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

 A: 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. 
A: 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 $i\mathbb R$.
A: We can prove it  with only elementary facts :
Since $A$ is a real matrix, for any eigenvalue $\lambda\in\mathbb{C}$, one has
$\mathrm{det}(A-\lambda I_n)=0$, so $\mathrm{det}(\overline{A-\lambda I_n})=\mathrm{det}(A-\overline{\lambda} I_n)=0$.
Hence $\overline{\lambda}$ is also an other eigenvalue. But as $\lambda+\overline{\lambda}\in\mathbb{R}$ is again an eigenvalue of $A$, $\lambda$ is a pure imaginary, since all real eigenvalues of a skew-symmetric matrix are zero.
