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Question: Find a basis for $M_{2\times 2}$, the space of $2 \times 2$ matrices.

My query: Can I say that all of the following are valid answers(out of many other possible answers),

  1. $\bigl[ \begin{smallmatrix} 1 & 1\\ 1 & 1 \end{smallmatrix} \bigr]$

  2. $\big< \bigl[ \begin{smallmatrix} 1 & 0\\ 1 & 0 \end{smallmatrix} \bigr]$, $\bigl[ \begin{smallmatrix} 0 & 1\\ 0 & 1 \end{smallmatrix} \bigr] \big>$

  3. $\big< \bigl[ \begin{smallmatrix}0 & 0\\ 1 & 1 \end{smallmatrix} \bigr]$, $\bigl[ \begin{smallmatrix} 1 & 1\\ 0 & 0 \end{smallmatrix} \bigr] \big>$

  4. $\big< \bigl[ \begin{smallmatrix} 1 & 1\\ 1 & 0 \end{smallmatrix} \bigr]$, $\bigl[ \begin{smallmatrix} 0 & 0\\ 0 & 1 \end{smallmatrix} \bigr] \big>$

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A basis is a set (or a sequence) so you should say what those angle brackets mean. –  GEdgar Sep 3 '12 at 13:43

4 Answers 4

up vote 3 down vote accepted

You are right in that you may have many possible bases for a vector space. But, the ones that you have listed are not bases for the vector space that you are considering. Note that the space of $2\times 2$ matrices (over say the real numbers) is the set of all matrices of the form $$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} $$ where $a,b,c,$ and $d$ each can be any real number. Now, to see that your first option doesn't work as a basis, you could consider the matrix

$$ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}. $$ The If the first option was a basis, then you would have to be able to find a real number $s$ such that $$ s\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}. $$ But here the left hand side is equal to $$ \begin{pmatrix} s & s \\ s & s \end{pmatrix}. $$ Noting that two matrices are equal if and only if each entry is equal, that would mean that $s$ would have to be $0$ and $1$, and there are no solutions for this. So you can say that the element $\left(\begin{smallmatrix}1 & 1 \\ 1 & 1 \end{smallmatrix}\right)$ does not span the space of $2\times 2$ matrices.

Now you could try to think about your other option in the same manner. Maybe this would lead you to a correct basis. As others have already noted, a correct basis would have $4$ elements.

As a sidenote and just to make sure that you understand. The span of the elements in, for example, your second option would be $$ s\begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix} + t\begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} s & t \\ s & t \end{pmatrix}. $$ But again, this does not span the space of $2\times 2$ matrices.

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Hint: In LineLand, instead of writing $2\times 2$ matrices like this $$\begin{pmatrix} a & b\\c & d\end{pmatrix}$$ people write them in a line like this $$(a,b; c,d).$$

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If $M_{2 \times 2}$ is 4-dimensional then you are probably going to want no fewer than 4 things in the basis.

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This space is four-dimensional. You need four matrices to obtain a basis.

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