# Thread: dimension of the subspace of a matrix size is N by N consisting of all symmetric ?

1. ## dimension of the subspace of a matrix size is N by N consisting of all symmetric ?

What is the dimension of the subspace of a matrix whose size is N by N consisting of all symmetric matrices? How do I find this out and justify the answer?

2. Originally Posted by xycode
What is the dimension of the subspace of a matrix whose size is N by N consisting of all symmetric matrices? How do I find this out and justify the answer?
Try it on a few small cases and then generalize the pattern.

for example

$\displaystyle \begin{bmatrix}a & b \\ b & c \end{bmatrix}=a\begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}+b\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}+c\begin{bmatrix}0 & 0 \\ 0 & 1 \end{bmatrix}$

$\displaystyle 2+1=3$

This gives $\displaystyle \begin{bmatrix}a & b & c \\ b & d & e \\ c &e & f \end{bmatrix} = a\begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} + b\begin{bmatrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 &0 & 0 \end{bmatrix} + c\begin{bmatrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 &0 & 0 \end{bmatrix}+ d\begin{bmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}+ e\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 &1 & 0 \end{bmatrix}+ f\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 &0 & 1 \end{bmatrix}$

$\displaystyle 3+2+1=6$

See if you can finish from here