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I am looking for a basis in M22. So I need to get a linearly independent set that also spans the vector space. I have worked out how to tell independence, but I am stuck on the spanning requirement.

For example, I have the set of matrices:

$$ \left[ \begin{array}{ c c } 3 & 6 \\ 3 & -6 \end{array} \right] ,\left[ \begin{array}{ c c } 0 & -1 \\ -1 & 0 \end{array} \right] , \left[ \begin{array}{ c c } 0 & -8 \\ -12 & -4 \end{array} \right] , \left[ \begin{array}{ c c } 1 & 0 \\ -1 & 2 \end{array} \right] $$

I found the set to be linearly independent by reducing it to row echelon form and having only trivial solutions. (I left out the augmented 0 row)

$$ \left[ \begin{array}{ c c } 1 & 2 & 0 & 1/3\\ 0 & 1 & 8 & 2\\ 0 & 0 & 1 & 1/2\\ 0 & 0 & 0 & 1 \end{array} \right] $$

But this seems to be to also be a spanning set, because if I continue to reduce the matrix to reduced row echelon form I will get a = b = c = d = 0 since there are no non-trivial solutions the set can only be a linear combination of scalars = 0. $a*M_1 + b*M_2 + c*M_3 + d*M_4 = 0$.

Is this correct or am I missing something?

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    $\begingroup$ As in any finite-dimensional vector space, if you have an independent set of vectors, and if the number of vectors in the set equals the dimension of the space, then the set is also a spanning set. In your example you have four independent vectors in the space of $2\times2$ matrices, so it is also a spanning set. $\endgroup$ – David Apr 24 '14 at 6:03
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    $\begingroup$ @David maybe post that as an answer so as to keep this from piling on to the already massive set of unanswered questions. I think your response is sufficient as an answer. Clearly the op is satisfied. $\endgroup$ – J. W. Perry Apr 24 '14 at 6:07
  • $\begingroup$ @J.W.Perry, thanks for the suggestion - done. $\endgroup$ – David Apr 24 '14 at 6:10
  • $\begingroup$ Your title needs serious improving. What is a $2\times 2$ set? $\endgroup$ – Andrés E. Caicedo Apr 24 '14 at 6:21
  • $\begingroup$ @AndresCaicedo: Just a transpose. Fixed especially for you ;) $\endgroup$ – Travis Apr 24 '14 at 8:45
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As in any finite-dimensional vector space, if you have an independent set of vectors, and if the number of vectors in the set equals the dimension of the space, then the set is also a spanning set. In your example you have four independent vectors in the space of $2\times2$ matrices, so it is also a spanning set.

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