vector space dimensions question

Question

Let $V$ be a vector space of dimension $4$, and $T:V\to V$ a linear transformation. Let $U=\ker(T)$ and suppose $T(x)\neq0$ for some $x\in V$. What are the possible values of dimension of $U$?

My attempt: so I can prove that $U$ is a subspace of $V$, so I know $\dim(U)$ must be less than or equal to $4$, but since clearly it isn't true that $U=V$ then it must be less than 4. What else am I missing? Thanks

Answer 1

Yes you are absolutely right, if $T(x)\neq 0$ for some $x$ then $dim(U) <4$.

As an example, if it was also given that “for some $x\neq0$ we had $T(x)=0$“ then we could also conclude that $dim(U) >0$.

Answer 2

Yes. The fact that $T(x)\ne0$ for some $x\in V$ implies that the rank of $T$ is $\ge1$. By the rank-nullity theorem, $$ \dim U=\dim V-\operatorname{rank}T=4-\operatorname{rank}T<4 $$ It's not difficult to show that any value between $0$ and $3$ can be taken.