How do you find the basis for the spanning set which contains matrices?

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Given a set of vectors that spans a vector space, the largest subset that contains only independent vectors is a basis. The condition for dependence is that
[tex]a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n= 0[/tex]
with at least on of the scalars non-zero. If it is, say [\itex]a_i\ne 0[/itex], we can solve for the vector [itex]v_i[/itex] as a function of the others and so can drop it from the set.

Here, that equation is
[tex]a_1\begin{bmatrix}0& 3\\ 1 & 1\end{bmatrix}+ a_2\begin{bmatrix}4 & 5 \\ 3 & 1 \end{bmatrix}+ a_3\begin{bmatrix}-4 & 1 \\ -1 & 1\end{bmatrix}+ a_4\begin{bmatrix}2 & 1 \\ 1 & 2\end{bmatrix}[/tex]
[tex]= \begin{bmatrix}4a_2- 4a_3+ 2a_4 & 3a_1+ 5a_2+ a_3+ a_4 \\ a_1+ 3a_2- a_3+ a_4 & a_1+ a_2+ a_3+ 2a_4\end{bmatrix}[/tex]
which is the same as the four equations
[tex]4a_2- 4a_3+ 2a_4= 0[/itex]<br /> [tex]3a_1+ 5a_2+ a_3+ a_4= 0[/tex]<br /> [tex]a_1+ 3a_2- a_3+ a_4= 0[/tex]<br /> [tex]a_1+ a_2+ a_3+ 2a_4= 0[/tex]<br /> <br /> An obvious solution is [itex]a_1= a_2= a_3= a_4= 0[/itex]. If that is the <b>only</b> solution, the four matrices are independent and so are a basis for there span. If there is a solution in which one of the coefficients is non-zero, we can solve for that matrix in terms of the other three and so drop it from the set.[/tex]
 
@HallsofIvy: I suspect that lawnmowjob asked this as a homework problem, then either figured it out or got help from somewhere else. Given the shortness of his original post, I doubt that he'll be back. It happens too often in these forums, and makes me reluctant to write out more explicit and helpful discussions like yours.

Anyway, I hid the answer in the middle full stop of my answer.
 
sad, but true. many people just want "the answer", and have little interest in gaining the knowledge.