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

Introduction My Semester just started and we have a new Professor for Linear Algebra II (replacing our former Professor). Apparently we are behind our schedule and thus we only had a brief introduction (5 Minutes on the Blackboard) about the Dual Space and canonical isomorphisms.

In the current problem set there is this optional (not mandatory and not accredited) exercise which I understand as good as nothing about:

Problem: Let $V$ and $W$ be finite dimensional $\mathbb{K}$-Vectorspaces$^{1)}$. Show that $\hom(V,W)$ is canonical isomorph to $\hom(W^*,V^*)$ and find the canonical isomorphism.

$^{1)}$ ($\mathbb{K}$ denotes a field here, I presume in english literature they write $\mathbb{F}$)

Hint (given by my tutor): $$\Phi: \hom(V,W) \longrightarrow \hom(W^*,V^*) \\ \Psi \longmapsto (\varphi \longmapsto \varphi\circ \Psi)$$ My problems are vast now:

• The literature I am reading does a bad job at explaining this topic
• The online literature I find seems to be way beyond my level, they usually include tensor algebra to solve this. So I fall down the math rabbit hole see, On "familiarity" (or How to avoid "going down the Math Rabbit Hole"?)
• I don't understand the Hint, I did not encounter such a function before, not even in Analysis and I don't know what's happening. I can only guess that it is a function that somehow completes a circle.

On the bright side:

• I believe to understand what a canonical isomorphism is. My Professor said it is an isomorphism that does not involve making a choice for a basis. In the above function we do not make such a decision and therefore it would be a candidate for a canonical isomorphism.

• If I would understand the above function, I'd merely have to show that it is injective (trivial Kern) and surjective to complete the task. Unfortunately, as far, we've only done such things using an explicit Matrix and not a function.

• I do know the definition of the Dual Space, also using the Kronecker Delta. (although I know nothing about it's meaning and geometric intrepreation, if there is any)

• The way my tutors sell this exercise it's supposed to be very easy, once the definitions are clear.

To-Do-List:

• Understand the function, know what's happening.

• Show that the function is surjective

• Show that the function is injective (trivial Kern)

• Have you tried asking your instructor? – Mariano Suárez-Álvarez Mar 2 '14 at 18:36
• @MarianoSuárez-Alvarez, I will certainly tomorrow, this was the problem Set during the weekend and I tried to work on it on my own. I will further look into the answers below in a few and try to make more sense out of it :-) – Spaced Mar 2 '14 at 18:44

## 4 Answers

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.

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.

• thanks a lot for your detailed description on how you think (or used to think) about such problems, it helps me a lot. I am not yet at the stage where I can claim to understand this problem, but I am sure if I let all this informations sink and try again tomorrow I will already do much better. Thanks again for your help, regards. – Spaced Mar 2 '14 at 20:49

$\hom(V,W)$ is canonically isomorphic to $V^*\otimes W$.

$\hom(W^*,V^*)$ is canonically isomorphic to $(W^*)^*\otimes (V^*)$ for the same reason.

Finally, $W^{**}\otimes V^*$ is canonically isomorphic to $V^*\otimes W$.

EDIT: Dustan is right, I shouldn't have used tensor algebra for this (missed the quoted part of OP). Here the tensor algebra part in hand-waving intuitive manner:

Each element $H$ of $\hom(V,W)$ can be considered as a collection of $\dim W$ linear functionals on $V$. If you fix a basis in $V$ and $W$ you can look at $H$ as a matrix, with $i$-th row denoting a functional on $V$ corresponding to the coefficient of the $i$-th coordinate in $W$. Such matrix would make a good intuitive picture of what's going on, although one should try to rephrase w/out a basis to make things canonical.

There is one thing we need to know about dual spaces: $W^{**}$ is canonically isomorphic to $W$. Here's why: for any pair $(v\in V,w\in V^*)$ we have the value $w(v)$ that linearly depends on both $v$ and $w$. Therefore $v\in V$ can be thought of as a functional on $V^*$. You can tell by comparing dimensions that such $v$ would produce all linear functionals on $V^*$.

If we fix the basis the isomorphism you are looking for is between the spaces of $\dim(V^*)\times \dim(W)$ matrices and $\dim(W^{**})\otimes dim(V^*)$ matrices, which is given by transposition because $W^{**}=W$.

• "The online literature I find seems to be way beyond my level, they usually include tensor algebra to solve this." – Dustan Levenstein Mar 2 '14 at 18:49

The isomorphism is simply the transpose...

This is as easy as things come, really :-)

• This is not really fair. Even thinking about functions on a space of functions can be really hard for beginning students, who are only used to thinking of functions of "numbers". It is also not immediately clear that the "transpose" map does not depend on the choice of basis, which is part of what OP is worried about. I do agree that it would be sad to sort through this exercise without relating it to the transpose of a matrix. – Steven Gubkin Mar 2 '14 at 19:31
• Transpose is really the wrong analogy: in coordinates, $V,W$ are spaces of column vectors, and $V^*,W^*$ are spaces of row vectors, and the isomorphism is the identity map on matrices: it's the shift in viewpoint between thinking of a matrix $A$ as a map $v \mapsto Av$ and as a map $\omega \mapsto \omega A$. – Hurkyl Mar 2 '14 at 20:04
• @Hurkyl Not sure it is an analogy. For $f\colon V\to W$, the map ${}^tf \colon \varphi \mapsto \varphi\circ f$ is often called the transpose of $f$ (when it isn't called the dual, or adjoint of $f$). – Daniel Fischer Mar 2 '14 at 20:10
• @Daniel: Fair enough. Transpose makes sense if you need to swap things from acting on the right to acting on the left, but IMO one really should be thinking of dual vectors as being in an algebraic role analogous to row vectors. – Hurkyl Mar 2 '14 at 20:18
• @Hurkyl I never grokked (gah, that un-word) the distinction between row-vectors and column vectors. I have a (linear) map $V\to W$, I can pull back linear forms on $W$ to linear forms on $V$ with it. That pull-back is given a name. The three names I'm familiar with are transpose, dual, and adjoint. The name transpose obviously (?) comes from the fact that its matrix is the transpose of the matrix of $f$ (with respect to the dual bases) in the finite-dimensional case, but in my linear algebra course it was called transpose in general. – Daniel Fischer Mar 2 '14 at 20:27