# direct sum of Hermitian and skew-Hermitian matrices

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

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

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

 $\displaystyle A_{+}$ $\displaystyle=$ $\displaystyle\frac{1}{2}\big{(}A+A^{\ast}\big{)},$ $\displaystyle A_{-}$ $\displaystyle=$ $\displaystyle\frac{1}{2}\big{(}A-A^{\ast}\big{)}.$

Here $A^{\ast}=\overline{A}\hskip 1.13811pt^{\mbox{\scriptsize{T}}}\hskip 0.569055pt$, and $\overline{A}$ is the complex conjugate  of $A$, and $A\hskip 1.13811pt^{\mbox{\scriptsize{T}}}\hskip 0.569055pt$ is the transpose  of $A$. It follows that $A_{+}$ is Hermitian and $A_{-}$ is anti-Hermitian. Since $A=A_{+}+A_{-}$, any element in $M$ can be written as the sum of one element in $M_{+}$ and one element in $M_{-}$. Let us check that 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

Title direct sum of Hermitian and skew-Hermitian matrices DirectSumOfHermitianAndSkewHermitianMatrices 2013-03-22 13:36:30 2013-03-22 13:36:30 mathcam (2727) mathcam (2727) 5 mathcam (2727) Example msc 15A03 msc 15A57 DirectSumOfEvenoddFunctionsExample