---- > [!definition] Definition. ([[dual map]]) > Suppose $V,W$ are [[vector space|vector spaces]] and $T \in \text{Hom}\big(V,W\big)$. The **dual map** of $T$ is the [[linear map]] $T^* \in \hom(W^*, V^*)$ defined by $T^*(\varphi) = \varphi \circ T$ for $\varphi \in W^{*}$. > If $V,W$ f.d. with [[basis|bases]] $(v_{i})$, $(w_{j})$ wrt which $T$ has [[matrix]] $\boldsymbol A$, then the [[matrix]] of $T^{*}$ wrt the [[dual basis|dual bases]] $(v^{i}), (w^{j})$ [[matrix transpose|equals]] $\boldsymbol A ^{\top}$. > [!proof]- Proof on Matrices. > First note: > - Have $Tv_{i}=\sum_{k}A_{ki} w_{k}$ for some scalars $A_{ki} \in \mathbb{F}$. Thus $w^{j}(Tv_{i})=A_{ji}$. > - Have $w^{j} \circ T= b_{ij} v^{i}$ for scalars $b_{ij} \in \mathbb{F}$. Thus $(w ^{j} \circ T)(v_{i})=b_{ij}$. > > > We see that the scalars $b_{ij}$ needed to write $T^{*}w^{j}=w^{j} \circ T$ as a [[linear combination]] of the $v^{i}$ equal the scalars $A_{ji}$ needed to write $Tv_{i}$ as a [[linear combination]] of the $w _{j}$. > [!basicproperties] > When $V,W$ are [[norm|(semi)normed]], $T^{*}$ is [[operator norm|bounded]] iff $T$ is, in fact with $\|T^{*}\|=\|T\|$.[^1] ^properties [^1]: Use [[norm|computing norm with linear functionals]]: both $\|T^{*}\|$ and $\|T\|$ are equal to $\sup_{\|v\|=1} \sup _{\|\varphi\|=1} |(T^{* }\varphi)(v)|.$ > [!basicexample] > Define $D: \PP(\rr) \to \PP(\rr)$ by the [[derivative]] map $Dp=p'$. Suppose $\varphi$ is the [[linear functional]] on [[vector space of all polynomials with coefficients in F]], where $\ff=\rr$ , defined by $\varphi(p)=p(3)$. Then $D^*(\varphi)$ is the [[linear functional]] on $\PP(\rr)$ given by $\big( D^{*}(\varphi) \big)(p) = (\varphi \circ D)(p) = \varphi(D(p)) = \varphi(p')=p'(3).$ In other words, $D^{*}(\varphi)$ is the [[linear functional]] on $\PP(\rr)$ that takes $p$ to $p'(3)$. ^basic-example > [!basicproperties] > - [[algebraic properties of dual maps]] > >- $T^{*}$ is [[operator norm|bounded]] when $T$ is. In fact, $\|T\|=\|T^{*}\|$ (the inequality $\geq$ is elementary; its reverse is an application of the [[Hahn-Banach Extension Theorem]]). > > > > [!proof]- Proof. > > Using submultiplicativity of [[operator norm]], we have $\|T^{*} \varphi\|=\|\varphi \circ T\| \leq \|\varphi\| \| T\|=\|T\|$ for all bounded linear functionals $\varphi \in W^{\vee}$ with $\|\varphi\|=1$. Taking the [[supremum]], it follows that $\|T^{*}\|=\sup_{\|\varphi\|=1} \|T^{*} \varphi\| \leq \|T\|.$ > > For the reverse inequality: pick $f_{0} \in V$ with $\|f_{0}\|=1$ and $\|Tf_{0}\|$ arbitrarily close to $\|T\|$ (can do by equivalent definition of [[supremum]]). Define a linear functional $\psi_{0}$ on the one-dimensional [[linear subspace]] $\span (Tf_{0})$ as $\psi_{0} (\alpha Tf_{0})=\alpha \|Tf_{0}\| \ (\alpha \in \mathbb{F}).$ > > Note that $\|\psi_{0}\|=1$. Using [[Hahn-Banach Extension Theorem|Hahn-Banach]], extend $\psi_{0}$ to a linear functional $\varphi$ on all of $W^{*}$ satisfying $\|\varphi\|=\|\psi_{0}\|=1$. Now $\| T^{*}\| \geq \|T^{*} \varphi\| \geq |(T^{*} \varphi)(f_{0}) |= | \varphi ( Tf_{0}) |=|\psi_{0}(Tf_{0})|=\|Tf_{0}\|.$ > > Now since $\|Tf_{0}\|$ is arbitrarily close to $\|T\|$, the result follows (if $\|T\|>\|T^{*}\|$ then we could get $\|Tf_{0}\|>\|T^{*}\|$, contradicting the typeset line above). > > > > (There is probably a slicker way to do this.) > > > [!generalization] > - [[dual transformation]] ^generalization > [!justification] > - For arbitrary $\phi$ we have that $T^*(\varphi) \in$ [[dual vector space]] because $T^{*}(\varphi)$ is a composition $\varphi \circ T$ of [[linear map]]s (hence linear) with source $V$ (since that's the source of $T$) and target [[field]] (since $\varphi$ a [[linear functional]]). > - The proof that $T^{*}$ is [[linear map]] is as follows: let $\varphi, \psi \in W^{*}$. We have $T^{*}(\varphi + \psi) = (\varphi+\psi) \circ T = \varphi \circ T + \psi \circ T = T^{*}(\varphi)+T^{*}(\psi)$. Note that few special properties were used here; for example, the assertion that $(\varphi+\psi) \circ T = \varphi \circ T + \psi \circ T$ is a property that holds for arbitrary function composition. Next we let $\lambda \in$ [[field]] and consider $T^{*}(\lambda \varphi)=T^{*} = (\lambda \varphi) \circ T = \lambda (\varphi \circ T) = \lambda T^{*}(\varphi)$. > ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```