Let $V$ is a finite vector space, $W$ is a subspace of $V$. Show that:
- Exist a endomorphism $f$ on $V$ such that $Im (f) = W$.
- Exist a endomorphism $g$ on $V$ such that $Ker (g) = W$.
My attempt: Consider a homomorphism $h$ on $V$. Let $f = h|_{W}: W \rightarrow V$, we have $Im(f) = f(W)$. The remainder of this problem is $f(W) = W$ implies $f = id(W)$.
On the one hand, consider $g = h|_{W}: W \rightarrow V$ such that $g(w)=0, \forall w \in W$. Q.E.D
Is my solution right?
Let $\{w_1,\ldots,w_m\}$ be a basis for $W.$ Extend this basis to a basis for $V,$ $\{w_1,\ldots,w_m,v_1,\ldots,v_p\},$ where $m+p = \dim V.$
Define $f : \{w_1,\ldots,w_m,v_1,\ldots,v_p\}\longrightarrow V$ by $f(w_i)=w_i$ for $1\leq i \leq m$ and $0$ otherwise. Then $f$ extends uniquely to a linear map $L_f,$ and $\text{Img}(L_f) = W.$
Define $g : \{w_1,\ldots,w_m,v_1,\ldots,v_p\}\longrightarrow V$ by $g(w_i)=0$ for $1\leq i \leq m$ and $g(v_j)=v_j$ for $1\leq p$ . Then $g$ extends uniquely to a linear map $L_g,$ and $\ker(L_g) = W.$