# Describe the smallest subspace of $M_{2\times 2}$ that contains matrices…

Describe the smallest subspace of $M_{2\times 2}$ that contains matrices

$$\begin{bmatrix}2&1\\0&0\end{bmatrix},\begin{bmatrix}1&0\\0&2\end{bmatrix},\begin{bmatrix}0&-1\\0&0\end{bmatrix}\;.$$

Find the dimension of this subspace.

It sounds like I would find the basis. I know how to do this with vectors, but how do I do this with a set of matrices?

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$M_{2\times 2}$ is isomorphic to $\Bbb R^4$ by the correspondence $$\begin{bmatrix}a&b\\c&d\end{bmatrix}\leftrightarrow\begin{bmatrix}a\\b\\c\\d\end{bmatrix}\;;$$ treat the matrices as $4$-vectors, just displayed in a different way.

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I actually did that, though I got [1 0 0 2; 0 1 0 -4; 0 0 0 -4]. I'm not sure what to make of the last row. Do I just disregard it? –  LearningPython Nov 19 '12 at 8:02
@LearningPython What's wrong with the last row? (I assume this result comes from a row reduction process.) –  Ted Nov 19 '12 at 8:06
I guess my first instinct is to think, "No solution!". So if it's fine, then how do I translate that to a basis? Is it a free variable column, and I just disregard the column of 0s before it? –  LearningPython Nov 19 '12 at 8:07
@LearningPython: I’ve not done the row reduction to check, but that could well be right; certainly there’s nothing obviously wrong with it, and it reduces further to $\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\end{bmatrix}$, which I know is right. What it shows is that the original three matrices are linearly independent, so what do you conclude? –  Brian M. Scott Nov 19 '12 at 8:07
@BrianM.Scott, how does this translate to a basis? I'm used to row reducing into free variables that I translate into a general solution. –  LearningPython Nov 19 '12 at 8:09