvalues of the square of a skew-symmetric matrix

Question

Let $S$ be a skew-symmetric matrix of size $n$. Is-it-true to say that $\lambda(S^2)=(\lambda(S))^2$, that is, the eigenvalue of $S^2$ is equal to the square of the eigenvalue of $S$ ?

Thank you in advance!

Answer 1

Your statement is true, and in fact more is true; we can show that the eigenvectors are the same. $S$ being skew-symmetric means $S = - S^{\ast}$, and hence $SS^{\ast} = S^{\ast} S$, hence $S$ is normal; by the Spectral Theorem, a normal matrix is unitarily diagonalizable.

So we have that $S = X \Lambda X^{\ast}$, where $X$ is unitary. Therefore $S^2 = X \Lambda^2 X^{\ast}$, and this is the diagonalization of $S^2$.

Answer 2

Hint. If $x$ is an eigenvector of $S$ with eigenvalue $\lambda$, then $$S^2x=S(Sx)=S(\lambda x)=\lambda(Sx)=\lambda(\lambda x)=\lambda^2 x$$