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
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1$\begingroup$ Are you talking about $2\times2$ matrices? And what have you tried? $\endgroup$– Simon SCommented Nov 30, 2014 at 14:26
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1$\begingroup$ 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. $\endgroup$– Travis WillseCommented Nov 30, 2014 at 14:28
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$\begingroup$ 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. $\endgroup$– GorTeXCommented Nov 30, 2014 at 14:29
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$\begingroup$ 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. $\endgroup$– Simon SCommented Nov 30, 2014 at 14:33
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1$\begingroup$ @GorteX it doesn't create the whole of $\Bbb R^{2\times 3}$, but it does create all of $V$. $\endgroup$– Ben GrossmannCommented Nov 30, 2014 at 14:42
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2 Answers
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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 $$ M = av_1 + bv_2 + cv_3 + dv_4 + ev_5 $$
answered Nov 30, 2014 at 14:45
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$\begingroup$ 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? :-) $\endgroup$– GorTeXCommented Nov 30, 2014 at 14:58
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$\begingroup$ 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 :-) $\endgroup$– GorTeXCommented Nov 30, 2014 at 15:11
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$\begingroup$ Yes, that's right. We know it creates all of $V$ because the only thing we knew about $M$ was that $M \in V$ $\endgroup$ Commented Nov 30, 2014 at 15:27
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$\begingroup$ and all of you guys thank you very much, really appreciate it...thx $\endgroup$– GorTeXCommented Nov 30, 2014 at 15:31
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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?
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$\begingroup$ 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? $\endgroup$– GorTeXCommented Nov 30, 2014 at 14:52
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$\begingroup$ @GorteX Yes, that's the idea. Those matrices belong to $V$; are they a basis? $\endgroup$– egregCommented Nov 30, 2014 at 14:53
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$\begingroup$ I would say they are a basis, beacause they are linearly independent $\endgroup$– GorTeXCommented Nov 30, 2014 at 14:54
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$\begingroup$ @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 $5$. $\endgroup$– egregCommented Nov 30, 2014 at 14:58