# Vector space, set that generates vector space

I have difficulties with this problem: $$V$$ is a set of all real matrices $$2\times3$$ such as that the sum of elements in the matrices is equal to zero. The set $$V$$ with addition and scalar multiplication creates a vector space $$\mathbf V$$. Find a five element set that generates vector space $$\mathbf V$$. Any ideas? Thank you very much

• Are you talking about $2\times2$ matrices? And what have you tried? Commented Nov 30, 2014 at 14:26
• Welcome to math.se, GorteX! You'll find you'll receive more (and better) answers when your question indicates your work so far, or at least what you've already tried. Commented Nov 30, 2014 at 14:28
• Sorry, it is 2x3 matrices, well I have tried matrices, where one lement is 1 and second one in a different position is -1 but that didnt workout. I have tried to combine where the ones and minus ones are, but it never generated the whole vector space. Commented Nov 30, 2014 at 14:29
• It does work out. You just need to show that any 2x3 matrix whose elements sum to zero can be written as a linear combination of such matrices. Commented Nov 30, 2014 at 14:33
• @GorteX it doesn't create the whole of $\Bbb R^{2\times 3}$, but it does create all of $V$. Commented Nov 30, 2014 at 14:42

Let $\{v_1,\dots,v_5\}$ denote the matrices $$v_1 = \pmatrix{1&0&0\\0&0&-1}, v_2 = \pmatrix{0&1&0\\0&0&-1},\dots\\ v_5 = \pmatrix{0&0&0\\0&1&-1}$$ Now, let $M \in V$ be given by $$M = \pmatrix{a&b&c\\d&e&f}$$ since $M \in V$, we note that $a+b+c+d+e+f = 0 \implies f = -(a+b+c+d+e)$. Now, show that we can write $$M = av_1 + bv_2 + cv_3 + dv_4 + ev_5$$

• I will try that :) that sounds very good :-)...and if I manage to prove that M can be written as the sum of av_1+b_v2+...+ev_5 that should be it, shouldn´t it? :-) Commented Nov 30, 2014 at 14:58
• The last equation I can rewrite as sum of matrices \begin{pmatrix} a&0&0\\ 0&0&-a \end{pmatrix}+...+\begin{pmatrix} 0&0&0\\0&e&-e \end{pmatrix}= \begin{pmatrix} a&b&c\\d&e&-a-b-c-d-e \end{pmatrix} and because a,b,c,d,e are real numbers it creates the whole V? Am I right? Thx...P.S. I will try to make my LaTeX writing better :-) Commented Nov 30, 2014 at 15:11
• Yes, that's right. We know it creates all of $V$ because the only thing we knew about $M$ was that $M \in V$ Commented Nov 30, 2014 at 15:27
• and all of you guys thank you very much, really appreciate it...thx Commented Nov 30, 2014 at 15:31

Hint: if the coefficients in five places are chosen arbitrarily, then the sixth coefficient is determined. Let the “sixth place” be the place $(2,3)$. For example, you have the matrix \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \end{bmatrix} Can you find other four and determine they are a basis for your vector space?

• So if I took your matrix lets call it matrix a) and then the rest of 4 matrices will look like the matrix a) but the +1 elemnt will be shifted one position to the right? Commented Nov 30, 2014 at 14:52
• @GorteX Yes, that's the idea. Those matrices belong to $V$; are they a basis? Commented Nov 30, 2014 at 14:53
• I would say they are a basis, beacause they are linearly independent Commented Nov 30, 2014 at 14:54
• @GorteX They certainly are, but you have also to prove that any element in $V$ is a linear combination of them. Alternatively, you have to prove that the dimension of $V$ is at most $5$. Commented Nov 30, 2014 at 14:58