4
$\begingroup$

Assume $T'\in \mathcal{L}(W',V')$ is an isomorphism. Also, assume that $V$ and $W$ are not finite dimensional.

My professor is asserting that if $T'$ is an isomorphism, then it can be shown that so is $T$. All of the theorems available to me in my textbook speak ONLY of finite vector spaces, and a seperate but related question I asked here got lots of comments essentially to the affect "there is no way to tell".

This is for an assignment, and our answers are supposed to come from theorems we have learned throughout the course. I don't see how to get to the end result using what I know (only finite vector spaces).

$\endgroup$
4
  • $\begingroup$ Are $V$ and $W$ Hilbert spaces? $\endgroup$
    – Michael L.
    May 2, 2017 at 17:19
  • $\begingroup$ I doubt it, since it wasn't mentioned in class ever. This is a basic linear algebra course for first year graduate students. $\endgroup$ May 2, 2017 at 17:20
  • $\begingroup$ Can you give us some more details on $V$ and $W$ then? Are they inner product spaces, at least? Banach spaces? Those two paradigms have different definitions for the adjoint $T'$. $\endgroup$
    – Michael L.
    May 2, 2017 at 17:24
  • $\begingroup$ The question as I posed it provides all the details that I was given. This is the dilemma I am running into. I can show it to be true if I get to work with finite dimensions, but my professor says that he didn't leave that out on accident, and that I can still show it to be true regardless of dimension. Plus, we never went over special spaces and just brushed inner product spaces on the last couple days of class. We never went over adjoints, either. I recognize the word from undergraduate applied linear algebra, but have no theoretical knowledge of them. Again, this class was very elementary. $\endgroup$ May 2, 2017 at 17:30

1 Answer 1

3
$\begingroup$

Suppose $T$ is not an isomorphism. Then either $T$ is not injective or is not surjective. If $T$ is not injective, let $v\neq 0$ be such that $T(v)=0$ and let $\alpha\in V'$ be such that $\alpha(v)\neq 0$. Note that for any $\beta\in W'$, $(T'\beta')(v)=\beta(Tv)=\beta(0)=0$, so $T'\beta'\neq\alpha$. Thus $T'$ is not surjective, and so is not an isomorphism.

Now suppose $T$ is not surjective. Then the quotient space $W/T(V)$ is nontrivial, so there exists a nonzero linear map $\alpha:W/T(V)\to k$ (where $k$ is the scalar field). Let $q:W\to W/T(V)$ be the quotient map and let $\beta\in W'$ be given by $\beta(w)=\alpha(q(w))$. Since $\alpha\neq 0$ and $q$ is surjective, $\beta\neq 0$. For any $v\in V$, $q(T(v))=0$, so $\beta(T(v))=\alpha(q(T(v)))=0$. Thus $T'(\beta)=0$. Since $\beta\neq 0$, $T'$ is not injective, and so is not an isomorphism.

$\endgroup$
2
  • $\begingroup$ So, do I follow you correctly by saying that the contrapositive of your proof is that if $T'$ is surjective then $T$ is injective and that if $T'$ is injective, then $T$ is surjective, therefore if $T'$ is a bijection, so is $T$ and is therefore $T$ is an isomorphism? $\endgroup$ May 3, 2017 at 16:11
  • $\begingroup$ That's correct. $\endgroup$ May 3, 2017 at 16:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .