# Any $n \times n$ matrix $A$ can be written as $A = B + C$ with $B$ is symmetric and $C$ skew-symmetric.

How can I proof the following statement?

Any $n \times n$ matrix $A$ can be written as a sum $$A = B + C$$ where $B$ is symmetric and $C$ is skew-symmetric.

I tried to work out the properties of a matrix to be symmetric or skew-symmetric, but I could not prove this. Does someone know a way to prove it?

Thank you.

PS: The question Prove: Square Matrix Can Be Written As A Sum Of A Symmetric And Skew-Symmetric Matrices may be similiar, in fact gives a hint to a solution, but if someone does not mind in expose another way, our a track to reach to what is mentioned in the question of the aforementioned link.

• look at $a$ and it transpose $a^\top$ – abel Jun 7 '16 at 17:09

Suppose $$A=B+C$$ If $$B^T=B,$$ $$C^T=-C,$$ then according to the known property of transposition of sum of matrices $$A^T=(B+C)^T=B^T+C^T=B+(-C)=B-C$$ Now we have $$A=B+C \tag 1\\$$ $$A^T=B-C\tag 2\\$$ Adding $(1)$ to $(2)$ gives $$B={(A+A^T)\over 2}\\$$ Subtracting $(2)$ from $(1)$ gives $$C={(A-A^T)\over 2}\\$$

Given a matrix $A$, let $B=\frac{A+A^T}{2}$ and $C=\frac{A-A^T}{2}$. Observe that $B^T=B$, so $B$ is symmetric. Also, $C^T = -C$, so $C$ is skew-symmetric.

Let $A$ be our matrix and

\begin{align}B&={A^T+A\over 2}\\C&={A-A^T\over 2}\end{align}

We have $B^T=B$ and $C^T=-C$ and $A=B+C$

For a matrix $A$,

$A=\frac{A+A^T}{2}+\frac{A-A^T}{2}$, where the first one is symmetric, and second one is skew-symmetric