Linear Operator Equations

Theorem 2.10: Let $A \in \mathcal{B}(\mathcal{H})$; then:

Proof: For $x \in \mathcal{H}$ we have the following sequence of equivalences:

Thus $\ker(A) = \text{Ran}(A^*)^{\perp}$. Since $A^{**} = A$, the second assertion follows from the first one if $A$ is replaced by $A^*$. $\blacksquare$

Theorem 2.11: Let $A \in \mathcal{B}(\mathcal{H})$. Then:

Proof: For every closed subspace $M \subset \mathcal{H}$, $\mathcal{H} = M \oplus M^{\perp}$. Furthermore for every subspace (not necessarily closed) $N \subset \mathcal{H}$, $N^{\perp \perp} = \overline{N}$. In combination with Theorem 2.10, this provides our claim. $\blacksquare$

Discussion: Theorem 2.11 is important when we aim to solve the linear equation:

Here $y$ is considered as given and $x$ is the unknown variable. Clearly, the necessary and sufficient condition for solvability of this equation is $y \in \text{Ran}(A)$. Thus for the above equation to be solvable it is necessary to have $y \perp \ker A^* \supset \text{Ran}(A)$. If $\text{Ran}(A)$ is closed, this condition is also sufficient. You can also note that when $A$ is self-adjoint, it is not necessary to compute an adjoint: the condition becomes $y \perp \ker A$.