Skew-Hermitian matrix

In linear algebra, a square matrix with complex entries is said to be skew-Hermitian or anti-Hermitian if its conjugate transpose is the negative of the original matrix.^[1] That is, the matrix $A$ is skew-Hermitian if it satisfies the relation

$A{\text{ skew-Hermitian}}\quad \iff \quad A^{\mathsf {H}}=-A$

where $A^{\textsf {H}}$ denotes the conjugate transpose of the matrix $A$ . In component form, this means that

$A{\text{ skew-Hermitian}}\quad \iff \quad a_{ij}=-{\overline {a_{ji}}}$

for all indices $i$ and $j$ , where $a_{ij}$ is the element in the $i$ -th row and $j$ -th column of $A$ , and the overline denotes complex conjugation.

Skew-Hermitian matrices can be understood as the complex versions of real skew-symmetric matrices, or as the matrix analogue of the purely imaginary numbers.^[2] The set of all skew-Hermitian $n\times n$ matrices forms the $u(n)$ Lie algebra, which corresponds to the Lie group U(n). The concept can be generalized to include linear transformations of any complex vector space with a sesquilinear norm.

Note that the adjoint of an operator depends on the scalar product considered on the $n$ dimensional complex or real space $K^{n}$ . If $(\cdot \mid \cdot )$ denotes the scalar product on $K^{n}$ , then saying $A$ is skew-adjoint means that for all $\mathbf {u} ,\mathbf {v} \in K^{n}$ one has $(A\mathbf {u} \mid \mathbf {v} )=-(\mathbf {u} \mid A\mathbf {v} )$ .

Imaginary numbers can be thought of as skew-adjoint (since they are like $1\times 1$ matrices), whereas real numbers correspond to self-adjoint operators.

Example edit

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 edit

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.^[3]
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).^[4]
If $A$ and $B$ are skew-Hermitian, then $aA+bB$ is skew-Hermitian for all real scalars $a$ and $b$ .^[5]
$A$ is skew-Hermitian if and only if $iA$ (or equivalently, $-iA$ ) is Hermitian.^[5]
$A$ 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 edit

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 $B$ : $C=A+B\quad {\mbox{with}}\quad A={\frac {1}{2}}\left(C+C^{\mathsf {H}}\right)\quad {\mbox{and}}\quad B={\frac {1}{2}}\left(C-C^{\mathsf {H}}\right)$