Bound on the rank of a matrix.

Question

Let $A$ be an $m \times n$ matrix. Prove that the rank of the Linear Transformation which multiplies elements in vector space $F^n$ to $A$ to give elements in vector space $F^m$ ($F$ being the field) is atmost $m$.

I thought of using the fact that rank is the dimension of the image.

Also, since the $n$ element set of columns span the image, and since cardinality of basis $\leq$ cardinality of span, we get Rank $\leq n$.

But this gives $n$ instead of $m$.

Any help will be appreciated.
Thanks in advance.

Answer 1

To summarise our discussion in the comments, there are three options:

1. Use the fact that $\operatorname{rk}M=\operatorname{rk}M^T$ for every matrix $M$
2. Use the First Isomorphism Theorem: for every linear transformation $T:V\to W$ there is a natural isomorphism $$\operatorname{Im}T\simeq V/\operatorname{Ker}T$$
3. Use the weaker version (or corollary) of the FIT known as the rank-nullity theorem: $$\dim \operatorname{Im}T= \dim V- \dim \operatorname{Ker}T$$