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.
 
Cute! I'll have to remember that!
 
sad, but true. many people just want "the answer", and have little interest in gaining the knowledge.