## A space of matrices

Let $$V$$ ve the set of all real symmetric matrices of order three with zeros on the diagonal.

Show that $$V$$ is a subspace of $$\mathbb R^{3\times 3}$$. Determine the dimension of $$V$$ and find some basis of this subspace.

• #### Resolution

When $$A,B\in V$$ then both matrices $$\alpha A$$ and $$A+B$$ have zeros on the diagonal: $$(\alpha A)_{i,i}=\alpha a_{i,i}=\alpha 0 =0$$, $$(A+B)_{i,i}=a_{i,i}+b_{i,i}=0+0=0$$.

The sum and a scalar multiple remain symmetric: $$(\alpha A)_{i,j}=\alpha a_{i,j}=\alpha a_{j,i}=(\alpha A)_{j,i}$$, $$(A+B)_{i,j}=a_{i,j}+b_{i,j}=a_{j,i}+b_{j,i}=(A+B)_{j,i}$$.

As the sum and the scalar multiple are preserved in the set $$V$$, it is a subspace.

When $$A \in V$$, then its elements satisfy: $$a_{1{,}1}=a_{2{,}2}=a_{3{,}3}=0$$, $$a_{1{,}2}=a_{2{,}1}$$, $$a_{1{,}3}=a_{3{,}1}$$ a $$a_{2{,}3}=a_{3{,}2}$$.

The solution of this system could be expressed by using three parameters as follows:
$$A= a_{1{,}2} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} + a_{1{,}3} \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \\ \end{pmatrix} + a_{2{,}3} \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{pmatrix}$$.

Hence $$\dim(V)=3$$ and the three matrices above could be chosen as a basis of $$V$$.