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.
Best Answer
As the OP asked for a method using quotient spaces, here we go.
Let $V$ be a vector space (no restriction on dimensions!), and consider a subspace $U\subset V$. We can form the quotient space $V/U$: by definition, as a set, it is the set of equivalence classes $v+U$ where $v+U=u+U$ if and only if $v-u\in U$. There is a natural notion of addition (we add the representatives) and a natural notion of scalar multiplication (we multiply the representative by the given scalar) making the set $V/U$ into a vector space.
Now, define the projection map $$T:V\to V/U$$ by $T(v)=v+U$. This is well defined (exercise!), and the kernel is precisely $U$.
One advantage of this method is that it is more tidy: no bases. Also, it does not depend on the dimension of your space.