Is $Im(T^\dagger) = Ker(T)^\perp$ true for infinite dimensional vector spaces?

Question

Let $T$ be some operator on an inner product space $(V, \langle\cdot,\cdot\rangle)$, and $T^\dagger$ be its adjoint. I found too many questions about the proof of $$Im(T^\dagger) = Ker(T)^\perp$$ for finite dimensional vector spaces. But I don't know if it is true for infinite dimensional vector spaces ?

Answer 1

For the sake of completeness of approach, I would like to present here the general context of purely algebraic adjunction, where we consider no metric-inducing inner products and no topological vector spaces.

In the most general frame, consider an arbitrary ring $A$ together with a left $A$-module $M$. We introduce the (algebraic) dual of $M$ as: $$M^*\colon=\mathrm{Hom}_{\operatorname{\mathit{A}-\mathbf{Mod}}}(M, A),$$ structure which carries a natural right $A$-module structure induced by the canonical right $A$-module structure on $A$. Furthermore we introduce the canonical pairing: $$\begin{align*} \langle{\ ,\ \rangle}_M \colon M \times M^* &\to A\\ \langle x, u\rangle_M&=u(x), \end{align*}$$ which is easily seen to be $(A, A)$-bilinear. By virtue of this bilinearity we can introduce the canonical product map $\pi_M \in \mathrm{Hom}_{\operatorname{\mathit{A}-\mathbf{Mod}-\mathit{A}}}\left(M \otimes_{\mathbb{Z}}M^*, A\right)$, which is the unique $(A, A)$-linear map satisfying the relation $\pi_M \circ \otimes_{MM^*}=\langle\ ,\ \rangle_M$, where $\otimes_{MM^*} \colon M \times M^* \to M \otimes_{\mathbb{Z}}M^*$ is the canonical map into the tensor product.

Given another left $A$-module $N$ together with an $A$-linear map $f \in \mathrm{Hom}_{\operatorname{\mathit{A}-\mathbf{Mod}}}(M, N)$, we introduce the transpose (or dual) of $f$ as: $$\begin{align*} f^*\colon=\mathrm{Hom}_{\operatorname{\mathit{A}-\mathbf{Mod}}}(f, \mathbf{1}_A)\colon N^* &\to M^*\\ f^*(v)&=v \circ f \end{align*}$$ and remark that it is a morphism of right $A$-modules: $f^* \in \mathrm{Hom}_{\operatorname{\mathbf{Mod}-\mathit{A}}}(N^*, M^*)$. As is well-known, the correspondence described by: $$\begin{align*} M &\mapsto M^*\\ \mathrm{Hom}_{\operatorname{\mathit{A}-\mathbf{Mod}}}(M, N) \ni f &\mapsto f^*\in \mathrm{Hom}_{\operatorname{\mathbf{Mod}-\mathit{A}}}(N^*, M^*) \end{align*}$$ implements a functor, the so-called dualisation functor $(\bullet)^* \colon \left(\operatorname{\mathit{A}-\mathbf{Mod}}\right)^{\circ} \to \operatorname{\mathbf{Mod}-\mathit{A}}$, from the opposite category of left $A$-modules to the one of right $A$-modules.

Let us now fix a morphism $f \in \mathrm{Hom}_{\operatorname{\mathit{A}-\mathbf{Mod}}}(M, N)$ and remark that by definition the following relation: $$\langle f(x), v \rangle_N=\langle x, f^*(v)\rangle_M$$ holds for any $x \in M$ and $v \in N^*$, or in a more succinct formulation the following:

is a commutative diagram, where the interior rectangle is a diagram consisting of morphisms in the category $\operatorname{\mathit{A}-\mathbf{Mod}-\mathit{A}}$ of $(A, A)$-bimodules.

