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Here is the question...

"Either find a basis for the vector space M2×2 of 2 × 2 matrices that includes the vectors

$$\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$$ $$\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$$ $$\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$$

or else explain why this is not possible. [4 marks]"

Do i need to find the reduced row echelon form of these 3 matrices? I know how to find the basis if I was given something like this "v1=(1,1),v2=(2,−1)" but not this style of question. Just an insight on how to start to answer this question would be much appreciated.

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  • $\begingroup$ First, a basis of $M(2,K)$ is composite by matrices. Second, do you want include $(0,0)$ in a base? $\endgroup$ – Rafael Holanda Aug 23 '16 at 7:55
  • $\begingroup$ I think you will have to figure out how to properly convey your ideas here (the easiest way: learn the easy directions to properly write mathematics with MathJaX in this site), or else it is going to be very hard to understand you. Where you wrote "MATRIX" there are three pairs of numbers: how are we to turn this into a $\;2\times2\;$ matrix?? $\endgroup$ – DonAntonio Aug 23 '16 at 8:03
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    $\begingroup$ Well, the correct way to format a matrix is to write $$\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}$$ to get $$\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}$$ $\endgroup$ – celtschk Aug 23 '16 at 8:04
  • $\begingroup$ BTW, even if you were not able to do it in MathJax, you could at least have written it as preformatted text. As is, I have no idea what your matrix looks like. $\endgroup$ – celtschk Aug 23 '16 at 8:06
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Hint: There isn't anything particularly special about the structure of how each of the entries are arranged in each $2 \times 2$ matrix, as long as we're consistent. In fact, since we're more comfortable with working with $m \times 1$ vectors in $\mathbb R^m$ instead of $n \times n$ vectors in $M_{n \times n}$, we can imagine unfolding each square matrix into a column vector in some arbitrary (but fixed) order, say: $$ \begin{bmatrix} b & c \\ a & d \end{bmatrix} \iff \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} $$ So what we can do is:

  • Convert each $2 \times 2$ matrix into a $4 \times 1$ column vector.
  • Apply the usual algorithm for throwing out any extraneous linearly dependent vectors until you get a basis.
  • Convert each $4 \times 1$ column vector in your basis back to a $2 \times 2$ matrix.
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