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For a square matrix having the property that every $\vec{v} \in \mathbb{R}^n$ is a linear combination of its columns, show that every $\vec{v} \in \mathbb{R}^n$ is also a linear combination of its rows.

I wasn't sure which direction to go with this one. I'm not sure if I see why it should be true either.

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    This is a special case of "row rank=column rank" You can find three prrofs here http://en.wikipedia.org/wiki/Rank_%28linear_algebra%29#Column_rank_.3D_row_rank_or_rk.28A.29_.3D_rk.28AT.292012-06-26
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    If you don't know about ranks yet, this also follows from the fact that a matrix having a one-sided inverse, has a two-sided inverse.2012-06-26

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