$\hom(V,W)$ is canonic isomorph to $\hom(W^*, V^*)$

Solution 1:

First notice that if $E$ is a linear space then its dual $E^*$ is the linear space of the linear forms $$\varphi: E\rightarrow \Bbb K$$

Now if $\Psi\in \operatorname{hom}(V,W)$ i.e. a linear transformation $\Psi:V\rightarrow W$ then $\Phi(\Psi)\in \operatorname{hom}(W^*,V^*)$ since $$\underbrace{\Phi(\Psi)(\varphi)}_{\in V^*}=\varphi\circ\Psi:V\xrightarrow{\Psi} W\xrightarrow{\varphi} \Bbb K\quad \forall \varphi\in W^*$$ and we prove easily that $\Phi$ is a linear transformation: $$\Phi(\Psi)(\varphi+\lambda\psi)=\Phi(\Psi)(\varphi)+\lambda\Phi(\Psi)(\psi)$$ and if $\Psi\ne0$ hence there's $x\in V$ such that $W\ni\Psi(x)\ne0$ hence with $\varphi=(\Psi(x))^*$ the linear form such that $$(\Psi(x))^*(\Psi(x))=1$$ we see that $$\Phi(\Psi)(\varphi)\ne0$$ hence $\Phi$ is injective, moreover, since $$\dim \operatorname{hom}(V,W)=\dim V\times\dim W=\dim V^*\times\dim W^*=\dim\operatorname{hom}(W^*,V^*)$$ we see that $\Phi$ is bijective.

Solution 2:

IMO, the only hard part about this problem is wrapping your head around the notion of a function-valued function. Or worse, this problem is about a function-valued-function-valued function of functions.

When I first started learning about these things, I managed by writing with a very precise syntax, so I could see exactly what type everything is, and exactly what substitutions I could make, and I would introduce variables so that I could work pointwise as much as possible.

In this case, we can introduce variables as

• $\Psi$ is a variable denoting a map $V \to W$
• $\varphi$ is a variable denoting a functional on $W$
• $v$ is a variable denoting a vector on $v$

and we can type things as:

• $\Phi : \hom(V, W) \to \hom(W^*, V*)$
• $\Phi(\Psi) : W^* \to V^*$
• $\Phi(\Psi)(\varphi) \in V^*$
• $\Phi(\Psi)(\varphi)(v) \in \mathbf{K}$.

The definition you cited is

$$ \Phi(\Psi)(\varphi)(v) := \varphi(\Psi(v)) $$

Once you're used to these things, it turns out this is "obvious", since the right hand side is pretty much the only way you can combine the three symbols $\varphi$, $\Psi$, and $v$ in the same expression, without mixing in other stuff.

Now, for example, to show that $\Phi$ is additive, we need to show

$$ \Phi(\Psi + \Psi') = \Phi(\Psi) + \Phi(\Psi') $$

but since equality and addition of functions is defined pointwise, this is the same thing as showing

$$ \Phi(\Psi + \Psi')(\varphi)(v) = \Phi(\Psi)(\varphi)(v) + \Phi(\Psi')(\varphi)(v) $$

which can be seen after making the substitution.

Once you're comfortable doing everything pointwise, you can start thinking in terms of the functions themselves... but every now and then I still find it useful to go back to this approach to things when problems get complicated.

Of course, you weren't asked to show that $\Phi$ was additive, but it is, and it's a good thing to show. Everything else you might want to prove goes in a similar way.