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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?2012-11-19
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    @LearningPython What's wrong with the last row? (I assume this result comes from a row reduction process.)2012-11-19
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    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?2012-11-19
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    @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?2012-11-19
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    @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.2012-11-19
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    Ah. In that case, the set IS the smallest subspace. It is the basis.2012-11-19
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    @LearningPython: There you go: the original set is **a** basis. Another is the set of matrices corresponding to the rows of the reduced matrix in my previous comment.2012-11-19
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Notice that they are all of the form, $E = \begin{bmatrix}a & b\\0 & d\end{bmatrix}$.

Thus, we can show that $E = a\begin{bmatrix}1 & 0\\0 & 0\end{bmatrix} + b\begin{bmatrix}0 & 1\\0 & 0\end{bmatrix} + d\begin{bmatrix}0 & 0\\0 & 1\end{bmatrix}$.

You can show that $E_1 = \begin{bmatrix}1 & 0\\0 & 0\end{bmatrix}$, $E_2 = \begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}$, $E_3 = \begin{bmatrix}0 & 0\\0 & 1\end{bmatrix}$ forms a linearly independent set and will span $E$, and the set $\{E_1, E_2, E_3\}$ forms a basis for $E$. Thus, the dimension of $E$ is 3.