Prove there is a linear transformation

Question

I would like some advice with one problem.

Let $L$ be a linear space with $\dim(L)=n$. Let $K$ be some linear subspace of $L$. I need to prove there exists a linear transformation $l: L \rightarrow L$ such that $Im(l)=K$.

My shot:

let $B=\{b_1...b_n\}$ be a basis of $L$. let $K=\{k_1...k_s\}$ be a basis of $K$, $n>s$. we can replaces some vectors $b_1...b_s$ with $k_1...k_s$ and $Span \ B$ won't change. Than $Im(l)=Span\{l(k_1)...l(k_s), l(b_{s+1})...l(b_n)\}$ I have no idea if it's even right or what to do next. Any hints? Thank you

Answer 1

Hint: You can use any ${\underline{\mbox{onto}}}$ mapping of $\{b_1,\dots,b_n\}$ to $\{k_1,\dots,k_s\}$. This will determine a linear transformation $L$ with the properties you need.

Answer 2

Define $l:L \to L$ by $$l(x) = x \text{ for } x \in K$$ $$l(x) = 0 \text{ for }x \in L \setminus K$$

Obviously $l$ is linear on $K$ and on $L \setminus K$. It remains to define $l$ on $K + (L \setminus K)$ and it is just by the linearity.