T is onto $\mathbb R^m$ iff the columns of A span [closed]

i need to proof why $T$ is onto $\mathbb R^m$ iff the columns of $A$ span if the columns are Linearly independent $T$ is onto..

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closed as off-topic by 900 sit-ups a day, daw, J. W. Perry, RecklessReckoner, Jyrki LahtonenJul 16 at 7:03

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It's not clear to me what your question is - iff the columns of $A$ span what? Also, what is $T$ and what is $A$. –  Joe Tait Mar 18 '13 at 17:37
I think he means that $T$ is given by an $m\times n$ matrix, and he needs to show that it is onto $\mathbb{R}^m$ iff the columns of $A$ span $\mathbb{R}^m$. But yes, a little more effort and clarity would be nice... –  Lepidopterist Mar 18 '13 at 17:44

It's useful to realize that a matrix times a vector, $Tx$, can be represented as $$Tx=x^{(1)}t_1+x^{(2)}t_2+\cdots+x^{(n)}t_n.$$Here $t_i$ represents the i-th column of $T$ and $x^{(i)}$ the i-th component of the vector $x$. Can you finish the proof knowing this?