Prove if $U$ is surjective, then $T$ is injective as $UT = 0$

Question

So I am stuck with this problem:

(i) To prove this by contradiction:

Suppose if $U$ is surjective, then $T$ is injective.

$U$ is surjective, then $\dim(\operatorname{Im}(U)) = \dim(R^2) = 2$.

So by Rank-Nullity:

$\dim(\ker(U)) = \dim(\mathbb{R}^4) - \dim(\operatorname{Im}(U)) = 4 - 2 = 2$.

And then what next?

Answer 1

(i) You started well. Now, if $T$ was injective $\dim\operatorname{im}T=3$ and therefore $\operatorname{im}T$ cannot be a subset of $\ker U$. In other words, $U\circ T\neq0$.

(ii) Take $T(x,y,z)=(x,y,z,0)$ and $U(x,y,z,t)=(0,t)$.