A problem from Linear Algebra Done Right (Third Ed): Suppose $W$ is finite dimensional and $T \in \mathcal{L}(V, W)$. Prove that $T=0$ if and only if its dual $T' \in \mathcal{L}(W', V')=0$.
I am confused by the information that $W$ is finite dimensional. Why is that required, while nothing is said about finite dimensionality of $V$ ?
There is a solution here: Does that fact that the dual map is zero imply that the map is zero? but I am more interested in the reason for the assumptions.
I was able to solve it (or so I thought) without using a contrapositive argument or the assumption on $W$ which is what worries me.
EDIT: Here is my argument. Since T' = 0, we get $\phi_w(Tv) = 0 \forall \phi_w$, which yields Tv = 0 for all v, hence T = 0. I can "reverse" the argument to get the converse. What is my error?
Note that the statement is correct also for infinite-dimensional $W$, as the key point is the
(This follows from the fact that if $w \ne 0$, we can extend $\{w\}$ to a basis $B$ of $W$ and define $f$ on the basis $B$ by $f(w) = 1$ and $f(x) = 0$ for $x \in B \setminus \{w\}$, and extend by linearity. Then $f \in W'$ and $f(w) \ne 0$).
But this argument relies on the axiom of choice (this is always needed if we want to construct bases on infinite-dimensional vector spaces), and may be wrong if the axiom of choice is not to be used. As far as I know, in the mentioned book, the basis extension theorem is not proved for infinite-dimensional vector spaces, and hence cannot be used.
As we do not need a basis of $V$ in our argument, not dimensionality assumption if needed for $V$.