# Finding a basis for vector space $U$

Let $U$ denote the subspace of $M_{2\times 2}(\mathbb{C})$ defined by

$$U=\left\lbrace\left(\begin{matrix}a&b\\ c&0\end{matrix}\right):a + b + c=0\right\rbrace.$$

How would one find a basis for that vector space? Any clues please.

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Well, the only requirement is that $a+b+c=0$, and since we don't require the matrix to be invertible, a general matrix is then $$\left(\begin{matrix}a&b\\ -a-b&0\end{matrix}\right)$$ Can you work out the basis from here?

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So then the basis will be B = {[1 0, -1 0],[0 1, -1 0]}.Did you assume that it's also linearly independent? –  John Jul 20 '13 at 10:03
That is the right basis. Also, the fact that they are linearly independent is easy to prove. If you are satisfied with the answer, mark it correct to remove it from the unanswered section. –  BlackAdder Jul 20 '13 at 10:08
Thanks!Out of curiosity, how would you set up the general matrix if the zero in U is replaced with d and the new condition is a + b + c + d = 0? –  John Jul 20 '13 at 10:13
Well, the same thing really. First, make one variable dependent on the rest of the variables, so we have $d=-a-b-c$, so work out the general matrix. Then notice that we have 3 "free" variables in $a,b,c$, so the dimension of the basis must be 3. Can you work it out? I'm happy to explain more if you need me to. –  BlackAdder Jul 20 '13 at 10:17
I understand now :)!!! I asked a question math.stackexchange.com/questions/447949/… and want to find the basis for that span, will it be B={[-1 0, 0 1]}? –  John Jul 20 '13 at 10:23