# 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..

-

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – 900 sit-ups a day, daw, J. W. Perry, RecklessReckoner, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.

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?