Show that set of all $2 \times 2$ matrices forms a vector space of dimension $4$ I have this question:

Show that the set of all $2 \times 2$ matrices with real coefficients forms a linear space over $\Bbb R$ of dimension $4$. 

I know that the set of the matrices will basically form a linear combination which will define the vector space and they satisfy the axioms defined for the vector space. 
I do not know how to show that this is possible. Do the vectors have to be linearly independent?
Any sort of help is appreciated.
Thanks.
 A: Begin by listing the axiom satisfied by a general vector space over $\mathbb{R}$. Now consider your set of matrices. What does you addition look like? What is your zero element? What does your scalar multiplication look like? Does your scalar multiplication and your addition behave as they should? Finally, show that all 2 by 2 matrices can be written as a linear combination of 4 special matrices. A good choice is to pick matrices who each have precisely 3 zero entries and 1 non zero entry. Can you show that these 4 matrices are linearly independent?
A: One can easily see that $\mathbb{R}^{2\times 2}$ is in fact exactly the same as the vectorspace $\mathbb{R}^4$, only the notation of the elements differs. And of course whether or not something is a vector space does not depend on the way some unimportant species of primates decides to write its elements.
A: Hint Can you find a basis of the set of $2 \times 2$ matrices consisting of four elements? (There is a natural choice of basis here that includes the matrix $\pmatrix{1&0\\0&0}$.)
Alternatively, can you find a vectorspace isomorphism from the space of $2 \times 2$ matrices to some vector space you know to be $4$-dimensional, e.g., $\Bbb R^4$?
A: Let
$v_{11} = \begin{bmatrix}
1 & 0\\
0 & 0
\end{bmatrix}$
,
$v_{12} = \begin{bmatrix}
0 & 1\\
0 & 0
\end{bmatrix}$
,
$v_{21} = \begin{bmatrix}
0 & 0\\
1 & 0
\end{bmatrix}$
,
$v_{22} = \begin{bmatrix}
0 & 0\\
0 & 1
\end{bmatrix}$
Suppose $av_{11} +bv_{12} +cv_{21} +dv_{22} = \begin{bmatrix}
0 & 0\\
0 & 0
\end{bmatrix}$.
$Then \begin{bmatrix}a & b\\c & d\end{bmatrix} = \begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}$.
from which it follows immediately that a = b = c = d = 0. Thus $v_{11}$,  $v_{12}$, $v_{21}$, $v_{22}$ are linearly independent.
Now let $\begin{bmatrix}a & b\\c & d\end{bmatrix}$ be any 2×2matrix.
Then $\begin{bmatrix}a & b\\c & d\end{bmatrix}$ = $av_{11} +bv_{12} +cv_{21} +dv_{22}$. Thus $v_{11}$,  $v_{12}$, $v_{21}$, $v_{22}$ span the space of 2 × 2 matrices.
Thus $v_{11}$,  $v_{12}$, $v_{21}$, $v_{22}$ are both linearly independent and they span the space of all 2 × 2 matrices. Thus $v_{11}$,  $v_{12}$, $v_{21}$, $v_{22}$
constitue a basis for the space of all 2 × 2 matrices.
Now you can see the dimension can be clearly deduced.
