Prove that there is a linear map $T:V\to V$ such that $\text{ker}(T) = U$?

Question

Let $V$ be a vector space and $U \subseteq V$ a sub-space of V.
How can I prove that there is a linear map $T:V\to V$ such that $\text{ker}(T) = U$?
(Please suggest answers; not hints.)

Attempt
I tried $T_1(v) = 0$ for $v\in U$ and $T_1(v) = w$ for $v\not\in U$ and a chosen $w\in V$ yet $w\not\in U$. But this isn't necessarily a linear map since $u, v \in U$ yet it might be that $u + v \not\in U$. And I got stuck.

Answer 1

We make use of the following theorem ("extending a map by linearity"):

Let $V$, $W$ be vector spaces. Let $\beta$ be a basis of $V$. For any map $f:\beta\to W$, there exists a unique linear map $T:V\to W$ such that $T(x)=f(x)$ for any $x\in\beta$.

Then we can simply choose a basis $\alpha$ for $U$ and extend it to a basis $\beta$ of $V$. Choose $f:\beta\to V$ such that $f(x)=0$ for all $x\in \alpha$ and $f(x)=x$ for all $x\in \beta\setminus\alpha$. Then there exists linear map $T:V\to V$ such that $T(U)=\{0\}$. The fact that $\text{ker}(T)\le U$ can be proved easily by noticing that $\beta\setminus\alpha$ is linearly independent.