Finding the vectors that form a basis of a span Let 
$$\begin{pmatrix}  
1 & 2\\
0 & 1 
\end{pmatrix},
\begin{pmatrix}
0 & 1\\ 3 & 0 
\end{pmatrix},
\begin{pmatrix}
3 & 3\\ 4 & 0 
\end{pmatrix},
\begin{pmatrix}
2 & 0\\ 1 & -1 
\end{pmatrix}$$
be vectors in some vector space $U$. Which vectors form a basis of $Sp(U)$?
In this case it is "easy to see" that $A_3 = A_1 + A_2 +A_4$, and $\sum_{i\in\{1,2,4\}} \alpha_{i}A_{i}=0$ has only the trivial solution. But what if I had $n>5$ matrices of order $k\times k$, $k>4$. How then can I found a basis and identify which matrices are a linear combination of the others?
Thank you. 
 A: In general, we can construct an isomorphism $T: M_{n\times n} \to R^{n\cdot n}$. In your example, we have that $T: M_{2\times 2} \to R^4$, where  $T$ is defined by:
$$T\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right)=(a,b,c,d).$$
From here, we can simply work with vectors in $R^4$ (i.e. we can put the four vectors together, row reduce them and count the pivots, etc.) to find a basis for the vectors you have in $M_{2\times 2}$.
A: Generally, a "basis" for a vector space must both span that space and be linearly independent.   If the give spanning set is already independent, then it is a basis.  If not you can delete one or more vectors until it is.  
Here, for example, to see if these matrices are independent, look at the equation $a\begin{pmatrix}1 & 2 \\ 0 & 1 \end{pmatrix}+ b\begin{pmatrix}0 & 1 \\ 3 & 0 \end{pmatrix}+ c\begin{pmatrix}3 & 3 \\ 4 & 0 \end{pmatrix}+ d\begin{pmatrix}2 & 0 \\ 1 & -1 \end{pmatrix}= \begin{pmatrix}0 & 0 \\ 0 & 0 \end{pmatrix}$.
That gives us the four equations a+ 3c+ 2d= 0, 2a+ b+ 3c= 0, 3b+ 4c+ d= 0, and a- d= 0.  An obvious, trivial, solution, to this is a= b= c= d= 0.  If that is the only solution, then these vectors are independent so a basis.  If we attempt to solve these equations, we can see that d= a from the last equation and then we hafe 3a+ 3c= 0, 2a+ b+ 3c= 0, and 4a+ 4c= 0.  The first and third of those equations give c= -a.  Putting that into the second equation, b- a= 0 so b= a.  We have a= b= -c= d so we and write our equation as $a\begin{pmatrix}1 & 2 \\ 0 & 1 \end{pmatrix}+ a\begin{pmatrix}0 & 1 \\ 3 & 0 \end{pmatrix}- a\begin{pmatrix}3 & 3 \\ 4 & 0 \end{pmatrix}+ a\begin{pmatrix}2 & 0 \\ 1 & -1 \end{pmatrix}= \begin{pmatrix}0 & 0 \\ 0 & 0 \end{pmatrix}$.
We can divide through by a to get $\begin{pmatrix}1 & 2 \\ 0 & 1 \end{pmatrix}+ \begin{pmatrix}0 & 1 \\ 3 & 0 \end{pmatrix}- \begin{pmatrix}3 & 3 \\ 4 & 0 \end{pmatrix}+ \begin{pmatrix}2 & 0 \\ 1 & -1 \end{pmatrix}= \begin{pmatrix}0 & 0 \\ 0 & 0 \end{pmatrix}$.
The point now is that we can solve for any one of those vectors in terms of the other three- four example, we can write $\begin{pmatrix}3 & 3 \\ 4 & 0 \end{pmatrix}= \begin{pmatrix}1 & 2 \\ 0 & 1 \end{pmatrix}+ \begin{pmatrix}0 & 1 \\ 3 & 0 \end{pmatrix}+ \begin{pmatrix}2 & 0 \\ 1 & -1 \end{pmatrix}= \begin{pmatrix}0 & 0 \\ 0 & 0 \end{pmatrix}$.  Since that can be written as a linear combination if the other three, we can span the space with those three alone.  If those three are independent, then they form a basis.  
In general, if a spanning set is also independent, then they are a basis.  If not then they can be written as $a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n= 0$ with at least some of the $a_i$ non-zero.  If, say, $a_i$ is non-zero, we can solve for $a_iv_i= -(a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n)$ (where the sum on the right does NOT incude $a_1v_1$) and then, since $a_1\ne 0$, we can divide by it to get $v_i= -\frac{1}{a_i}(a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n)$ so that we can drop $v_i$ from the set and still span the space.
