Generalisation of fundamental theorem of linear algebra

Question

Does something similar to "Fundamental theorem of linear algebra" hold for infinite dimensional Hilbert spaces?

https://en.wikipedia.org/wiki/Fundamental_theorem_of_linear_algebra

Answer 1

Regarding the relations between the kernel and the image :

Given a separable Hilbert space $(H,\langle.,.\rangle)$ and a bounded linear operator $A : H \to H$, we have that $$ \ker(A)={Im(A^*)}^{\perp} $$ $$ \overline{Im(A)}=\ker(A^*)^{\perp} $$ where $A^*$ denotes the adjoint of $A$ (a generalization of the notion of transpose). The result should also hold for an operator between two different Hilbert spaces $H_1 \to H_2$.

Proof of the first equality :

(*) Consider $x \in \ker(A)$ and $y \in Im(A^*)$. Rewriting $y$ as $y=A^*z$ for some $z \in H$, we have that

\begin{align} \langle x,y \rangle &= \langle x,A^*z\rangle \\ &=\langle Ax,z\rangle \\ &=0 \end{align} since $x \in \ker(A)$. Thus $\ker(A) \subset Im(A^*)^{\perp}$.

(**) On the other hand, take $x\in Im(A^*)^{\perp}$. Then for all $y \in H$,

\begin{align} \langle Ax,y \rangle &= \langle x,A^*y \rangle \\ &= 0 \end{align} since $A^*y \in Im(A^*)$. Therefore we must have $ Ax=0 $ so $x \in \ker(A)$, which completes the proof.

Proof of the second equality : Using the first equality, and noting that $(A^*)^*=A$, we have $$ \ker(A^*)={Im(A)}^{\perp} $$ Taking the orthogonal complement, we have that $$ \ker(A^*)^{\perp}=(Im(A)^{\perp})^{\perp}=\overline{Im(A)} $$ which completes the proof.