You have the set of all 2 by 2 matrices of the form $\displaystyle \begin{bmatrix}a & b- d \\ b+ a & d\end{bmatrix}$ and you want to
1) Show that this is a subspace of the space of all 2 by 2 matrices.
You must show it is closed under matrix addition and scalar multiplication. The sum of two such matrices can be written
$\displaystyle \begin{bmatrix}u & v- w \\ v+ u & w\end{bmatrix}+\begin{bmatrix}x & y- z \\ y+ x & z\end{bmatrix}= \begin{bmatrix}u+ x & v- w+ y- z \\ v+ u+ y+ x & w+ z\end{bmatrix}$
Can that be written in the form $\displaystyle \begin{bmatrix}a & b- d \\ b+ a & d\end{bmatrix}$? What must a, b, and d be?
The product of the number "k" and the matrix $\displaystyle \begin{bmatrix}x & y- z\\ y+ x & z\end{bmatrix}$ is $\displaystyle \begin{bmatrix}kx & k(y- z) \\ k(y+ x) & kz\end{bmatrix}$.
Can that be written in the form $\displaystyle \begin{bmatrix}a & b- d \\ b+ a & d\end{bmatrix}$? What must a, b, and d be?
The third condition, that the subset be non-empty, can be done by showing that the 0 matrix is in the set. Take a= b= d= 0.
2) Find a basis for this subspace.
Well, $\displaystyle \begin{bmatrix}a & b- d \\ b+ a & d\end{bmatrix}$$\displaystyle = \begin{bmatrix}a & 0 \\ a & 0\end{bmatrix}+ \begin{bmatrix}0 & b \\ b & 0 \end{bmatrix}+ \begin{bmatrix}0 & -d \\ 0 & d\end{bmatrix}$$\displaystyle = a\begin{bmatrix}1 & 0 \\ 1 & 0 \end{bmatrix}+ b\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}+ d\begin{bmatrix}0 & -1 \\ 0 & 1\end{bmatrix}$. Does that give you any ideas?
3) Find the dimension.
After (2), this should be trivial.