For $\mathbb{S}$ subspace of $\mathbb{V}$, prove $L:\mathbb{V}\to\mathbb{V}$ such that Ker$(L)=\mathbb{S}$

Question

Let $\{\vec{v}_1,\ldots,\vec{v}_k\}$ be a basis for a subspace $\mathbb{S}$ of an $n$-dimensional vector space $\mathbb{V}$. Prove that there exists a linear mapping $L:\mathbb{V}\to\mathbb{V}$ such that $\text{Ker}(L)=\mathbb{S}$.

Answer 1

Extend the given basis for $S$ to a basis $(v_1,...,v_k,...,v_n)$ for $V$. Define a linear map $T$ that maps the basis vectors of $S$ to $0$ and the other $n-k$ vectors to themselves.

Let $v\in S$. Then there are scalars $c_i$ such that $v=c_1v_1+...+c_kv_k$. Then $Tv=c_1Tv_1+...+c_kTv_k=0+...+0=0.$ Thus, $S\subset \text{Ker}T$.

Let $u\in V$. Then there exist scalars $a_i$ such that $u=a_1v_1+...+a_nv_n$. $Tu=0\implies a_1Tv_1+...+a_kTv_k+...+a_nTv_n=0\implies a_{k+1}v_{k+1}+...+a_nv_n=0\implies a_{k+1},...,a_{n}=0$ since these vectors are linearly independent. Therefore $u\in S$. So, $\text{Ker}T\subset S$. Therefore, $\text{Ker}T=S$.