Prove that a set of matrices is a subspace I'm self studying linear algebra and now I'm starting with proofs and so on. I found this exercise and this is the way I prove it. I think it's correct but I'm not sure I mean, what do you think?


Is the set of matrices 
    $
\begin{pmatrix}
x && x+y \\
x-y && y
\end{pmatrix}
$ where $x, y \in R $ a subspace of $M_{2\times 2}$


I know that for the set to be a subspace it needs to be closed under vector addition and under scalar multiplication. So:
Given $M_1, M_2 \in M_{2\times 2}, x,y, \in R$. Then $M_1 + M_2$ will be:
$$\begin{align}
\begin{pmatrix}
x_1 && x_1+y_1\\
x_1-y_1 && y_1
\end{pmatrix}
&+
\begin{pmatrix}
x_2 && x_2+y_2\\
x_2-y_2 && y_2
\end{pmatrix}
=
\\&
\begin{pmatrix}
(x_1 + x_2) && (x_1+x_2)+(y_1+y_2)\\
(x_1+x_2)-(y_1+y2) && (y_1 + y_2)
\end{pmatrix}
\end{align}$$
Which has the same structure so it's closed under vector addition.
Now, Given $M \in M_{2\times 2}, x,y,r \in R$. Then $rM_1$ will be:
$$
r\begin{pmatrix}
x && x+y\\
x-y && y
\end{pmatrix}
=
\begin{pmatrix}
rx && r(x+y)\\
r(x-y) && ry
\end{pmatrix}
$$
Which, given that $x,y,r \in R$ is also closed under vector multiplication.
So yes. The set is a subspace of $M_{2\times 2}$

Note
$$
\begin{pmatrix}
x && x+y \\
x-y && y
\end{pmatrix}
=
x \begin{pmatrix}
1 && 1 \\
1 && 0
\end{pmatrix}
+
y \begin{pmatrix}
0 && 1 \\
-1 && 1
\end{pmatrix}
$$
so it is the subspace generated by the two matrices
$$
\begin{pmatrix}
1 && 1 \\
1 && 0
\end{pmatrix},
\qquad
\begin{pmatrix}
0 && 1 \\
-1 && 1
\end{pmatrix}.
$$
