if $T : R^n \rightarrow R^m$ onto then $rank A = m$

Question

Let $T : R^n \rightarrow R^m$ linear transformation

$T(x) = Ax$ , $A \in M_{m \times n}(\mathbb{R})$, $x \in \mathbb{R}$

prove or disprove :

1. if $T$ onto then $\mathrm{rank} A = m$
2. if $T$ one to one then $\mathrm{rank} A = n$
3. if $n\ge m$ then $T$ is onto
4. if $n-m=2$ then $\dim(\ker T) = 2$

$T$ is one to one $\iff \ker T = {\{0}\} \rightarrow Ax=0$ so to get the trivial solution of $Ax=0$ we need that $\mathrm{rank}(A)=n$

$T$ is onto $\iff \mathrm{Im} T = R^m \implies \dim(\mathrm{Im}T) = m \implies \mathrm{rank}(A)=m$

Is this how to prove the first two

and about 3/4

if $T$ is onto then $n\ge m$ but how about the opposite ?

and any hint how to solve the last one ?

thanks

Answer 1

(3) and (4) are false. Zero matrix of dimension $3\times 5$ disproves both. Nullity of $m\times n$ null matrix is $n$ and it is not onto.

Answer 2

For any linear map $T: V\rightarrow W$, if $T$ is onto then $imT=W$ which means $dim(imT)=dimW=rankT$. If $T$ is injective, then by the rank nullity theorem, $dimimT=V$. 3 and 4 are false... try to find an example to see why (3) is false. For (4), dim(kerT) is not usually equal to $dim(V)-dim(W)$. Try to find an injective map where $dimV \neq dimW$