# How do you symmetrize a 3x3 matrix?

Let's say that I have the following matrix: $\begin{bmatrix} 5 &3 &0 \\ 0 &0 &5 \\ 7 &3 &1 \end{bmatrix}$

How do I symmetrize this matrix, so that I get the following? I am confused about where the original matrix components get moved around to.

$\left (\begin{bmatrix} 5 &3 &0 \\ 0 &0 &5 \\ 7 &3 &1 \end{bmatrix}+\begin{bmatrix} 5 &0 &7 \\ 3 &0 &3 \\ 0 &5 &1 \end{bmatrix} \right ) = \begin{bmatrix} 10 &3 &7 \\ 3 &0 &8 \\ 7 &8 &2 \end{bmatrix}$

• That's just adding the transpose of the original matrix to itself. – symplectomorphic Feb 18 '17 at 23:05

$$M=\dfrac{M+M^T}{2}+\dfrac{M-M^T}{2}$$ $\dfrac{M+M^T}{2}$is symmetric part
This work is look like ,to rewrite function $f(x)$
as $g(x)+h(x)$
$g(x)$ is even part of $f(x)$
$h(x)$ is odd part of $f(x)$
and $$f(x)=g(x)+h(x)=\dfrac{f(x)+f(-x)}{2}+\dfrac{f(x)-f(-x)}{2}$$