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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 en.wikipedia.org/wiki/… –  Simon Markett Jun 26 '12 at 7:37
    
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. –  mrf Jun 26 '12 at 8:01

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Let the square matrix be denoted by $A$. That every vector $v$ is a linear combination of the columns of $A$ means that you can always solve the equation $Ax=v$, where $x$ is the unknown. This is only possible if $A$ is invertible, so we have $\det(A)\neq 0$. The determinant of $A'$ (the transpose of $A$) is the same as the determinant of $A$, so $A'$ is also invertible. This means you can always solve the equation $A'x=v$, or that $v$ is a linear combination of the columns of $A'$. Since the columns of $A'$ are the same as the rows of $A$, we are done.

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Saying the same thing shorter: linear independence of columns is equivalent to determinant different than zero is equivalent to linear independence of rows. –  Krastanov Jun 26 '12 at 12:17

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