I'm trying to show that, given two finite-dimensional vector spaces $V,W$, and any subspace V' of $V$, that there is a linear map $T:V\to W$, whose kernel is precisely V', given the condition that $\dim V-\dim(\ker T)<\dim W$. I would like to know if the same is true for infinite-dimensional spaces.
Because of Rank-Nullity, we have restrictions on the respective dimensions; we need
$\dim W =\dim V-\dim(\ker T), \qquad\mbox{ (I think) }.$
This is my work: let $\dim V=m $, $\dim W=r$; $r=m-n $, for $\dim(\ker T)=n$. We start by taking a basis
B_V':=\{v'_1,\ldots,v'_n\}, and extend B_V' into a basis B_V:=\{v'_1,\ldots,v'_n,v'_{n+1},\ldots,v'_m\} for $V$. Let $B_W:=\{w_1,w_2,\ldots,w_r\}$.
Now, we define $T$: T(B_V'):=0, i.e., $T$ is zero for every vector in B_V', and $T$ is linear. By linearity, $T$ is zero on V'.
Now:
This is the part that seems harder: how to define $T$ outside of V', so that $T(w) \neq0$ for w \in V\setminus V'.
My idea is:
i)We set up a bijection between the basis vectors in B_V\setminus B_V', and the basis vectors in $B_W$, say:
T(v'_{n+1})=w_1, T(v'_{n+2})=w_2, $\vdots$ T(v'_m)=w_r, and extend $T$ linearly.
ii) Since a bijection between basis vectors extended linearly gives rise to a Vector Space isomorphism, the kernel of T|_{V\setminus V'}\rightarrow W is an isomorphism, so that its kernel is $0$.
Does this work? Can we extend it to the infinite-dimensional case?
Thanks.