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Assume that $A \in M_{m,n}(F)$ and define $T_A:F^n \to F^m$ like this :

$T_A(X)=AX$

Prove that the column-space of $A$ is the same as $Im(T_A)$ .

Note 1 : My problem is that i don't know which part of $F^m$ is spanned by $Im(T_A)$ .

Note 2 : There is a similar question here. But it's just an example, Not a proof.

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    What does $T_A$ map the standard base $\{(1,0,\ldots,0),\ldots,(0,\ldots,0,1)\}$ of $F^n$ into? – 2017-01-03
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    See https://youtu.be/st6D5OdFV9M for a detailed explanation. – 2017-01-03

1 Answers 1

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Suppose $\;\mathbf r=(r_1,...,r_m)\in F^m\;$ is in the image of $\; T_A\;$ , then there exists $\;\mathbf x=(x_1,...,x_n)\in F^n\;$ s.t.

$$T_A\mathbf x=\mathbf r\implies A\mathbf x^t=\mathbf r^t$$

Try now to take it from here...