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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?2012-04-11
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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.2012-04-11
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    No biggie, that's nothing to worry about. Fortunately many here will help out with the editing process.2012-04-11
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    Thanks, Mr. Moreland, do you refer to my initial paragraph?2012-04-11
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    I think your idea is sound, but proving that it works might have to go differently. Certainly you don't want to talk about the kernel of a map out of $V - V'$, since that isn't a vector space. I guess the point is that if you have an element of $V - V'$ then it's a sum $\sum_{i = 1}^m a_iv_i'$ where some $a_i$ is non-zero for an $i \geq n + 1$. You apply $T$ and get $\sum_{i = n + 1}^m a_iw_{i - n}$, and you argue that this is non-zero.2012-04-11
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    I see, so maybe I could instead refer to the linear map defined on the subspace generated by $B_{V-V'}$?2012-04-11
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    Thanks; does the (new and improved!--by all) idea extend to the infinite-dimensional case?2012-04-11
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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.2012-04-11
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    Dylan: thanks; I guess re your first suggestion, that if $\sigma a_iv_i'$ (my apologies, I don't know how to do upper- and lower- indices for the sum), then this would be a non-zero linear combination which is equal to zero, contradicting that $w_i$ is a basis. Is that it?2012-04-11
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    @JayK Sounds good to me.2012-04-11
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    @Dylan: thanks for the help.2012-04-11
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    @Jay K: Now that the condition is part of the statement of the theorem, everything is fine. You don't need $<$, things are OK with $\le$.2012-04-11
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    Thanks, both. Sorry to insist on this, but, can we extend the idea to the infinite-dimensional case?2012-04-11
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    @JayK For infinite-dimensional vector spaces, the natural way to do it would be to use Dylan's suggestion: $V'$ is the kernel of the quotient map $V/V'$ (but if you don't know about quotient spaces, that's not much help. If you want to uses bases, you have to be careful (in particular, you will need the axiom of choice, as noted below by André Nicolas).2012-04-12
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    And by the way, I fixed some of the Latex. If you want to know what command I used, just look at the editing history.2012-04-12
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    Thanks all. I will look into the editing.2012-04-12
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    Actually, I do know a little about quotient spaces: given V' subspace of V, w,w' in V, then w~w' iff (def.) w-w' is in V'. So for V' the kernel of a map, we collapse elements with same image. Can you suggest a method/source for quotients, please?2012-04-12
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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

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