I am doing the exercise in Linear Algebra Done Right Chapter 3.E.

  1. Suppose $V_1, ...,V_m$ are vector spaces. Prove that $\mathcal{L}(V_1 \times ... \times V_m, W)$ and $\mathcal{L}(V_1, W) \times...\times \mathcal{L}(V_m, W)$ are isomorphic vector spaces.

Can I say they are isomorphic because their dimensions are both $\text{dim } W \times (\text{dim }V_1 + ... + \text{dim }V_m)$

  1. Suppose $W_1, ..., W_m$ are vector spaces. Prove that $\mathcal{L}(V, W_1 \times ... \times W_m)$ and $\mathcal{L}(V, W_1) \times ... \times \mathcal{L}(V, W_m)$ are isomorphic vector spaces.

Can I use the same argument that they have the same dimension?

  1. For $n$ a positive integer, define $V^{n}$ by $V^n = V \times ...\times V$. Prove that $V^n$ and $\mathcal{L}(\mathbb{F}^n, V)$

Can I say that since the dimensions are both $n \times \text{dim }V$, therefore, they are isomorphic?

The reason I ask these questions is that in the book, it said:

Two finite-dimensional vector spaces over $\mathbb{F}$ are isomorphic if and only if they have the same dimension.

and I found the online solutions used different method.

  • $\begingroup$ notice how the book said "two FINITE-DIMENSIONAL..." so, when the spaces $V_1, \dots V_m,W$ are all finite-dimensional, then yes you can make use of dimension arguments. But note that you can have infinite-dimensional vector spaces which are not isomorphic, so your dimension argument wouldn't work in this case. However, what you're being asked holds in general, even without any finite-dimensionality assumptions, hence to show the spaces are isomorphic, you will have to explicitly construct an isomorphism. $\endgroup$
    – peek-a-boo
    Jun 23, 2019 at 7:22
  • $\begingroup$ If you are concerned about the case when $V_i$ are finite dimensional, then sure you can argue using dimensions. But there are more to the story. When two vector spaces are of the same dimension, they are isomorphic but in general there are many different isomorphisms. But in the question, they are "canonically" isomorphic, in the sense that there is an explicit way to obtain/write down an isomorphism, no matter what $V_i$ are. $\endgroup$
    – lEm
    Jun 23, 2019 at 9:21

1 Answer 1


Technically, yes, it is a valid proof, if the vector spaces are assumed to be finite dimensional (which they don't seem to be), and the principle you quoted is correct.

However the results work for infinite dimensions too, and the dimension counts you made are no longer valid, so without the finite-dimensionality argument you need another proof.

Let me try to convince you, thoigh, that even if the vector spaces were assumed to be finite dimensional, you should do another proof.

Indeed, a very important first point is that this proof doesn't give you an explicit isomorphism, and in the examples you gave, there actually are very interesting explicit isomorphisms. They are interesting for a bunch of reasons :

First of all, they are explicit, and having explicit isomorphisms is always a good thing in practice. Here, one is often brought to identify, say $L(V\times W, E)$ with $L(V,E)\times L(W,E)$ and to make this identification you need the specific isomorphism.

Second of all, the explicit isomorphisms I am thinking of (which are almost surely the ones the author had in mind) are very nice, in that they're "natural". We could get into the technicalities to know what this means precisely, but essentially it means they behave well with the whole of linear algebra. By that I mean that for instance, regarding the last one, if you have two vector spaces $V,W$ and identify $V^n$ and $L(\mathbb F^n, V)$ via that isomorphism, and $W^n$ and $L(\mathbb F^n, W)$ via that isomorphism too, then nothing “goes wrong“ when you look at maps between $V$ and $W$. You can try to fiddle around with it a bit to see what I mean.

And lastly, note that although the course only mentions the principle for finite-dimensional spaces, the isomorphisms I have in mind work for infinite-dimensional spaces too, and the proof remains unchanged. That is often something that happens : doing proofs with dimensions is often more restrictive than doing "synthetic" proofs, and so the latter is better. Indeed, it often happens that there is extra structure on your spaces, and the dimension proofs don't see that at all, whereas proofs that actually exhibit the isomorphism are more likely to help you with that extra structure.

All of this is very common in linear algebra and other disciplines : students tend to use bases, dimensions, charts in differential geometry etc. whenever they have the opportunity, whereas it's often much better, for various reasons, to stay at the "synthetic" level, and the gain is often huge, although sometimes hard to see for students.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.