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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.

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:O Have you heard of paragraphs? –  anon Apr 11 '12 at 22:49
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I got a little nervous in my editing, since my Latex is poor, and it is my first post here; I will edit it. Thanks for the warm welcome. –  Jay K Apr 11 '12 at 22:53
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No biggie, that's nothing to worry about. Fortunately many here will help out with the editing process. –  anon Apr 11 '12 at 22:57
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My favorite way of thinking about this fact, by the way, is that $V'$ is precisely the kernel of the quotient map $V \to V/V'$. And if the dimensions work out then you can just embed $V/V'$ in $W$. But that probably isn't the spirit of the exercise. –  Dylan Moreland Apr 11 '12 at 23:29
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Actually this thing generalises into many things: For groups, every normal subgroup is the kernel of some group homomorphism, for rings every ideal is the kernel of some ring homomorphism, for modules every submodule is the kernel of some module homomorphism, etc!!!! –  user38268 Apr 12 '12 at 13:04

2 Answers 2

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.

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Let the spaces be finite dimensional. If $\dim(W) \ge \dim(V)-\dim(V')$, then it can be done.

Extend, as you did, a basis $v_1',v_2',\dots, v_m'$ for $V'$ to a basis $v_1',\dots,v_m', v_1,\dots, v_n$ for $V$. Let $\{w_1,w_2,\dots, w_n\}$ be a linearly subset of $W$, and map the $v_i'$ to $0$, and $v_j$ to $w_j$, and extend by linearity. Because of the linear independence of the $w_j$, the kernel of the resulting linear transformation is $V'$.

For infinite dimensional spaces, then, assuming the Axiom of Choice, the same argument works. Since subtraction is problematical, the appropriate condition is that $\dim(V/V') \le \dim(W)$.

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But I set the condition DimW=DimV-DimV', and DimW=2, DimV'=0, and DimV=3, so 2-0 $\neq$ 3 –  Jay K Apr 11 '12 at 23:05
    
@JayK Maybe it would be best to modify the initial statement of the result? It seems like you want the map $V \to W$ to be surjective, among other things. –  Dylan Moreland Apr 11 '12 at 23:08
    
@Jay K: Sorry, I typed my reply immediately after reading the statement of the theorem. –  André Nicolas Apr 11 '12 at 23:11
    
@Dylan Moreland: You're right, I guess I should state that $r>(m-n)$ to allow for the possibility of maps being non-surjective. –  Jay K Apr 11 '12 at 23:15
    
O.K, I edited the initial paragraph. Is this O.K now? –  Jay K Apr 11 '12 at 23:18

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