[Math] Double dual is isomorphic to the vector space – problem with injectivity

Question

This is my proof that $V$ is isomorphic to $V^{**}$:

When I say: "Since this is valid for any $w\in V^*$, then $v=0$", is it correct? I do not see why. I wrote it "just because", expecting to figure it out later.

Answer 1

What you want is the following lemma:

If $v \in V$ is such that for all $w \in V^*$, one has $w(v) = 0$, then $v = 0$.

To prove the lemma, it's easier to prove the contrapositive: if $v \neq 0$, we can explicitly build a function $w \in V^*$ that doesn't vanish on $v$. To do this, note that since $v \neq 0$, the set $\{v\}$ is linearly independent, so, by a result you have hopefully already seen in the course, it can be extended to a basis $\{v, v_1, v_2, \ldots, v_{n-1}\}$.

Now there is a unique $w \in V^*$ with $w(v) = 1$ and $w(v_i) = 0$ for the other basis vectors. (It is uniquely determined from these rules by linearity, since the above set is a basis.) In particular, $w$ doesn't vanish on $v$, as desired.