Is $V$ a subspace of $M_{2,2}$?

Question

Consider the subset

$$ V = \{A \in M_{2,2} | v^TA = (Av)^T\} $$

of vector space $M_{2,2}$ fo 2x2 matrices. $v^T$ inidcates the transpose of v. Is V a subspace of $M_{2,2}$?

Assuming the set is non-empty, I should only have to prove the following two conditions:

1. V is closed under addition.
2. V is closed under scalar multiplication.

How do I prove this? Any help is appreciated.

Answer 1

If $D,E \in V$ then $v^T D = (D v)^T $ and $v^T E = (E v)^T$. Then $v^T(D+E)=v^T D + v^T E$. On the ther hand, $((D+E) v) ^T = (Dv + Ev)^T =(Dv)^T + (Ev)^T$, putting all togheter: $$ v^T(D+E)=v^T D + v^T E = (Dv)^T + (Ev)^T = (Dv + Ev)^T ,$$ so $D+E$ is in $V$.

Let's check the second condition. If $D \in V$ then $v^T D = (Dv)^T$. Let $c$ be a scalar then $$v^T (cD)=c v^T D= c (Dv) ^T = ((cD)v)^T, $$ this means that $cD$ is also in $V$.