---- > [!definition] Definition. ([[adjoint]]) > Let $(V,U, \langle -,- \rangle)$ and $(\widehat{V}, \widehat{U}, \widehat{\langle -,- \rangle})$ be [[orthogonal complement|dual pairs]] over a [[field]] $k$ and let $T:V \to \widehat{V}$ be a [[linear map|linear map]]. There is at most one [[linear map|linear map]] $T^{\top}: \widehat{U} \to U$ satisfying $\widehat{\langle Tv, \hat{u} \rangle} =\langle v, T^{\top} \hat{u}\rangle \text{ for all } v \in V, \hat{u} \in \widehat{U}.$ We call $T^{\top}$ the **adjoint of $T$**, if it exists.[^2] A sufficient condition for existence is to have ${\langle -,- \rangle}$ be [[perfect pairing|perfect]].[^1] > The notation $T^{*}$ is also used. There is no confusion with [[dual map]] because the adjoint $T^{\top}$ is always [[equivalent matrices|equivalent]] to the [[dual map]] $T^{\vee}$ (see below). [^2]: See explicit construction in the justification. But usually this characterizing property is what one wants to work with. [^1]: As it is e.g. when $\widehat{V}$ is [[dimension|finite-dimensional]] (since [[injectivity is equivalent to surjectivity in finite dimensions]]) or when it is an [[inner product]] making ${V}$ into a [[Hilbert space]] (since the [[Riesz Representation Theorem]] applies). > [!justification] > **Uniqueness.** Suppose $S$ also satisfies the identity. Then $0=\langle v, S \hat{u}-T^{\top} \hat{u} \rangle$ for all $v, \hat{u}$. Since $\langle -,- \rangle$ separates points, this implies $S \hat{u}-T^{\top} \hat{u}=0$ for all $\hat{u}$. So $S=T^{\top}$. > > **Existence construction.** Assume $\langle -,- \rangle$ is [[perfect pairing|perfect]], so in particular that the [[linear map|linear map]] $u \xmapsto{J} \langle -,u \rangle$ is an [[isomorphism]]. Claim: $T^{\top}$ equals the [[dual map|composition]] $\widehat{U} \xrightarrow{\hat{J}} \widehat{V}^{\vee} \xrightarrow{T^{\vee}}V^{\vee} \xrightarrow{J ^{-1}} U.$ > Indeed, ${J}^{-1} T^{\vee}\hat{J} \hat{u}=J ^{-1} \widehat{\langle T(\_), \hat{u} \rangle}$. The LHS is the element $u \in U$ for which $\langle -,u \rangle=\widehat{\langle T(\_), \hat{u} \rangle}$. If $v \in V$, then $\langle v, u \rangle = \widehat{\langle T(\_), \hat{u} \rangle}$ > which is precisely the claimed identity. > > [!specialization] The canonical special case: [[dual map|dual maps]]. > For the usual dual pairs $(V, V^{\vee})$, $(U,U^{\vee})$, the composition defining $T$ becomes $U ^{\vee} \xrightarrow{\hat{J}} U^\vee \xrightarrow{T ^{\vee} }V^{\vee} \xrightarrow{J^{-1}} V^{\vee}$ where $\hat{J},J$ are the literal identities. So $T^{\top}=T ^{\vee}$. Every other adjoint operator is [[equivalent matrices|equivalent]] to this one. ^specialization > [!basicproperties] The matrix of $T^{\top}$. > Choose [[basis|bases]] $(v_{i}), (u_{j}), (v_{k}), (\hat{u}_{\ell})$ for $V,U,\widehat{V},\widehat{U}$ respectively. Let $\boldsymbol A$ denote the [[matrix]] of $T$ wrt $(v_{i})$ and $(\widehat{v}_{k})$. [[dual map|Recall]] the [[matrix transpose]] $\boldsymbol A^{\top}$ is then the [[matrix]] of $T^{\vee}$ wrt the [[dual basis|dual bases]] $(v^{i})$ and $(\hat{v}^{k})$. > > The [[matrix]] of $T^{\top}$ wrt $(\hat{u}_{\ell})$ and $(u_{j})$ is then $\mathcal{M}\big(T^{\top}, (\hat{u}_{\ell}), (u_{j})\big)=\widehat{\boldsymbol J} \boldsymbol A^{\top}\boldsymbol J^{-1},$where $\widehat{\boldsymbol J}$ and $\boldsymbol J$ are the matrices of $J, \hat{J}$ wrt the relevant bases. > > Important special cases: > - When $T^{\top}=T^{\vee}$ is the dual map, $\hat{\boldsymbol J}=\boldsymbol I=\boldsymbol J$ in the dual bases and the matrix of $T^{\vee}$ is $\boldsymbol A^{\top}$, [[dual map|as we know]]. > - When $(V,\langle -,- \rangle)$ and $(\widehat{V}, \widehat{\langle -,- \rangle})$ are [[inner product space|inner product spaces]], [[matrix of a bilinear form|we know]] $\hat{\boldsymbol J}$, $\boldsymbol J$ are the appropriate [[Gram matrix|Gram matrices]]. If the chosen [[basis|bases]] are [[orthonormal basis|orthonormal]], then $\hat{\boldsymbol J}=\boldsymbol I=\boldsymbol J$ and the matrix of $T^{\top}$ is $\boldsymbol A^{\top}$, [[adjoint matrix w.r.t. orthonormal bases equals conjugate transpose|as we know]]. > [!basicproperties] > - *([[involution]])* $(T^{\top})^{\top}=T$[^5] > - [[kernel