Generalisation of fundamental theorem of linear algebra 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
 A: 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. 
