1. 1.

   The trace of a skew-Hermitian matrix is imaginary.

2. 2.

   The eigenvalues of a skew-Hermitian matrix are imaginary.

Proof. Property (1) follows directly from property (2) since the trace is the sum of the eigenvalues. But one can also give a simple proof as follows. Let $x_{{ij}}$ and $y_{{ij}}$ be the real respectively imaginary parts of the elements in $A$. Then the diagonal elements of $A$ are of the form $x_{{kk}}+iy_{{kk}}$, and the diagonal elements in $A^{\ast}$ are of the form $-x_{{kk}}+iy_{{kk}}$. Hence $x_{{kk}}$, i.e., the real part for the diagonal elements in $A$ must vanish, and property (1) follows. For property (2), suppose $A$ is a skew-Hermitian matrix, and $x$ an eigenvector corresponding to the eigenvalue $\lambda$, i.e.,

Here, $x$ is a complex column vector. Since $x$ is an eigenvector, $x$ is not the zero vector, and $x^{\ast}x>0$. Without loss of generality we can assume $x^{\ast}x=1$. Thus

Hence the eigenvalue $\lambda$ corresponding to $x$ is imaginary. $\Box$