Find the basis for the space of $2\times 2$ matrices. Question: Find a basis for $M_{2\times 2}$, the space of $2 \times 2$ matrices.
My query: Can I say that all of the following are valid answers(out of many other possible answers), 


*

*$\bigl[ \begin{smallmatrix} 1 & 1\\ 1 & 1
    \end{smallmatrix} \bigr]$

*$\big< \bigl[ \begin{smallmatrix} 1 & 0\\ 1 & 0
    \end{smallmatrix} \bigr]$, $\bigl[ \begin{smallmatrix} 0 & 1\\ 0 & 1
    \end{smallmatrix} \bigr] \big>$

*$\big< \bigl[ \begin{smallmatrix}0 & 0\\ 1 & 1
    \end{smallmatrix} \bigr]$, $\bigl[ \begin{smallmatrix} 1 & 1\\ 0 & 0
    \end{smallmatrix} \bigr] \big>$

*$\big< \bigl[ \begin{smallmatrix} 1 & 1\\ 1 & 0
    \end{smallmatrix} \bigr]$, $\bigl[ \begin{smallmatrix} 0 & 0\\ 0 & 1
    \end{smallmatrix} \bigr] \big>$
 A: Hint: In LineLand, instead of writing $2\times 2$ matrices like this 
$$\begin{pmatrix} a & b\\c & d\end{pmatrix}$$ 
people write them in a line like this 
$$(a,b; c,d).$$ 
A: This space is four-dimensional. You need four matrices to obtain a basis.
A: You are right in that you may have many possible bases for a vector space. But, the ones that you have listed are not bases for the vector space that you are considering. Note that the space of $2\times 2$ matrices (over say the real numbers) is the set of all matrices of the form
$$
\begin{pmatrix}
a & b \\ c & d
\end{pmatrix}
$$
where $a,b,c,$ and $d$ each can be any real number. Now, to see that your first option doesn't work as a basis, you could consider the matrix
$$
\begin{pmatrix}
1 & 0 \\ 0 & 0
\end{pmatrix}.
$$
The If the first option was a basis, then you would have to be able to find a real number $s$ such that
$$
s\begin{pmatrix}
1 & 1 \\ 1 & 1
\end{pmatrix} = 
\begin{pmatrix}
1 & 0 \\ 0 & 0
\end{pmatrix}.
$$
But here the left hand side is equal to
$$
\begin{pmatrix}
s & s \\ s & s
\end{pmatrix}.
$$
Noting that two matrices are equal if and only if each entry is equal, that would mean that $s$ would have to be $0$ and $1$, and there are no solutions for this. So you can say that the element $\left(\begin{smallmatrix}1 & 1 \\ 1 & 1
\end{smallmatrix}\right)$ does not span the space of $2\times 2$ matrices.
Now you could try to think about your other option in the same manner. Maybe this would lead you to a correct basis. As others have already noted, a correct basis would have $4$ elements.
As a sidenote and just to make sure that you understand. The span of the elements in, for example, your second option would be
$$
s\begin{pmatrix}
1 & 0 \\ 1 & 0
\end{pmatrix} + 
t\begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix}
s & t \\ s & t
\end{pmatrix}.
$$
But again, this does not span the space of $2\times 2$ matrices.
A: If $M_{2 \times 2}$ is 4-dimensional then you are probably going to want no fewer than 4 things in the basis.
