Answer
The following matrices
\begin{align*}
\left[\begin{array}{ccc}1 & 0&0 \\0 & 0&0\\0 & 0&0\end{array}\right]&,\left[\begin{array}{ccc}0 & 1&0 \\ 1 & 0&0\\0 & 0&0\end{array}\right],\left[\begin{array}{ccc}0 & 0&1 \\0 & 0&0\\1 & 0&0\end{array}\right],\\
\left[\begin{array}{ccc}0 & 0&0 \\0 & 1&0\\0 & 0&0\end{array}\right]&,\left[\begin{array}{ccc}0 & 0&0 \\0 & 0&1\\0 & 1&0\end{array}\right],\left[\begin{array}{ccc}0 & 0&0 \\0 & 0&0\\0 & 0&1\end{array}\right]
\end{align*}
form a basis for the vector space of all $3 \times 3$ symmetric matrices. The dimension is $6$.
Work Step by Step
Any $3 \times 3$ symmetric matrix has the form
$$\left[\begin{array}{ccc}a & b&c \\ b & d&e\\c & e&f\end{array}\right]$$
Rewriting the above matrix as follows
Then, we have the following system of equations
\begin{align*}
\left[\begin{array}{ccc}a & b&c \\ b & d&e\\c & e&f\end{array}\right]&=a\left[\begin{array}{ccc}1 & 0&0 \\0 & 0&0\\0 & 0&0\end{array}\right]+b\left[\begin{array}{ccc}0 & 1&0 \\ 1 & 0&0\\0 & 0&0\end{array}\right]+c\left[\begin{array}{ccc}0 & 0&1 \\0 & 0&0\\1 & 0&0\end{array}\right]\\
&+d\left[\begin{array}{ccc}0 & 0&0 \\0 & 1&0\\0 & 0&0\end{array}\right]+e\left[\begin{array}{ccc}0 & 0&0 \\0 & 0&1\\0 & 1&0\end{array}\right]+f\left[\begin{array}{ccc}0 & 0&0 \\0 & 0&0\\0 & 0&1\end{array}\right].
\end{align*}
The following matrices
\begin{align*}
\left[\begin{array}{ccc}1 & 0&0 \\0 & 0&0\\0 & 0&0\end{array}\right]&,\left[\begin{array}{ccc}0 & 1&0 \\ 1 & 0&0\\0 & 0&0\end{array}\right],\left[\begin{array}{ccc}0 & 0&1 \\0 & 0&0\\1 & 0&0\end{array}\right],\\
\left[\begin{array}{ccc}0 & 0&0 \\0 & 1&0\\0 & 0&0\end{array}\right]&,\left[\begin{array}{ccc}0 & 0&0 \\0 & 0&1\\0 & 1&0\end{array}\right],\left[\begin{array}{ccc}0 & 0&0 \\0 & 0&0\\0 & 0&1\end{array}\right]
\end{align*}
form a basis for the vector space of all $3 \times 3$ symmetric matrices. The dimension is $6$.
