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Let $V, W$ be vector spaces over a field $F$ and let $\psi: V \to W$. Show that $\psi$ induces a linear map $\psi^{*}: W^{*} \to V^{*}$ naturally.

Although the question asks for a naturally induced linear map, it does not seem at all that easy to me. Any suggestions will be greatly appreciated!

4 Answers4

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First, realize that the word 'natural' carries about as much meaning as the word 'obvious'. Don't let the word intimidate you.

Remember that $\psi^*:W^* \to V^*$, so if $w^* \in W^*$, then $\psi^*(w^*) \in V^*$.

Now select $v \in V$, and apply $\psi^*(w^*) $, ie, make sense of $(\psi^*(w^*))(v)$, remembering that $\psi(v) \in W$.

The only way it 'fits' together is $w^*(\psi(v))$.

Rudin uses the suggestive notation $\langle v, \psi^*(w^*) \rangle$ for $(\psi^*(w^*))(v)$.

The above then reads $\langle v, \psi^*(w^*) \rangle = \langle \psi(v), w^* \rangle $, which fits well with the Hilbert space notation.

copper.hat
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  • Within this context, the word "natural" has a very precise and, imfho, important meaning: the sought isomorphism does not depend on any bases whatsoever. This says tons when we're used to almost any isomorphism we can think of depends on choosing bases... – DonAntonio Mar 06 '13 at 17:58
  • What is natural about it? If something is deemed 'natural' in retrospect, it hardly warrants the moniker natural. I think the word choice can be bothersome/intimidating when one is learning? – copper.hat Mar 06 '13 at 18:01
  • Natural in the sense that you define on abstract objects without any reference to bases, and I can't see how it can be deemed defined "in retrospect"... – DonAntonio Mar 06 '13 at 18:05
  • I guess using natural for the abstract hints at the root of the issue... – copper.hat Mar 06 '13 at 18:09
  • Why the downvote? After 2.5 years? – copper.hat Nov 04 '16 at 02:26
  • I've no idea and I've no idea either why was I pinned: I don't usually downvote answers and much less after two and a hlaf years. – DonAntonio Nov 04 '16 at 05:12
  • What does pinned mean? I didn't think it was you :-). I'm just trying to make 100k by answering unchallenging questions :-). – copper.hat Nov 04 '16 at 05:15
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    It's when you do the @ thing and after it the name of a user. I thought that what happened here since this isn't my answer. It doesn't matter. Good luck with your efforts. – DonAntonio Nov 04 '16 at 11:42
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    @gary Can you fix the MathJax in your comment please? – copper.hat Jun 09 '24 at 22:03
  • @copper.hat: I'm right on it, apologies. – gary Jun 09 '24 at 22:04
  • @copper.hat : Why doesn't the induced map $L1: W^{} \rightarrow V^{}$ extend to the infinite-dimensional case for $W,V$ ? I understand the Algebraic dual $W^{*}$ is not isomorphic to $W$, but I don't see why this should be an obstacle. – gary Jun 09 '24 at 22:09
  • @copper.hat : Apologies for the mangled post/question. – gary Jun 09 '24 at 22:13
  • @gary I don't understand your question, where does dimension enter the above? – copper.hat Jun 09 '24 at 22:27
  • @copper.hat: I mean, the induced map where $L:V\rightarrow W $ gives rise to a map $ L': W^{} \rightarrow V^{}$ applies only when $V,W$ are finite dimensional I'm just curious as to the requirement of $V,W$ being finite dimensional. But maybe I'm wrong? Could you verify this is well-defined for infinite-dimensional Vector Spaces? – gary Jun 10 '24 at 01:33
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    @gary I'm not sure what you are asking about, perhaps this might shed some light? https://math.stackexchange.com/q/35779/27978 – copper.hat Jun 10 '24 at 03:16
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Here is a more pictorial hint:

You have $V\overset{\psi}{\to} W$ and then given $W\overset{f}{\to} F$ you want to find $V\overset{f^{\prime}}{\to} F$

Here $F$ is the ground field.

mv3
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Hint:

$$\forall\,\,f\in W^*\;\;,\;\;\psi^*(f)(v):=f(\psi(v))$$

DonAntonio
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  • Antonio: Sorry for the necro. What happens if we iterate this process, going from $V \rightarrow W$ to $W^{} \rightarrow V^{}$, to $V^{}$ to $W^{}$ , would we arrive at the initial map from $V$ to $W$? V,W both finite dimensional ? Obviously yes, if $V=W$, but is it true otherwise? – gary Jun 04 '24 at 05:42
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DonAntonio's hint is good. Here is perhaps a more intuitive explanation:

We start with a given $\psi$, which tells us a specific road going from $V$ to $W$.

The elements of $W^*$ comprise all the different routes from $W$ to $F$.

Now, your goal is to come up with a general rule (called $\psi^*$) for taking routes from $W$ to $F$ and turning them into routes from $V$ to $F$.

Do you see how to use $\psi$ to accomplish this?

Zev Chonoles
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  • Hi, sorry for the necro. Why doesn't this result of a linear map $ L: V \rightarrow W $ giving rise to a map $L':W^{} \rightarrow V^{}$ hold for the infinite-dimensional case? I understand in the infinite-dimensional case the Algebraic dual $W^{*}$ of $W$ is not isomorphic to $V$, but that shouldn't be an obstacle? – gary Jun 09 '24 at 21:58
  • I'm a bit confused by your question - it does hold in the infinite-dimensional case. – Zev Chonoles Jun 10 '24 at 02:49
  • Sorry, I meant if we iterated the induced map, we would end up with an induced map fromV$\rightarrow$ W , which isn't well defined, as neither V** not iso. to $V$, same for $W$, in the infinite-dimensional case. – gary Jul 05 '24 at 06:14
  • But maybe I'm confused here. If so, apologies, please ignore. – gary Jul 06 '24 at 03:18
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    Remember that $V^={\text{linear maps }f:V\to k}$, where $k$ is the base field. A linear map $L:V\to W$ induces a map $L':W^\to V^$ defined by $L'(f)=f\circ L$, and for the same reason $L'$ induces a map $L'':V^{}\to W^{}$. This is all perfectly well-defined, regardless of dimension or whether $V^{*}$ is isomorphic to $V$ or not. – Zev Chonoles Aug 07 '24 at 12:45