direct sum of Hermitian and skew-Hermitian matrices

In this example, we show that any square matrix with complexentries can uniquely be decomposed into the sum of one Hermitian matrix andone skew-Hermitian matrix. A fancy way to say this is thatcomplex square matrices is the direct sum of Hermitian and skew-Hermitianmatrices.

Let us denote the vector space (over $ℂ$) ofcomplex square $n\times n$ matrices by $M$.Further, we denote by $M_{+}$ respectively $M_{-}$ the vectorsubspaces of Hermitian and skew-Hermitian matrices.We claim that

$M$$=$$M_{+}\oplus M_{-}.$(1)

Since $M_{+}$ and $M_{-}$ are vector subspaces of $M$, it is clearthat $M_{+}+M_{-}$ is a vector subspace of $M$. Conversely, suppose$A\in M$. We can then define

$A_{+}$$=$${\displaystyle \frac{1}{2}}\left(A+A^{\ast }\right),$

$A_{-}$$=$${\displaystyle \frac{1}{2}}\left(A-A^{\ast }\right).$

Here $A^{\ast }={\overline{A}}^{\text{T}}$, and $\overline{A}$ is the complex conjugate of $A$,and $A^{\text{T}}$ is the transpose of $A$. It follows that $A_{+}$ is Hermitianand $A_{-}$ is anti-Hermitian. Since $A=A_{+}+A_{-}$, any elementin $M$ can be written asthe sum of one element in $M_{+}$ and one element in $M_{-}$. Let us checkthat this decomposition is unique. If $A\in M_{+}\cap M_{-}$, then$A=A^{\ast }=-A$, so $A=0$.We have established equation 1.

Special cases

• In the special case of $1\times 1$ matrices, we obtain thedecomposition of a complex number into its real and imaginary components.

• In the special case of real matrices, we obtain the decomposition ofa $n\times n$ matrix into a symmetric matrix and anti-symmetric matrix.