[Math] If two vector spaces $V$ and $W$ are isomorphic and $V$ is F-D then $W$ is F-D. Furthermore, $\text{dim} V = \text{dim} W$.

linear algebra

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

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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

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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 $\mathcal{L}(V,W)$.

Best Answer

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.

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