Could you please explain the following and give me a meticulous solution:"What is the dimension of the vector space consisting of all 3 by 3 symmetric matrices? What is a basis for it?" Solution till now:
Let Matrix $$A=\begin{pmatrix}a&b&c\\d&e&f\\h&i&j\end{pmatrix}$$
Since it's symmetric $c=h$, $d=b$, $f=i$
$$A=\begin{pmatrix}a&b&c\\b&e&i\\c&i&j\end{pmatrix}$$
$A$ could be written as $$A=a \times\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix} +b\times\begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}+c\times\begin{pmatrix}0&0&1\\0&0&0\\1&0&0\end{pmatrix}+e\times\begin{pmatrix}0&0&0\\0&1&0\\0&0&0\end{pmatrix}+i\times\begin{pmatrix}0&0&0\\0&0&1\\0&1&0\end{pmatrix}+j\times\begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix}$$
What's the conclusion? Is it six dimentional matrix?
Hint:
Any symmetric matrix $\mathbf{A}\in M_{3,3}(\mathbb{R})$ can be written as
Can you write $\mathbf{A}$ as a sum of some matrices multiplied by some scalars?
You have found that any symmetric matrix may be written $$A=a\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix} +b\begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}+c\begin{pmatrix}0&0&1\\0&0&0\\1&0&0\end{pmatrix}+e\begin{pmatrix}0&0&0\\0&1&0\\0&0&0\end{pmatrix}+i\begin{pmatrix}0&0&0\\0&0&1\\0&1&0\end{pmatrix}+j\begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix}\tag{1}$$ The trick is to realize that the set of all symmetric matrices forms a vector space, and that you may consider each of the $6$ matrices as vectors. To emphasize this, $(1)$ may be written as $$A=aM_1+bM_2+cM_3+eM_4+iM_5+jM_6$$ Next thing is to note that $M_1,M_2,M_3,M_4,M_5,M_6$ are all linearly independent, because the only way to have $A$ be the zero-matrix is precisely when $a=b=c=e=i=j=0$. It therefore follows that $\{M_1,M_2,M_3,M_4,M_5,M_6\}$ is a basis for your vector space, and the dimension is $6$.