Let $\mathbf{R}^{3\times3}$ be the linear space of all $3 \times 3$ matrices. Let $W$ be the set of all symmetric $3\times3$ matrices. Then $W$ is a linear subspace of $\mathbf{R}^{3×3}$. Find a basis for $W$ and identify $\dim(W)$.
Would the $\dim(W)$ be $3$ since it is a $3\times3$ matrix. Is that correct? How do you go about finding the basis? From my understanding, if the span of $W$ are linearly independent then they would be the basis of the $3\times3$ matrix.
Please help!
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HINT
You can threat 3-by-3 matrices as 9 dimensional vectors.
For example for the case of a 2-by-2 matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ we can write
$$\begin{bmatrix}a&b\\c&d\end{bmatrix}=a\begin{bmatrix}1&0\\0&0\end{bmatrix}+b\begin{bmatrix}0&1\\0&0\end{bmatrix}+c\begin{bmatrix}0&0\\1&0\end{bmatrix}+d\begin{bmatrix}0&0\\0&1\end{bmatrix}$$
thus in this case the dimension is 4 and the (a) standard basis is given by the 4 elementary matrix.
For 3-by-3 matrices you can follow the same idea take into account the symmetry.
The $3\times3$ symmetric matrices are those of the form$$\begin{pmatrix}a&b&c\\b&d&e\\c&e&f\end{pmatrix}.$$They form a $6$-dimensional space. A basis of this space is$$\left\{\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix},\begin{pmatrix}0&0&0\\0&1&0\\0&0&0\end{pmatrix},\begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix},\begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix},\begin{pmatrix}0&0&0\\0&0&1\\0&1&0\end{pmatrix},\begin{pmatrix}0&0&1\\0&0&0\\1&0&0\end{pmatrix}\right\}.$$Can you prove it?