Skew-Hermitian matrix

For example, the following matrix is skew-Hermitian A = \begin{bmatrix} -i & +2 + i \\ -2 + i & 0 \end{bmatrix} because -A = \begin{bmatrix} i & -2 - i \\ 2 - i & 0 \end{bmatrix} = \begin{bmatrix} \overline{-i} & \overline{-2 + i} \\ \overline{2 + i} & \overline{0} \end{bmatrix} = \begin{bmatrix} \overline{-i} & \overline{2 + i} \\ \overline{-2 + i} & \overline{0} \end{bmatrix}^\mathsf{T} = A^\mathsf{H} ==Properties==

• The eigenvalues of a skew-Hermitian matrix are all purely imaginary (and possibly zero). Furthermore, skew-Hermitian matrices are normal. Hence they are diagonalizable and their eigenvectors for distinct eigenvalues must be orthogonal. • All entries on the main diagonal of a skew-Hermitian matrix have to be pure imaginary; i.e., on the imaginary axis (the number zero is also considered purely imaginary). • If A and B are skew-Hermitian, then is skew-Hermitian for all real scalars a and b. • A is skew-Hermitian if and only if i A (or equivalently, -i A) is Hermitian. • A is skew-Hermitian if and only if the real part \Re{(A)} is skew-symmetric and the imaginary part \Im{(A)} is symmetric. • If A is skew-Hermitian, then A^k is Hermitian if k is an even integer and skew-Hermitian if k is an odd integer. • A is skew-Hermitian if and only if \mathbf{x}^\mathsf{H} A \mathbf{y} = -\overline{\mathbf{y}^\mathsf{H} A \mathbf{x}} for all vectors \mathbf x, \mathbf y. • If A is skew-Hermitian, then the matrix exponential e^A is unitary. • The space of skew-Hermitian matrices forms the Lie algebra u(n) of the Lie group U(n). ==Decomposition into Hermitian and skew-Hermitian==

• The sum of a square matrix and its conjugate transpose \left(A + A^\mathsf{H}\right) is Hermitian. • The difference of a square matrix and its conjugate transpose \left(A - A^\mathsf{H}\right) is skew-Hermitian. This implies that the commutator of two Hermitian matrices is skew-Hermitian. • An arbitrary square matrix C can be written as the sum of a Hermitian matrix A and a skew-Hermitian matrix B: C = A + B \quad\mbox{with}\quad A = \frac{1}{2}\left(C + C^\mathsf{H}\right) \quad\mbox{and}\quad B = \frac{1}{2}\left(C - C^\mathsf{H}\right) ==See also==