5

Is the following statement true?

Conjecture. If two vector spaces $V$ and $W$ are isomorphic and $V$ is finite dimensional (F-D) then $W$ is finite dimensional. Furthermore, $\text{dim} V = \text{dim} W$.

and if YES, then how it can be proved?

I can prove the following which is slightly different from the conjecture

Theorem. Two finite dimensional vector spaces $V$ and $W$ are isomorphic if and only if they have the same dimension.

but it does not help to deduce the conjecture from it. However, I strongly feel that the conjecture should be true.

I would be thankful if you provide a hint. :)

Motivation of the Question

While I was reading Linear Algebra Done Right by Sheldon Axler I encountered this theorem

enter image description here

However, I was not able to prove the red underlined part by the references made! It seems that a little point was overlooked by the author! The references are

enter image description here enter image description here enter image description here

where the notations used are

  1. $\Bbb{F}$ is the field $\Bbb{R}$ or $\Bbb{C}$.
  2. $\Bbb{F}^{m,n}$ is the vector space of $m \times n$ matrices.
  3. $\mathcal{L}(V,W)$ is the vector space of linear maps from $V$ to $W$.
  4. $\mathcal{M}$ is a linear map from $\mathcal{L}(V,W)$ to $\Bbb{F}^{m,n}$ which gives the corresponding matrix of a linear map belonging to $\mathcal{L}(V,W)$.
Hosein Rahnama
  • 14,882
  • 2
    Yes. Pick a basis for $V$ and carry it through the isomorphism. Use the fact that it is an isomorphism to show that it is a basis of $W$. Surjectivity gives that it generates, and injectivity gives that it is linearly independent. – Aloizio Macedo Aug 08 '16 at 18:45
  • @AloizioMacedo: That was far from a hint! ;) Thanks. :) Would you please write an answer so that I can accept it. :) – Hosein Rahnama Aug 08 '16 at 18:47
  • Gladly : ).${}{}{}{}$ – Aloizio Macedo Aug 08 '16 at 18:47
  • @celtschk: The theorem assumes in advance that $V$ is finite dimensional while I want to prove it. :) In fact, when you write $\dim V=\dim W$ you assume that both $V$ and $W$ are finite dimensional. – Hosein Rahnama Aug 08 '16 at 18:55
  • 1
    @H.R.: Ah, OK, I've overlooked the "finite-dimensional" in the theorem. But why do I assume in the equation that they are finite-dimensional? You can also compare infinite cardinalities. – celtschk Aug 08 '16 at 18:59
  • 1
    The dimension of a space is defined in the language of vector spaces. Thus it is preserved by isomorphism. – Jacob Wakem Aug 08 '16 at 19:16
  • @SheldonAxler: Please take a look at this. :) – Hosein Rahnama Aug 08 '16 at 19:19
  • @celtschk: I am studying Linear Algebra Done Right and cardinality is not defined in the whole book. :) The concept of dimension is just defined for the finite-dimensional vector spaces. :) I just answered you by what I had in my head from the book. :) – Hosein Rahnama Aug 08 '16 at 19:32

1 Answers1

5

Yes. Pick a basis for $V$ and carry it through the isomorphism. Use the fact that it is an isomorphism to show that it is a basis of $W$. Surjectivity gives that it generates, and injectivity gives that it is linearly independent.

In more details, let $\{e_1,\cdots, e_n\}$ be a basis for $V$, and $f$ be the isomorphism.

Let's show $\{f(e_1),\cdots, f(e_n)\}$ is a basis.

First, let's show it generates. Take $w \in W$. Since $f$ is an isomorphism, it is surjective, and there exists $v \in V$ such that $f(v)=w$.

But $v$ can be written as $v=c_1e_1+\cdots+c_ne_n$ for some scalars $c_1,\cdots, c_n$, since $\{e_1,\cdots,e_n\}$ is a basis. Then, $$w=c_1f(e_1)+\cdots+c_nf(e_n),$$ since $f$ is linear.

Let's show it is linearly independent. Let

$$k_1f(e_1)+\cdots+k_nf(e_n)=0.$$ Then, $$f(k_1e_1+\cdots+k_ne_n)=0.$$ Therefore, since $f$ is injective, $k_1e_1+\cdots+k_ne_n=0$, which gives that each $k_i$ is zero, since $\{e_1,\cdots,e_n\}$ is a basis.

Aloizio Macedo
  • 34,292
  • Excellent like always! ;) So the conjecture is a stronger form of the theorem I mentioned! :) right? – Hosein Rahnama Aug 08 '16 at 18:52
  • 1
    @H.R. I don't think it is stronger. Note that the theorem you mentioned can conclude that two vector spaces have the same dimension, whereas the conjecture (as stated) only asserts that the other one is finite-dimensional. However, the proof given shows that they must have the same dimension as well.

    OBS: Note that the proof uses only injectivity to show that the set is linearly independent, and only surjectivity to show that it generates. Hence, one has other useful facts: that an injective function takes LI sets to LI sets and a surjective takes generator sets to generator sets.

     – Aloizio Macedo Aug 08 '16 at 18:58