Prove the existence of a linear mapping $L: V \to V $ such that $\ker(L)=S$ I am having problems in proving the following. should I find out the linear mapping so as to prove there exists a linear mapping? Thanks for any help.
Let $\{v_1, v_2, \ldots, v_k\}$ be a basis for a subspace $S$ of an $n$-dimensional vector space $V$.
Prove that there exists a linear mapping $L: V \to V $ such that $\ker(L) = S.$
 A: The trick is that a a linear map is fully defined by what it is worth on a basis. You need a basis for $V$, with the property desired.
For instance, you can complete the basis $\{v_1, \ldots, v_k \}$ of $S$ to a basis of $V$, say $\{v_1, \ldots, v_k,w_1, \ldots, w_{n-k} \}$. Then you can define $L$ as follow:
$L(v_j)=0$ for all $1\leq j \leq k$, whereas $L(w_i)=w_i$ for all $1 \leq i \leq n-k$. Then extend $L$ by linearity.
Obs: The important part is that $L(v_j)$ map to $0$, so $ker L = S$, the other part of the basis you can map to anywhere except the $0$ vector.
A: Yes, finding an example of such a linear mapping is a good way to show that such a linear mapping exists, and this is probably the intended approach to the problem.
Hint: Suppose that $\{v_1,\dots,v_n\}$ is a basis of $V$. Given any choice of $w_1,\dots,w_n \in V$, there exists a unique linear map $L$ satisfying $L(v_j) = w_j$ for each $j = 1,\dots,n$.
A: 
Lemma: Let $V$ be a $n$-dimensional vector space over field $F$ and let $\{\alpha_1,…,\alpha_n\}$ be a basis of $V$. Let $W$ be a vector space over field $F$. If $(\beta_1,…,\beta_n)$ is sequence in $W$, then $\exists !$ $T:V\to W$ such that $T$ is linear map and $T(\alpha_j)=\beta_j$, $\forall j\in J_n$.

Proof:  since $\{\alpha_1,…\alpha_n\}$ is basis of $V$, we have $\forall \alpha \in V$, $\exists !$$x_\alpha=(x_{\alpha ,1},…,x_{\alpha ,n})$ $\in F^n$ such that $\sum_{i=1}^n x_{\alpha,i}\cdot_V \alpha_i=\alpha$. Define $f:V\to F^n$ such that $f(\alpha)=x_\alpha$ and $g:F^n \to V$ such that $g(y)=\sum_{i=1}^n y_i \cdot_W \beta_i$. So $T=g\circ f:V\to V$ such that $T(\alpha)=g\circ f(\alpha)$ $=g(f(\alpha))$ $=g(x_\alpha)$ $=\sum_{i=1}^n x_{\alpha ,i}\cdot_W \beta_i$, is our desired function. It’s easy to check, $T$ is linear, $T(\alpha_j)=\beta_j$ for all $ j\in J_n$ and $T$ is unique.


Problem: Let $\{v_1, v_2, ... , v_k\}$ be a basis for a subspace $S$ of an $n$-dimensional vector space $V$. Prove that $\exists$ a linear mapping $T: V\to V$ such that $N_T = S$.

Proof: This kind of problem, most of the time, use above lemma to show existence of a linear map. We first extend basis of $S$ to $V$. Since $\{v_1,…,v_k\}\subseteq V$ is independent and $V$ is $n$-dimensional vector space, we have $\exists B\subseteq V$ such that $B$ is finite basis of $V$ and $\{v_1,…,v_k\}\subseteq B$. Let $B=\{v_1,…,v_k,v_{k+1},…,v_n\}$ be basis of $V$. The “main idea” of proof is to choose $(\beta_1,…,\beta_n)$ sequence in $V$ such that $N_T=S$. If you try to prove $N_T\subseteq S$ and $S\subseteq N_T$, you will notice that $\beta_i=0_V$, $\forall 1\leq i\leq k$ and $\beta_i=v_i$, $\forall k\lt i\leq n$ works perfect! By lemma, $\exists T:V\to V$ such that $T$ is a linear map and  $T(v_i)=\beta_i=0_V$, $\forall i\in J_k$ and $T(v_i)=\beta_i=v_i$, $\forall i\in J_n\setminus J_k$. It’s easy to check, $N_T=S$.