We introduce the following notions of orthogonality:

• for any subset $X \subseteq M$, we define the right orthogonal of $X$ as: $$X^{\perp}\colon=\left\{u \in M^*|\ (\forall x)(x \in X \Rightarrow \langle x, u \rangle=0_A)\right\}$$
• for any subset $Y \subseteq M^*$, we define the left orthogonal of $Y$ as: $${}^{\perp}Y\colon=\left\{x \in M|\ (\forall u)\left(u \in Y \Rightarrow \langle x, u \rangle=0_A\right)\right\}.$$ It is easy to ascertain that the right orthogonal of any subset $X \subseteq M$ is an $A$-submodule of $M^*$ -- $X^{\perp} \leqslant_A M^*$ -- and analogously the left orthogonal of any subset $Y \subseteq M^*$ is an $A$-submodule of $M$, ${}^{\perp}Y \leqslant_A M$. Furthermore, considering the sets $\mathscr{S}$ respectively $\mathscr{T}$ of submodules of $M$ respectively $M^*$ as ordered by inclusion, the pair of maps given by: $$\begin{align*} \mathscr{S} &\to \mathscr{T}\\ P &\mapsto P^{\perp}\\ \mathscr{T} &\to \mathscr{S}\\ Q &\mapsto {}^{\perp}Q \end{align*}$$ constitutes a Galois connection, fact which is justified by making the following three observations:
• for any subsets $X \subseteq Y \subseteq M$ we have $Y^{\perp} \leqslant_A X^{\perp} \leqslant M^*$
• for any subsets $U \subseteq V \subseteq M^*$ we have ${}^{\perp}V \leqslant_A {}^{\perp}U \leqslant_A M$
• for any subsets $X \subseteq M$ and $Y \subseteq M^*$ we have the equivalence $Y \subseteq X^{\perp} \Leftrightarrow X \subseteq {}^{\perp}Y$. Thus, the following two maps: $$\begin{align*} \mathscr{S} &\to \mathscr{S}\\ P &\mapsto {}^{\perp}\left(P^{\perp}\right)\\ \mathscr{T} &\to \mathscr{T}\\ Q &\mapsto \left({}^{\perp}Q\right)^{\perp} \end{align*}$$ constitute closure operators on each of their respective domains, and it is by means of these objects that we can have a notion of closedness or closure in this general algebraic context.

We conclude this expound with the following classical duality relations:

Proposition. Let $f \in \mathrm{Hom}_{\operatorname{\mathit{A}-\mathbf{Mod}}}(M, N)$ be a linear map and $P \leqslant_A M$, $Q \leqslant_A N^*$ two $A$-submodules. The relations $\left(f[P]\right)^{\perp}=\left(f^*\right)^{-1}\left[P^{\perp}\right]$ and ${}^{\perp}\left(f^*[Q]\right)=f^{-1}\left[{}^{\perp}Q\right]$ are valid.

Proof. It suffices to set up the straightforward sequences of equivalences: $$\begin{align*} v \in \left(f[P]\right)^{\perp} &\Leftrightarrow v \in N^* \wedge (\forall y)\left(y \in f[P] \Rightarrow \langle y, v\rangle=0_A\right)\\ &\Leftrightarrow v \in N^* \wedge (\forall x)(x \in P \Rightarrow \langle f(x), v\rangle=0_A)\\ &\Leftrightarrow v \in N^* \wedge (\forall x)(x \in P \Rightarrow \langle x, f^*(v)\rangle=0_A)\\ &\Leftrightarrow v \in N^* \wedge f^*(v) \in P^{\perp}\\ &\Leftrightarrow v \in \left(f^*\right)^{-1}\left[P^{\perp}\right] \end{align*}$$ respectively: $$\begin{align*} x \in {}^{\perp}\left(f^*[Q]\right) &\Leftrightarrow x \in M \wedge (\forall u)(u \in f^*[Q] \Rightarrow \langle x, u \rangle=0_A)\\ &\Leftrightarrow x \in M \wedge (\forall v)(v \in Q \Rightarrow \langle x, f^*(v)\rangle=0_A)\\ &\Leftrightarrow x \in M \wedge (\forall v)(v \in Q \Rightarrow \langle f(x), v\rangle=0_A)\\ &\Leftrightarrow x \in M \wedge f(x) \in {}^{\perp}Q\\ &\Leftrightarrow x \in f^{-1}\left[{}^{\perp}Q\right], \end{align*}$$ which directly prove our assertions. $\Box$

