For a bounded operator $A:X\to Y$ between <Hilbert spaces>, the <Moore--Penrose inverse of a Hilbert-space operator> is defined on
$$
\mathcal D(A^\dagger)=\mathcal R(A)\oplus\mathcal R(A)^\perp
$$
and has range $\mathcal R(A^\dagger)=\mathcal N(A)^\perp$. Its <Penrose equations> are
$$
AA^\dagger A=A,
\qquad
A^\dagger AA^\dagger=A^\dagger,
\qquad
(AA^\dagger)^*=AA^\dagger,
\qquad
(A^\dagger A)^*=A^\dagger A.
$$
Equivalently, $A^\dagger A=P_{\mathcal N(A)^\perp}$ and $AA^\dagger$ is the restriction of $P_{\overline{\mathcal R(A)}}$ to $\mathcal D(A^\dagger)$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 326 / 1 / a / Solution

{c}

For a bounded operator $A:X\to Y$ between <Hilbert spaces>, $A^\dagger$ has domain $\mathcal R(A)\oplus\mathcal R(A)^\perp$, maps into $\mathcal N(A)^\perp$, and inverts the restriction of $A$ to $\mathcal N(A)^\perp$. It is bounded exactly when $\mathcal R(A)$ is closed.

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 Incoming links: Moore--Penrose inverse of a Hilbert-space operator