-----
> [!proposition] Proposition. ([[kernel and image of the adjoint]])
> Let $k$ be a [[field]], and let $(X, Y, \langle -,- \rangle)$ and $(\widehat{X}, \widehat{Y}, \widehat{\langle -,- \rangle})$ be $k$-[[vector space|vector spaces]] endowed with [[orthogonal complement|dual pairs]]. Suppose $T:X \to \widehat{X}$ is a [[linear map|linear map]] whose [[adjoint]] $T^{\top}:\widehat{Y} \to Y$ exists. [[orthogonal complement|Then]] we have the following identities:
$\operatorname{ker }T^{\top}=({\operatorname{im } T})^{\perp} \text{ and }^{\perp}(\operatorname{ker }T^{\top})=\overline{\operatorname{im } T}^{\sigma(\widehat{X}, \widehat{Y})}$
>
and dually
>
$\operatorname{ker } T = ^{\perp}\!\!(\operatorname{im }T^{\top} ) \text{ and } (\operatorname{ker } T)^{\perp} =\overline{\operatorname{im }T^{\top}}^{\sigma(Y,X)}.$
>
Here, $\sigma(\cdot,\cdot)$ refers to [[weak topology]], and $\overline{(\cdot)}^{\sigma(\cdot, \cdot)}$ to [[closure]] therein.
>
>
> [!specialization]
>
> When our [[vector space|vector spaces]] are [[norm|normed]] and we are considering the canonical dual pairs $(X, X^{\vee})$ and $(Y, Y^{\vee})$, then [[Mazur's Lemma]] says $\overline{\operatorname{im } T}^{\sigma(Y, Y ^{\vee})}=\overline{\operatorname{im }T}$, i.e. we can just use regular (norm) [[closure]] instead of weak [[closure]] here: $(\operatorname{ker }T^{\vee})_{\circ} = \overline{\operatorname{im }T}^{}.$
>
Mazur's Lemma doesn't say anything about weak$^*$ vs. norm-closure. So it can still happen that $\overline{\operatorname{im }T^{\vee}} \subsetneq (\operatorname{ker }T)^{\circ}$. If we further assume [[Banach space|Banach]], we do get $\operatorname{im }T^{\vee}=(\operatorname{ker } T)^{\circ} \text{ in } \mathsf{Ban} \text{ iff } \operatorname{im } T \text{ is closed}$
([[characterization of quotienting a Banach space|i.e., iff T is a quotient operator onto its image]])[^9].
>
> > [!proof]- Proof.
> > $\impliedby$. We prove assuming $T$ is [[surjection|surjective]]. This is [[complete|WLOG]] due to [[Hahn-Banach Extension Theorem|Hahn-Banach]].
> >
> > $\subset$ is obvious/always holds. For $\supset$, let $f \in (\operatorname{ker } T)^{\circ}$. we claim the [[diagram]] below is valid not only in $\mathsf{Vect}$, but *in $\mathsf{Ban_{T}}$*:
> > ```tikz
> > \usepackage{tikz-cd}
> > \usepackage{amsmath}
> > \usepackage{amsfonts}
> > \begin{document}
> > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAE0QBfU9TXfIRQBGclVqMWbABrdeIDNjwEiAJjHV6zVohAAdPQFs6OABYAjc8ABiXOXyWCio4eK1Td0gPQGA1jAAnAAIAFW5xGCgAc3giUAAzAIhDJFEQHAgkdQltNnj7EETk1OoMpDIcjxAw6gY6cxgGAAV+ZSEQAKwo0xwQalMYOig2HAB3CAGhhB4EpJTENLLEAGYZwrmkZdLMxGyGLDAdECg6OAHhzUkjgxgADyw4HDgggEIggwgaQP3D4Hi7Wr1RotRwqXSdbq9NZFeYVJZbEA-I4nM6RPqVa56UZYWB4BiwYAhLgAPWAAFphADEUDmq0nOCuj10ZNhogwEwGAxSnQsAw2JBDuEuEA
> > \begin{tikzcd}
> > Y \arrow[rd, "\widetilde{T}^{-1}"', dashed] & X \arrow[r, "f"] \arrow[l, "T"', two heads] \arrow[d] & \mathbb{F} \\
> > & X/\ker T \arrow[ru, "\exists ! \overline{f}"', dashed] &
> > \end{tikzcd}
> > \end{document}
> > ```
> > Indeed:
> > - (Left) [[characterization of quotienting a Banach space|Recall]] that since $T$ is [[surjection|surjective]] between [[Banach space|Banachs]], $T$ is a [[characterization of quotienting a Banach space|quotient operator]] onto $Y$. Thus the [[characterization of quotienting a Banach space|first isomorphism theorem for Banach spaces]] applies to give $\widetilde{T}^{-1}$ a [[operator norm|bounded linear map]].
