When I have two vector spaces $W, V$ over $k$ a field. And I know that the algebraic dual spaces of $V$ and $W$ are isomorphic. Can I conclude, (in the infinite dimensional case) that $V$ and $W$ are isomorphic?
I am saying algebraic dual, because i don't want to be confused with the dual of continuous linear functionals. But I am just using the definition of dual space, that everyone uses in linear algebra.
Say I know, that I have a linear map $\varphi: V \to W$ such that its dual map is an isomorphism, can I then conclude it was an isomorphism all along? I know this is true if $\varphi$ is assumed to be injective, because then the injectivity of the dual map, implies surjectivity of the original map. Can I say something, if my map $\varphi$ was assumed to be surjective and the dual map is an isomorphism?
Best Answer
Yes. If $\phi^\lor$ is an isomorphism, then $\phi$ is an isomorphism.
The most straightforward way of seeing this is through category theory. We must show that the dual functor is faithful. That is, if we have vector spaces $W, V$ and linear maps $f, g : W \to V$ such that $f^\lor = g^\lor$, then $f = g$.
Indeed, suppose we have such $W, V, f, g$, and consider some $x \in W$. Suppose $f(x) \neq g(x)$. Then take some basis $B$ of $V$ such that $f(x) - g(x) \in B$, and consider the unique linear map $h : V \to k$ such that for all $b \in B$,
$$h(b) = \begin{cases} 1 & b = f(x) - g(x) \\ 0 & otherwise \end{cases}$$
Then $h \circ f = f^\lor(h) = g^\lor(h) = h \circ g$, so in particular, $h(f(x)) = h(g(x))$. So $h(f(x) - g(x)) = 0$; contradiction.
Now that we’ve established the dual functor is faithful, we note that the category of vector spaces is balanced - a linear map which is both a monomorphism and an epimorphism is an isomorphism. It follows from faithfulness that since $\phi^\lor$ is an isomorphism, it must be an epimorphism, and hence $\phi$ is a monomorphism. Similarly, it follows that $\phi^\lor$ is a monomorphism, and hence $\phi$ is an epimorphism. Therefore, $\phi$ is an isomorphism.
To replicate the previous paragraph without category theory, we first show $\phi : V \to W$ is injective. Consider some $x \in V$ such that $\phi(x) = 0$. Then consider the unique linear map $f : k \to V$ such that $f(1) = x$. We see that $\phi \circ f = 0$, so $f^\lor \circ \phi^\lor = (\phi \circ f)^\lor = 0^\lor = 0$. Take the equation $f^\lor \circ \phi^\lor = 0$ and compose $(\phi^\lor)^{-1}$ on the right to get $f^\lor = 0 = 0^\lor$. It follows from faithfulness that $f = 0$, so $x = f(1) = 0$. Thus, $\phi$ is injective. You said in your question that you can take it from there.
Finally, this argument is not sufficient to show that if $V^\lor$ is isomorphic to $W^\lor$, then $V \cong W$. In general, dual spaces are far larger than their underlying spaces, and there are many maps between them that don’t arise as a dual map. I don’t know whether this implication holds, but I have a strong sense it doesn’t (or may even be independent from ZFC). An analogous question arises in set theory: if there is a bijection between $P(S)$ and $P(R)$, is there a bijection between $S$ and $R$? It turns out the question cannot be answered in ZFC.
But I think it’s likely there’s a well-known answer somewhere. You just have to look harder.