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
a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n= 0
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 v_i as a function of the others and so can drop it from the set.

Here, that equation is
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}
= \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}
which is the same as the four equations
4a_2- 4a_3+ 2a_4= 0[/itex]<br /> 3a_1+ 5a_2+ a_3+ a_4= 0<br /> a_1+ 3a_2- a_3+ a_4= 0<br /> a_1+ a_2+ a_3+ 2a_4= 0<br /> <br /> An obvious solution is a_1= a_2= a_3= a_4= 0. 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.
 
@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.
 
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