and image of the adjoint|kernel and image of the adjoint (the four fundamental subspaces)]] ^properties [^5]: Because for all $\hat{u} \in \widehat{U}$, $\langle (T^{\top})^{\top} v, \hat{u}\rangle=\langle v, T^{\top} \hat{u} \rangle=\langle Tv, \hat{u} \rangle$ for all $v \in V$ and the pairing separates points. (Thus we in fact only needed to assume the source pair has left-separation. [[orthogonal complement|dual pair]] assumption is more than sufficient.) ---- #### #algebra/linear-algebra # Legacy Let [[inner product space]] and $\big(W,\langle \cdot,\cdot \rangle\big)$ be [[field]]-[[inner product space|inner product spaces]]. Suppose $T \in$ [[vector space of linear maps between two vector spaces]]. The **adjoint** of $T$ is the ([[Riesz Representation Theorem|unique]]) [[linear map]] $T^{\dagger} \in \hom(W,V)$ such that $\langle Tv,w \rangle = \langle v,T^{\dagger}w \rangle $ for every $v \in V$ and every $w \in W$. Properties: [[basic algebraic properties of the adjoint]], [[kernel and image of the adjoint]] # Justification 1. To see that $T^{\dagger}$ must exist, suppose $T \in$ [[vector space of linear maps between two vector spaces]]. Fix $w \in W$. Consider the [[linear functional]] $\varphi \in V^*$ given by $\varphi(v)=\langle Tv,w \rangle$. By the [[Riesz Representation Theorem|Riesz Representation Theorem]], there exists a unique [[vector]] $u \in V$ such that $\varphi(v)=\langle v,u \rangle$. We denote this vector $T^{\dagger} w :=u$. 2. **Proof that the adjoint is a [[linear map]].** Suppose $T \in$ [[vector space of linear maps between two vector spaces]]. Fix $w_{1},w_{2} \in W$. If $v \in V$, then $\begin{align} \langle v, T^{\dagger}(w_{1}+w_{2}) \rangle = & \langle Tv,w_{1}+w_{2} \rangle \\ = & \langle Tv,w_{1} \rangle + \langle Tv,w_{2} \rangle \\ = & \langle v,T^{\dagger}w_{1} \rangle + \langle v,T^{\dagger}w_{2} \rangle \\ = & \langle v, T^{\dagger}w_{1} + T^{\dagger}w_{2} \rangle. \end{align}$ Hence $T^{\dagger}$ respects addition. Next let $\lambda \in \ff$ and $w \in W$. If $v \in V$, then $\begin{align} \langle v, T^{\dagger}(\lambda w) \rangle = & \langle Tv, \lambda w \rangle \\ = & \overline{\lambda}\langle Tv,w \rangle \\ = & \overline{\lambda} \langle v,T^{\dagger}w \rangle \\ = & \langle v,\lambda T^{\dagger} w \rangle, \end{align} $ thus $T^{\dagger}$ respects [[scalar]] multiplication. Thus it is [[linear map|linear]]. $\qedin$ # Basic Examples ###### 1 Define $T: \rr^2 \to \rr^3$ by $T(x_{1},x_{2},x_{2})=(x_{2}+3x_{3},2x_{1}).$ Let's find a formula for $T^{\dagger}: \rr^2 \to \rr^3$, with the [[dot product]]. Recall that the [[Riesz Representation Theorem|Riesz representation theorem]] gives us all of the existence, uniqueness, and form of $T^{\dagger}$. Let $(y_{1},y_{2}) \in \rr^2$. We have $\begin{align} \overbrace{\langle (x_{1},x_{2},x_{3}), T^{\dagger}(y_{1},y_{2}) \rangle}^{\langle v,u \rangle =\langle v,T^{\dagger}w \rangle } = & \overbrace{\langle T(x_{1},x_{2},x_{3}), (y_{1},y_{2}) \rangle }^{\varphi(v)=\langle Tv,w \rangle } \\ = & \langle (x_{2}+3x_{3},2x_{1}), (y_{1},y_{2}) \rangle \\ = & x_{2}y_{1}+3x_{3}y_{1}+2x_{1}y_{2} \\ = & \underbrace{\langle (x_{1},x_{2},x_{3}),(2y_{1},y_{2},3y_{1}) \rangle }_{\langle v,u \rangle = \langle v,T^{\dagger}w \rangle }. \end{align}$ Hence $T^{\dagger}$ is given by $T^{\dagger}(x_{1},x_{2},x_{3})=(2y_{1},y_{2},3y_{1})$. ###### 2 Fix $u \in V$ and $x \in W$. Define $T \in$ [[vector space of linear maps between two vector spaces]] by $Tv=\langle v,u \rangle x$ for every $v \in V$. Let's find a formula for $T^{\dagger}$. Fix $w \in W$. Using the [[Riesz Representation Theorem|Riesz Representation Theorem]] we have $\begin{align} \langle v,T^{\dagger}w \rangle = & \varphi(v) \\ = & \langle Tv,w \rangle \\ = & \big\langle\langle v,u \rangle x, w \big\rangle \\ = & \langle v,u \rangle \langle x,w \rangle \\ = & \langle v, \overline{\langle x,w \rangle }u \rangle \\ = & \langle v, \langle w,x \rangle u \rangle . \\ \end{align}$ Thus $T^{\dagger}v=\langle w,x \rangle u$. # Properties - [[basic algebraic properties of the adjoint]] #notFormatted ---- #### 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 > ```