Existence of endomorphism $f$ such that $b(v,\phi)=\phi(f(v))$ for bilinear map $b$

Question

Let $V$ be a finite dimensional vector space over a field $F$ and let $b:V\times V^*\rightarrow F$ be a bilinear map (with $V^*$ the dual of $V$). How do I show that there exists an endomorphism $f$ of $V$ such that for all $v\in V$ and for all $\phi\in V^*$ we get $$b(v,\phi)=\phi(f(v))?$$ I don't really understand how we know what $b$ is, when $b$ is a random bilinear map. I would appreciate it if you could show me the proof of the above so I can understand it better. Thanks in advance.

Answer 1

It is a standard technique to characterize a bilinear pairing $b: V \times W \to F$, for any vector spaces $V$ and $W$, by a linear transformation

$$g: V \to W^*$$

$$v \mapsto \{w \mapsto b(v, w)\}.$$

That is, for $v \in V$, $g(v)$ is the element of $W^*$ which carries $w \in W$ to the scalar element $b(v, w) \in F$, and of course, it needs to be verified that, for each $v \in V$, $g(v)$ defines a linear transformation $W \to F$, and that $g$ itself is linear. Both of these come down to the bilinearity condition on $b$.

Now when $W = V^*$, as in this problem, we get

$$g: V \to (V^*)^*$$

$$v \mapsto \{\phi \mapsto b(v, \phi)\}.$$

The definition of $g$ can be characterized by the fact that, for $v \in V$, $\phi \in V^*$, we have $b(v, \phi) = [g(v)](\phi)$.

Now this linear transformation $g$ isn't quite what your problem asks for, but I claim that it is almost what your problem asks for; the problem is that $(V^*)^*$ is the wrong vector space here. What we need is $V$ instead of $(V^*)^*$ there, but those two vector spaces are actually isomorphic, in a manner which doesn't require choosing a basis for $V$. I'll define this isomorphism for you: $$\psi: V \to (V^*)^*$$ is given by $$v \mapsto \{\theta \mapsto \theta(v)\}.$$ Things for you to verify, if you haven't seen this map before:

• The definition of $\psi$ makes sense, and is linear (that is, $\psi$ does indeed take elements of $V$ to elements of $(V^*)^*$ and $\psi$ is a linear transformation).
• When $V$ is finite dimensional, as is assumed in this problem, $\psi$ is an isomorphism, so $\psi^{-1}: (V^*)^* \to V$ is well-defined.

Note that I've reversed my notation from my comment, because it is easier (read: possible) to define $\psi$ this way.

Now we put these two linear maps together to define $$f := \psi^{-1} \circ g.$$ This is certainly a linear transformation $V \to V$, i.e., an endomorphism of $V$. Now we're left to verify that $f$ satisfies the condition asked for in the problem. This will simply come down to unwrapping the definition of $\psi$ and $g$.

Let $v \in V, \phi \in V^*$, and consider $w := f(v) \in V$, that is, $w = \psi^{-1}(g(v))$. Remember that what we want to show is that $b(v, \phi) = \phi(f(v))$, i.e., $b(v, \phi) = \phi(w)$.

We can rewrite $w = \psi^{-1}(g(v))$ as $\psi(w) = g(v)$. The characterization of $g$ was that we have $b(v, \phi) = [g(v)](\phi)$; substituting $\psi(w)$ in, we have $b(v, \phi) = [\psi(w)](\phi)$. But $[\psi(w)](\phi) = \phi(w)$ by definition of $\psi$, so we have $b(v, \phi) = \phi(w)$. Since $w$ was defined as $f(v)$, this verifies that

$$b(v, \phi) = \phi(f(v)) \text{ for all } v \in V, \phi \in V^*,$$

and we're done.