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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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    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!!!!2012-04-12

2 Answers 2

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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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    O.K, I edited the initial paragraph. Is this O.K now?2012-04-11