Proving the transpose / dual map is well defined.

Question

The definition for a dual map is as follows:

The dual map, or transpose of linear $f:V \rightarrow W $ is given by $f^t(g)(v) = g(f(v)) $ for $\forall g \in W^* , v \in V $.

In my lecture notes, I have the following proof to show that the definition is well defined:

$f^t (g)(a_1v_1 + a_2v_2) = g(f(a_1v_1 + a_2v_2)) $

$ = ag(f(v_1)) + a_2 g(f(v_2))$

$= a_1 f^t(g)(v_1) + a_2f^t(g)(v_2)$ and so $f^t(g) \in V^* $.

How does this last step show $f^t(g) \in V^* $? How can I get my head around this proof?

My understanding is that $g(f(v))$ takes in an element of $V$, applies $f$ to obtain an element of $W$ and then the functional $g$ to produce an element of the field ($K$), ultimately going from $V \rightarrow K$. Thus the function itself is an element of $V^*$. How is this intuition shown in the proof above?

Answer 1

You want to show:

1. $f^t$ is a map from $W^*$ to $V^*$

2. $f^t$ is linear

For $g\in W^*$, you want to define an element of $V^*$, that is, a linear map $V\to K$. The definition is quite natural: $$ f^t(g)\colon v\mapsto g(f(v)) $$ Clearly, $f^t(g)$, as a map $V\to K$, is linear, because it's just $g\circ f$.

I see no reason for doing that complicated proof, which is just that the composition of linear maps is linear.

More interesting is showing that $f^t$ is linear. If $g_1,g_2\in W^*$ and $v\in V$, we have $$ f^t(a_1g_1+a_2g_2)\colon v\mapsto (a_1g_1+a_2g_2)(f(v)) $$ Now, $$ (a_1g_1+a_2g_2)(f(v))=a_1g_1(f(v))+a_2g_2(f(v)) $$ by definition. On the other hand $$ a_1f^t(g_1)+a_2f^t(g_2)\colon v\mapsto a_1f^t(g_1)(v)+a_2f^t(g_2)= a_1g_1(f(v))+a_2g_2(f(v)) $$ which is what was desired.

Answer 2

Given any function $f \colon V \rightarrow W$ (not necessarily linear), you can define a linear map $f^{t} \colon \mathrm{Map}(W,K) \rightarrow \mathrm{Map}(V,K)$ from the vector space of all (not necessarily linear) maps $W \rightarrow K$ to the vector space of all maps $V \rightarrow K$. Inside the vector space of all maps $\mathrm{Map}(W,K)$ you have the subspace $\mathrm{Hom}_{K}(W,K) = W^{*}$ of all linear maps $W \rightarrow K$, and similarly for $V$. Thus, you can always restrict $f^t$ and consider $f^{t}|_{W^{*}} \colon W^{*} \rightarrow \mathrm{Map}(V,K)$.

The point is that if $f \colon V \rightarrow W$ itself is linear and $g \in W^{*}$, then $f^{t}(g) = g \circ f$ is not just any map in $\mathrm{Map}(V,K)$ but it must be a linear map and so you can also restrict the range and obtain a well-defined linear map $f^{t} \colon W^{*} \rightarrow V^{*}$. To check this, you need to verify that $g \circ f$ is linear and this is what the proof does.

More generally, you can show that the composition of two linear maps is always linear and thus $f^{t}(g) = g \circ f$ is linear.