= Penrose equations
{c}
The Moore--Penrose inverse is characterized by
$$
AA^\dagger A=A,
\quad
A^\dagger AA^\dagger=A^\dagger,
\quad
(AA^\dagger)^*=AA^\dagger,
\quad
(A^\dagger A)^*=A^\dagger A.
$$
Here $A^\dagger=A^+$ denotes the <Moore-Penrose inverse>, while the star denotes the <conjugate transpose>. The first two identities express consistency when applying and reversing the <linear map>; the last two make the resulting <orthogonal projection matrices> <Hermitian>. They must hold together. For example, $A=(1\ \ 2)$ and $B=(1\ \ 0)^T$ satisfy $ABA=A$ and $BAB=B$, but
$$
BA=\begin{pmatrix}1&2\\0&0\end{pmatrix}
$$
is not <Hermitian>. This $B$ is a <right inverse>, yet $Bb=(3,0)^T$ is not the <minimum-norm least-squares solution> for $b=3$. The last two <Penrose equations> impose the <orthogonality> that selects the <pseudoinverse>.
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