If vector spaces $V$ and $W$ are isomorphic, are their dual spaces isomorphic?

Question

If $V$ and $W$ are two isomorphic vector spaces (they may not be finite-dimensional), is it always true that their dual spaces $V^*$ and $W^*$ are isomorphic?

If so, please give a proof. Otherwise, please give a counterexample.

Answer 1

Yes, duals of isomorphic spaces are isomorphic. This follows, for example, from the fact that the dual space construction forms a contravariant functor.

For a linear operator $A: V \rightarrow W$ we define its transpose (or dual) $A^*: W^* \rightarrow V^*$ by $A^*f = (x \mapsto f(Ax))$. Here $x \in V$, $Ax \in W$, $f \in W^*$ and $A^*f \in V^*$.

Forming transpose preserves composition. That is, for $A: V \rightarrow W$ and $B : W \rightarrow U$, we have $(BA)^* = A^* B^*$ (note the order!).

Furthemore, transpose preserves identity operators. Define the identity operator $\operatorname{id}_V : V \rightarrow V$ by $\operatorname{id}_V(x)=x$. Then, $\operatorname{id}_V^* = \operatorname{id}_{V^*}$.

Now, if two vector spaces are isomorphic, that is, there exist $A: V \rightarrow W$ and $A^{-1} : W \rightarrow V$, where $A A^{-1} = \operatorname{id}_W$ and $A^{-1} A = \operatorname{id}_V$, then, using the previous results, we have $A^* : W^* \rightarrow V^*$, $(A^{-1})^*: V^* \rightarrow W^*$, $A^* (A^{-1})^* = \operatorname{id}_{V^*}$ and $(A^{-1})^* A^* = \operatorname{id}_{W^*}$. This means that $A^*$ and $(A^{-1})^*$ are inverses to each other. In particular, they are bijections, thus, isomorphisms.