# Notation on isomorphism showing that a vector space is isomorphic to its double dual.

Let $$V$$ be a finite dimensional vector space, and let $$V^{\prime\prime}$$ be the double dual space.

From S. Axler's linear algebra text, he defines a function $$\Lambda : V \rightarrow V^{\prime\prime}$$ by $$(\Lambda v)(\varphi) = \varphi(v).$$

But I cannot make heads nor tails of this definition. I am confused on both sides of the equation.

1. For the LHS, $$\Lambda v$$ makes sense because that is what I would expect to appear when defining a function; for example, let $$f(x) = \text{whatever}$$. But then we tack on $$\varphi$$, which I know is in $$V^{\prime}$$; but if the LHS is supposed to deal with the domain $$V$$, then I don't understand how it is valid to juxtapose this $$\varphi$$.
2. The RHS is also concerning because $$\varphi(v)$$ is a scalar. Assuming that $$\varphi(v)$$ is supposed to be the output of $$\Lambda,$$ it should follow that $$\varphi(v) \in V^{\prime}$$.

I appreciate all clarification and insight concerning this notation and, perhaps, an explanation as to why such notation is valid (I think this last bit will follow once clarification is provided).

Thank you =)

Lets write out the map in question in more detail. It is $$\begin{array}{rcl} \Lambda: V & \longrightarrow & V‘‘=\operatorname{Hom}(\operatorname{Hom}(V,K),K)\\ v & \longmapsto & \left(\begin{array}{rcl} \operatorname{Hom}(V,K) & \xrightarrow{\Lambda v} & K\\ \varphi & \mapsto & \varphi(v) \end{array}\right) \end{array}$$

So we map a vector $$v$$ to a map $$\Lambda v$$, which takes in maps $$\varphi$$ and evaluates them at $$v$$. As a function is determined on its values the shortest form to write this is the equation $$(\Lambda v)(\varphi) = \varphi(v)$$ but with this it is hard to keep the overview over which map does what...

• Thank you for the clear explanation! The way you presented the mapping from the domain to the codomain, as well as what the mapping does on vectors helped tremendously. Commented Jan 30, 2021 at 21:47
• You are very welcome :) Commented Jan 30, 2021 at 23:25

For each $$v \in V$$ define a map $$\Lambda v : V' \to \mathbf F$$ whose value in a functional $$\varphi \in V'$$ is $$(\Lambda v)(\varphi) = \varphi(v)$$. Then it is easy to see that $$\Lambda v$$ is linear$$^1$$, and then $$\Lambda v$$ is an element of $$V'' := (V')'$$. Thus, $$\Lambda$$ is a map from $$V$$ to $$V''$$ that sends each $$v \in V$$ to the map $$\Lambda v \in V''$$.

Hope this helps :)

$$^1$$By definition of the operations in $$V'$$ we have that for any $$\varphi_1,\varphi_2 \in V'$$ and $$\lambda \in \mathbf F$$, $$(\lambda \varphi_1+\varphi_2)(v) = \lambda \varphi_1(v)+\varphi_2(v)$$ that is, $$(\Lambda v)(\lambda \varphi_1+\varphi_2) = \lambda (\Lambda v)(\varphi_1)+(\Lambda v)(\varphi_2)$$.

• Thank you for the clear explanation! Commented Jan 30, 2021 at 21:45

Here, $$v$$ is some vector and $$\Lambda v$$ is in $$V''$$, meaning that it sends some element of $$V'$$ (which is called $$\varphi$$ here) to a scalar. This is exactly what is happening on the left hand side: $$\Lambda v$$ sends $$\varphi$$ to a scalar.

On the right hand side, you already noted correctly that we see $$\varphi$$ sending the vector $$v$$ to a scalar.

Since we are talking about a fixed vector $$v$$, and the functional $$\varphi$$ is just a variable, this relation will define $$\Lambda$$ uniquely