#math/linear-algebra #flashcards/math Suppose $T\in \mathcal{L}(V,W)$. The `dual map` of $T$ is the linear map %%::%% $T'\in \mathcal{L}(W',V')$, where $W'=\mathcal{L}(W,\mathbf{F})$ and $V'=\mathcal{L}(V,\mathbf{F})$, defined for each $\varphi \in W'$ by $ T'(\varphi)=\varphi \circ T. $ $ T'(\varphi) =\varphi \circ T: V \xrightarrow{T} W \xrightarrow{\varphi} \mathbf{F} $ So, $T'(\varphi):V\to \mathbf{F}$ is a linear functional, hence $T'(\varphi)\in \mathcal{L}(V,\mathbf{F})=V'$. ## Example: Dual Map of Differentiation Linear Map Let $D:\mathcal{P}(\mathbf{R})\to \mathcal{P}(\mathbf{R})$ be the differentiation linear map where $D':\mathcal{P}(\mathbf{R})'\to \mathcal{P}(\mathbf{R})'$ is the dual map of $D$. Let $p'$ denote the derivative of polynomial $p$. We define $D$ by $Dp=p'$. Now, suppose $\varphi \in \mathcal{P}(\mathbf{R})'$ is the linear functional on $\mathcal{P}(\mathbf{R})$ defined by $\varphi(p)=p(3)\in \mathbf{F}$, i.e. the polynomial $p$ evaluated at $3$. With these we have the following: 1. $D:\mathcal{P}(\mathbf{R})\to \mathcal{P}(\mathbf{R})$; a function mapping polynomial to polynomial. Domain and codomain are the set of all polynomials over the reals. 2. $D':\mathcal{P}(\mathbf{R})'\to \mathcal{P}(\mathbf{R})'$; a function mapping a linear functional on $\mathcal{P}(\mathbf{R})$ to a linear functional on $\mathcal{P}(\mathbf{R})$. Domain and codomain are the set of all linear functionals. 3. $\varphi \in \mathcal{P}(\mathbf{R})'$; a linear functional $\mathcal{P}(\mathbf{R})\to \mathbf{F}$. 4. $D'(\varphi)=\varphi \circ D\in \mathcal{P}(\mathbf{R})'$; a linear functional on $\mathcal{P}(\mathbf{R})$. Then $D'(\varphi)$ is the linear functional on $\mathcal{P}(\mathbf{R})$ given by $ (D'(\varphi))(p)=(\varphi \circ D)(p)=\varphi (D(p))=\varphi(p')=p'(3). $ Thus $D'(\varphi):V\to \mathbf{F}$ is the linear functional on $\mathcal{P}(\mathbf{R})$ taking $p$ to $p'(3)$. > [!note] A function whose input is a function > $D'(\varphi)=\varphi \circ D$ is not a consequence of [[2B2 - Function Compositions|function composition notation]]. It is the defining choice for the dual map. After making that definition, the ordinary rules of composition let us conclude $(D'(\varphi))(p)=\varphi(D(p)).$