Question: Show that there is no linear map $T:\mathbb F^5 \to \mathbb F^2 $ (the field of either complex or real numbers, as in Axler's book) such that $\text{null} ~ T=\{(x_1,x_2,x_3,x_4,x_5)\}\in \mathbb F^5 : x_1=3x_2,x_3=x_4=x_5$
I am not sure of my proof, so i am posting it here.
$\dim \mathbb F^5=5, \dim \mathbb F^2=2$, and thus $\dim \text{range} ~ T\leq 2 $, and by the rank nullity theorem,one has $5\leq 2+ \dim \text{null} T$ or that $\dim \text{null} ~ T \geq 3$.
But note that the basis of null space of the map can be at most $2$ due to two free variables ($x_1,x_3$) and if there were more than two, then linear independence will be lost. This gives the contradiction.
I would be thankful in refinement of my proof. Thanks.