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Math Help - dimension of the subspace of a matrix size is N by N consisting of all symmetric ?

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    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?
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    Quote Originally Posted by xycode View Post
    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

    \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}

    2+1=3

    This gives \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} +<br />
c\begin{bmatrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 &0 & 0  \end{bmatrix}+<br />
d\begin{bmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0  \end{bmatrix}+<br />
e\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 &1 & 0  \end{bmatrix}+<br />
f\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 &0 & 1  \end{bmatrix}<br />

    3+2+1=6

    See if you can finish from here
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