Linear map between duals induced by linear maps between vector spaces

Question

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!

Answer 1

Hint:

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

Answer 2

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.

Answer 3

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?

Answer 4

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.