Considering the particular case $P=M$ in the first relation yields $\left(\mathrm{Im}f\right)^{\perp}=\mathrm{Ker}\left(f^*\right)$ and analogously considering the particular case $Q=N^*$ in the second yields ${}^{\perp}\left(\mathrm{Im}\left(f^*\right)\right)=f^{-1}\left[{}^{\perp}\left(N^*\right)\right]$. Let us remark that the curious submodule ${}^{\perp}\left(N^*\right)$ occurring on the right-hand side of the latter particular relation is none other than the closure of the null submodule $\{0_N\}$. This closure is not in general null (in other words, the null submodule of a given $A$-module will not be closed in general), however in particular cases this does occur. This closure of $\{0_N\}$ can also be characterised as the kernel of the canonical reflexivity morphism $\rho_N\colon N \to N^{**}$, which takes every element $x \in N$ to the form on $N^*$ given by evaluation in $x$. The null submodule will thus be closed for any module whose reflexivity morphism is injective, hence in particular for vector spaces of arbitrary dimension, where our particular relation takes the even more explicit form ${}^{\perp}\left(\mathrm{Im}\left(f^*\right)\right)=\mathrm{Ker}f$ (although we will not go into detail here, I want to mention that any free module has an injective reflexivity morphism and finitely generated free modules are furthermore reflexive, in the sense that for them the reflexivity morphism is an isomorphism). Taking right orthogonals in this form of the relation entails the formulation we had been discussing in the comment section, namely that the (right) orthogonal of the kernel is the closure of the image of the adjoint.

As a final note, I would like to point out that the general duality relations can be given a very brief diagrammatic formulation, which will require a few additional definitions. Given arbitrary linear map $g \in \mathrm{Hom}_{\operatorname{\mathit{A}-\mathbf{Mod}}}(P, Q)$, let us introduce the direct and inverse maps between the respective sets of submodules as follows: $$\begin{align*} \hat{\mathscr{S}}_{\operatorname{\mathit{A}-\mathrm{Mod}}}(g)\colon \mathscr{S}_{\operatorname{\mathit{A}-\mathrm{Mod}}}(P) &\to \mathscr{S}_{\operatorname{\mathit{A}-\mathrm{Mod}}}(Q)\\ \hat{\mathscr{S}}_{\operatorname{\mathit{A}-\mathrm{Mod}}}(g)(R)&=g[R]\\ \check{\mathscr{S}}_{\operatorname{\mathit{A}-\mathrm{Mod}}}(g)\colon \mathscr{S}_{\operatorname{\mathit{A}-\mathrm{Mod}}}(Q) &\to \mathscr{S}_{\operatorname{\mathit{A}-\mathrm{Mod}}}(P)\\ \check{\mathscr{S}}_{\operatorname{\mathit{A}-\mathrm{Mod}}}(g)\left(R’\right)&=g^{-1}\left[R’\right]. \end{align*}$$ We then have the following commutative diagram, in which the “exterior” arrows refer to the first duality relation, whereas the “interior” ones describe the second one:

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Analogous treatments of the notion of duality and adjunction can be given in either the more specific case of Hilbert spaces -- by virtue of the Riesz representation theorem characterising functionals of Hilbert spaces -- or the more general one of locally convex topological vector spaces. These settings carry natural topological structures and it is a remarkable fact that the closure operators induced by these topologies coincide with the closure operators stemming from the Galois connections analogous to the one described above.

Answer 2

Take a vector $T^\dagger v$ in the image of $T^\dagger$, take a vector $w$ in the kernel of $T$, and take their dot product, then use the definition of adjoint.

This shows that the image of $T^\dagger$ and the kernel of $T$ are orthogonal. In the finite dimensional case, an appeal to the rank-nullity theorem is enough to conclude your equality.

However, the dot product is a continuous function. Thus in the infinite dimensional case, any element in the closure of the image of $T^\dagger$ would also be orthogonal to $\ker T$. So if the image of $T^\dagger$ isn't closed, then there can't be equality.

Answer 3

What is true in any case is that the closure of $\text{im}(T^*)$ is equal to $\ker(T)^\bot$. This is superfluous in the finite dimensional case, since all subspaces are automatically closed. In the general case though this is necessary. I cannot think of a counterexample right now, but there are many. Proving that

$$\overline{\text{im}(T^*)}=\ker(T)^\bot $$ is done almost in the same way as in the finite dimensional case.