Prove that a set of matrices is a linear space Prove that the set of matrices 
$$v:=\left\{ \begin{pmatrix} 2x-y+z & x-2y-2z \\ x+y-z & 3x+y+2z \end{pmatrix} \middle|\, x,y,z \in R\right\}$$
Is a linear space above $R$ and find it's base.
As far as I know that for the set to be a linear space it needs to be closed under vector addition and under scalar multiplication, am I right?
but still I'm having a bit trouble structuring the proof
Hints, suggestions?
Thanks.
 A: Hints:
1) For $A,B\in v$ and $c\in\mathbb{R}$, show that
$$
A+B\in v\,\,\,\mbox{ and }\,\,\,cA\in v\,.
$$
That is, in each case, deduce the triplet $x, y, z\in \mathbb{R}$ that ensures the matrices also lie in $v$.
2) For $a,b,c,d\in\mathbb{R}$, an arbitrary member of $v$ has the form 
$$
\left( \begin{array}{cc}
a & b \\
c & d
\end{array} \right)\,\,,
$$
with restrictions on $a, b, c, d$. What plane in 4-dimensional space must $(a, b, c, d)$ lie in? Deduce this from  solving the linear system
$$
\left( \begin{array}{ccc}
2 & -1 & -1 \\
1 & -2 & -2 \\
1 & 1 & -1 \\
3 & 1 & 2
\end{array} \right)\left(\begin{array}{c} x \\ y \\ z\end{array}\right){}={}\left(\begin{array}{c} a \\ b \\ c \\ d\end{array}\right)\,.
$$
3) Use a basis of this plane to deduce a basis for $v$.
A: Another way to avoid all calculations is to know that the space of $m\times n$ matrices, i.e. $R^{m\times n}$ is in fact isomorphic to the space of a vector of length $mn$, i.e. $R^{mn}$.
Since we know that a vector space of vectors is linear, therefore, a space isomorphic to it is also linear. Hence, proved.
