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Having a determined set: $(a) =\{\left[ {\begin{array}{ccccc}1 & 1\\1 & 1\\\end{array} } \right] \left[ {\begin{array}{ccccc}0 & 1\\0 & 1\\\end{array} } \right] \left[ {\begin{array}{ccccc}1 & 0\\1 & 0\\\end{array} } \right] \left[ {\begin{array}{ccccc}0 & 2\\0 & 2\\\end{array} } \right] \}$

How can I check if it generates $M_{2*2}$?

$M_{2*2}$ represents any space with a 2 by 2 matrix, right?

I know the answer is that the set does not generate the space, but why not, even with only one 2 by 2 matrix making the set, shouldn't it generate, even though not completely, a part of the $M_{2*2}$?

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    $\begingroup$ By generating, you mean using linear combinations? Then notice that for all generated matrices the two rows are identical. $\endgroup$ – Jean-Claude Arbaut Dec 26 '18 at 13:53
  • $\begingroup$ @Jean-ClaudeArbaut Yes! The problem is, if $M_{2*2}$ is just a space with 2 by 2 matrices, even if the set doesn't generate all of the space, it still generates it, doesn't it? Thank you for the precious help! $\endgroup$ – Miguel Ferreira Dec 26 '18 at 14:05
  • $\begingroup$ "even if the set doesn't generate all of the space, it still generates it" How does this make any sense? It does not generate the space of $2\times2$ matrices, period. You can find what is generated: it's exactly the space $E$ of $2\times2$ matrices with equal rows. Note the 2nd and 3rd are linearly independent and generate $E$, the other two are linear combinations. $\endgroup$ – Jean-Claude Arbaut Dec 26 '18 at 14:11
  • $\begingroup$ @Jean-ClaudeArbaut Thank you! $\endgroup$ – Miguel Ferreira Dec 26 '18 at 14:27
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Well, the space $M_{2\times 2}$ has dimension $4$, you have $4$ matrices, and the second and the fourth are clearly l.d.

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  • $\begingroup$ Yes! Sure! But if we remove the second and the fourth matrices, will the set still not generate a space $M_{2*2}$? I mean, not the entire space, sure, but some of it? Thank you for the help! $\endgroup$ – Miguel Ferreira Dec 26 '18 at 14:03
  • $\begingroup$ @MiguelFerreira If you remove matrices, you obviously can't generate a larger space. $\endgroup$ – Jean-Claude Arbaut Dec 26 '18 at 14:12
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Your set has $4$ matrices, but the fourth one is twice the second on. So, forget the fourth one. Now, the first one is the sum of the second and the third ones. So, forget the first one too. So, what you're after is ths space$$\operatorname{span}\left\{\begin{bmatrix}0&1\\0&1\end{bmatrix},\begin{bmatrix}1&0\\1&0\end{bmatrix}\right\}.$$This is the space$$\left\{\begin{bmatrix}a&b\\a&b\end{bmatrix}\,\middle|\,a,b\in\mathbb R\right\},$$which clearly is not $M_{2,2}(\mathbb{R})$.

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