Is a symmetric matrix a subspace of nxn matrices? Let $M$ be a vector space for all 2x2 matrices.
Show that the set of all the symmetric matrices $M2$={$B ∈ M2 : B=B^t$} is a subspace of M.

My solution:

*

*The null matrix is symmetric

$$B=\begin{pmatrix}0&0\\0&0\end{pmatrix}$$



*Let $B$, $C$ $∈$ $M2$, where $B = B^t$, $C=C^t$. Then, $B+C=B^t + C^t$.

$$B = \begin{pmatrix}a&c\\c&b\end{pmatrix}$$
$$C = \begin{pmatrix}d&f\\f&e\end{pmatrix}$$
$$B+C = \begin{pmatrix}a+d&c+f\\c+f&b+e\end{pmatrix}$$
$$B^t = \begin{pmatrix}a&c\\c&b\end{pmatrix}$$
$$C^t = \begin{pmatrix}d&f\\f&e\end{pmatrix}$$
$$B^t+C^t = \begin{pmatrix}a+d&c+f\\c+f&b+e\end{pmatrix}$$
Therefore, $B + C ∈ M2$.



*Let $B ∈ M2$, $θ ∈ R$. Then, $θB = θB^t$
$$B = \begin{pmatrix}a&c\\c&b\end{pmatrix}$$ $$θB = \begin{pmatrix}θa&θc\\θc&θb\end{pmatrix}$$
$$B^t = \begin{pmatrix}a&c\\c&b\end{pmatrix}$$ $$θB^t = \begin{pmatrix}θa&θc\\θc&θb\end{pmatrix}$$
Therefore, $θB ∈ M2$.

So, $M2$ is, indeed, a subspace of $M$.
I am just learning linear algebra, so I apologize if I've made any mistakes.
Also, I just got into this community, so I am still learning the formatting, so I apologize for any mistakes as well.
Appreciate all the answers.

Your answer is fine.
Or you could have noticed that there is a basis for your space:
$$m_1=\begin{pmatrix}1&0\\0&0\end{pmatrix}, m_2=\begin{pmatrix}0&1\\1&0\end{pmatrix},
m_3=\begin{pmatrix}0&0\\0&1\end{pmatrix}$$
Show a $2\times 2$ matrix is symmetric if and only if it can be written as $$am_1+bm_2+cm_3$$
for some real $a,b,c.$
But this answer requires knowledge of bases. Your answer is more direct.
But the general rule makes it unnecessary to prove many individual case:

If $V$ is a real vector space, and $v_1,v_2,\cdots,v_k\in V$ then:
$$W=\{v\in V\mid \exists a_1,\dots,a_k\in \mathbb R(v=a_1v_1+a_2v_2+\cdots+a_kv_k\}$$

