Why are all nonzero eigenvalues of the skew-symmetric real matrices pure imaginary?

Question

Assume that $A$ is an $n\times n$ skew-symmetric real matrix, i.e. $$A^T=-A.$$

Since $\det(A-\lambda I)=\det(A^T-\lambda I)$, $A$ and $A^T$ have the same eigenvalues. On the other hand, $A^T$ and $-A$ also have the same eigenvalues. Thus if $\lambda$ is an eigenvalue of $A$, so is $-\lambda$. If $n$ is odd, $\lambda = 0 $ is an eigenvalue.

A curious search in Google returns that the nonzero eigenvalues of $A$ are all pure imaginary and thus are of the form $iλ_1, −iλ_1, iλ_2, −iλ_2,$ … where each of the $λ_k$ are real.

Here is my question:

How can I prove the fact that "the nonzero eigenvalues of $A$ are all pure imaginary"?

Answer 1

Consider $A$ as a matrix over $\mathbb{C}$. Then we have that for all $\mathbf{x},\mathbf{y}\in\mathbb{C}^n$, $$\langle A\mathbf{x},\mathbf{y} \rangle = \langle \mathbf{x},A^*\mathbf{y}\rangle,$$ where $\langle-,-\rangle$ is the standard complex inner product, and $A^*$ is the adjoint (which relative to the standard complex inner product is given by the conjugate transpose of $A$). Since $A$ is a real matrix, the adjoint is equal to the transpose, so for every $\mathbf{x},\mathbf{y}\in\mathbb{C}^n$, you have $$\langle A\mathbf{x},\mathbf{y}\rangle = \langle \mathbf{x},A^T\mathbf{y}\rangle = \langle \mathbf{x},-A\mathbf{y}\rangle = -\langle \mathbf{x},A\mathbf{y}\rangle.$$

Now suppose that $\mathbf{x}$ is an eigenvector with eigenvalue $\lambda$. Setting $\mathbf{y}=\mathbf{x}$, we have $$\langle A\mathbf{x},\mathbf{x}\rangle = \langle \lambda\mathbf{x},\mathbf{x}\rangle = \lambda \lVert\mathbf{x}\rVert^2.$$ On the other hand, $$-\langle \mathbf{x},A\mathbf{x}\rangle = -\langle\mathbf{x},\lambda\mathbf{x}\rangle = -\overline{\lambda}\langle\mathbf{x},\mathbf{x}\rangle = -\overline{\lambda}\lVert\mathbf{x}\rVert^2.$$ These two are equal, and since $\mathbf{x}$ is an eigenvector, then $\lVert\mathbf{x}\rVert\neq 0$. Therefore, we have that $\lambda=-\overline{\lambda}$, and hence $\lambda$ is either $0$ or a pure imaginary number.

Answer 2

Disclaimer: I saw the following argument in some other thread on this site but cannot find it right now.

$$A^T = -A \implies A A^T = -A^2$$

but for any real matrix $A$, the "Gramian" $A A^T$ is positive semidefinite (Is a matrix multiplied with its transpose something special?, Proof for why a matrix multiplied by its transpose is positive semidefinite); hence, all eigenvalues of $A^2$ are in $\mathbb R_{\le 0}$, i.e. all eigenvalues of $A$ are in $i\mathbb R$.

Answer 3

If $A$ is real and skew-symmetric, then $\frac{1}{i} \cdot A$ is hermitian.

$\bf{Added:}$ Recall that every square complex matrix $Z$ can be written as

$$Z = X + i Y$$

where $X$, $Y$ are hermitian ( the real and the imaginary part of $Z$). If $Z$ has real entries, then $X$ is real symmetric, and $i Y$ is real skew-symmetric.