Definition. A square matrix $A$ with complex entries is skew-Hermitian, if $$ A^* = -A. $$ Here , is the transpose of $A$ , and $\ccj{A}$ is is the complex conjugate of the matrix $A$ .

Properties.

1. The trace of a skew-Hermitian matrix is imaginary.
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 $x_{kk} + i y_{kk}$ , 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., \begin{eqnarray} \label{gugg} Ax &=& \lambda x. \end{eqnarray}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 \begin{eqnarray*} \ccj{\lambda} &=& x^\ast \ccj{\lambda} x\\ &=& ( x^\ast \lambda x )^\ast \\ &=& (x^\ast A x )^\ast \\ &=& x^\ast A^\ast x \\ &=& x^\ast (-A) x \\ &=& -x^\ast \lambda x \\ &=& - \lambda . \end{eqnarray*}Hence the eigenvalue $\lambda$ corresponding to $x$ is imaginary.