> > - (Right) $f(\operatorname{ker } T)=0$, meaning $\operatorname{ker }T \subset \operatorname{ker } f$. Hence the [[characterization of quotienting a Banach space|universal property of quotienting a Banach space]] gives $\overline{f}$ a [[operator norm|bounded linear map]].[^7]
> >
> > Set $g:=\overline{f} \circ \widetilde{T}^{-1}$. Then $f=g \circ T=T^{\vee}g$, thus $f \in \operatorname{im } T^{\vee}$.
> >
> > $\implies$. Assume $\operatorname{im }T^{\vee}=(\operatorname{ker } T)^{\circ}$. Because $(\operatorname{ker }T)^{\circ}$ is [[closed set|closed]] ([[orthogonal complement|it's closed]] in [[weak topology]], hence in [[norm]] [[topological space|topology]]), $\operatorname{im }T^{\vee}$ is [[closed set|closed]]. By the [[closed range theorem]], $\operatorname{im }T$ is closed.
>
[^7]: Note also $\operatorname{ker }f=\overline{\operatorname{ker }f}$ because $f$ is [[continuous]].
[^9]: Put differently, can say "iff $T$ corestricts to a morphism in $\mathsf{Ban}
quot;.
> [!proof]- Proof.
>
> **$\operatorname{ker }T^{\top}=(\operatorname{im } T)^{\perp}.$** No technicalities to fear here. We have $\begin{align}
> \hat{y} \in \operatorname{ker }T^{\top} &\iff T^{\top} \hat{y}=0 \\
> & \iff \langle x, T^{\top} \hat{y} \rangle =0 \text{ for all } x \in X \\
> & \iff \langle Tx, \hat{y} \rangle =0 \text{ for all } x \in X \\
> & \iff \hat{y} \in (\operatorname{im } T)^{\perp}.
> \end{align}$
>
> **$^{\perp}(\operatorname{ker } T^{\top})=\overline{ \operatorname{im }T}^{\sigma(\widehat{X}, \widehat{Y})}$.** Pass to left-annihilators in the above, using the [[orthogonal complement|general fact]] $^{\perp}(W^{\perp})=\overline{W}^{\sigma(V,U)}$.
>
> **$\operatorname{ker }T=^{\perp}\!\!(\operatorname{im }T^{\top})$.** Swap $T$ and $T^{\top}$ in the proof that $\operatorname{ker }T^{\top}=(\operatorname{im }T)^{\perp}$.
>
> **$(\operatorname{ker } T)^{\perp}= \overline{\operatorname{im } T^{\top}}^{\sigma(Y,X)}$.** Pass to right-annihilators in the above, using the [[orthogonal complement|general fact]] $(^{\perp}Z)^{\perp}=\overline{Z}^{\sigma(\cdot, \cdot)}$.
>
Special cases:
- Banach spaces (dual pairing)
- Hilbert spaces (including finite-dimensional inner product spaces)
- Finite-dimensional dual spaces
- etc
[[weak topology]]
$\operatorname{ker }(T^{\top})=(\operatorname{im } T)^{\perp} \text{ and } \operatorname{im } T^{\top}= (\operatorname{ker } T)^{\perp}.$
Accordingly, $\operatorname{ker } T = (\operatorname{im }T^{\top})^{\perp} \text{ and } \operatorname{im }T = (\operatorname{ker }T^{\top})^{\perp}.$
[[weak topology]]
-----
####
#algebra/linear-algebra
# Statement
Suppose $T \in$ [[vector space of linear maps between two vector spaces]]. Then
###### 1
[[kernel of a linear map]] $T^{\dagger}= (\im T)^{\perp}$ ;
#### 2
[[image]] $T^{\dagger} = (\ker T)^{\perp}$ ;
###### 3
$\ker T = (\im T^{\dagger})^{\perp}$ ;
###### 4
$\im T = (\ker T^{\dagger})^{\perp}$.
![[CleanShot 2023-10-07 at 22.27.33@2x 1.jpg]]
# Proof
We first prove [[kernel and image of the adjoint#1|1]]:
Let $w \in W$. Then $\begin{align}
w \in \ker T^{\dagger} \iff & T^{\dagger}w=0 \\
\iff & \langle v, T^{\dagger}w \rangle=0 \text{ for all } v \in V \and w \in W \\
\iff & \langle Tv,w \rangle=0 \text{ for all } v\in V \\
\iff & w \in (\im T)^{\perp}.
\end{align}$
Taking the [[orthogonal complement|orthogonal complement]] of both sides of [[kernel and image of the adjoint#1|1]] immediately yields [[kernel and image of the adjoint#4|4]]. Using [[basic algebraic properties of the adjoint#3]] and replacing $T$ with $T^{\dagger}$ in [[kernel and image of the adjoint#1|1]] gives [[kernel and image of the adjoint#3|3]], and applying the same strategy to [[kernel and image of the adjoint#4|4]] gives [[kernel and image of the adjoint#2|2]].
$(1)$ is useful in [[least squares]] approximation.
#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
> ```