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

The statement is true without an additional rank assumption. If $A=U\Sigma V^H$ is a singular value decomposition, then

This statement is also always true. The same singular value decomposition gives

This statement is true. The matrix $A^{†}A$ is the orthogonal projection onto the row space $\mathcal{R}(A^H)$, so its eigenvalues are zero and one. Consequently

The Reverse-order law for the Moore--Penrose inverse is false in general. TakeThenA sufficient condition isso that $A$ has full column rank and $B$ has full row rank. Indeed $A^{†}A=I_n$ and $B{B}^{†}=I_n$; these identities make $B^{†}{A}^{†}$ satisfy all four Penrose equations for $AB$.