Let $S=\{A\in M_2: A^{T}=A\}$ in $M_2$. Check that $S$ is a subspace of $M_2$ and find a basis and dimension.

Question

Let $S=\{A\in M_2: A^{T}=A\}$ in $M_2$. Check that $S$ is a subspace of $M_2$ and find a basis and dimension.

I managed to check it's a subspace, and the basis I found is: $$\begin{pmatrix}1 & 0\\0 & 0\end{pmatrix},\begin{pmatrix}0 & 1\\1 & 0 \end{pmatrix},\begin{pmatrix}0 &0\\0 & 1\end{pmatrix}$$.

However the solution manual says it should be: $$\begin{pmatrix}1 & 0\\0 & 0\end{pmatrix},\begin{pmatrix}0 & 1\\0 & 0 \end{pmatrix},\begin{pmatrix}0 &0\\0 & 1\end{pmatrix}$$.

Which one is correct?

Answer 1

The solution manual is wrong, as their second matrix clearly isn't symmetric. But note that generally, there are many different bases for a given vector space.