Let $A \in \mathbb{R}^{n \times p}$, let $D$ be a diagonal matrix with positive entries. $\dagger$ denotes the Moore-Penrose pseudoinverse.

Is it true in general that: $$(A^\top D A)^\dagger A^\top D = (A^\top D^2 A)^\dagger A^\top D^2 \enspace ? $$ (and if not, is it easy to determine when this holds?)

If $D$ is proportional to identity, ok. Multiplying by $A$ on the right yields an identity which looks reasonable when $A^\top D A$ is invertible.

Numerical simulations with $p > n$ give frequent equality, but not always, and I'm unable to determine in which case the equality holds.


1 Answer 1


Using the identity $(X^\top X)^\dagger X^\top=X^\dagger$, the equality in question can be rewritten as $(\sqrt{D}A)^\dagger\sqrt{D}=(DA)^\dagger D$.

It is true when $A$ has full row rank (but numerical verification may fail because of rounding errors). In this case, $(BA)^\dagger=A^\dagger B^\dagger$ for any $B$ with full column rank (see Wikipedia). Hence $$ (\sqrt{D}A)^\dagger\sqrt{D}=A^\dagger \sqrt{D}^\dagger\sqrt{D}=A^\dagger=A^\dagger D^\dagger D=(DA)^\dagger D. $$ Alternatively, when $A$ has full row rank, $(\sqrt{D}A)^\top$ and $(DA)^\top$ have the same column spaces. So, using the fact that $X^\dagger X$ is the orthogonal projection onto the column space of $X^\top$, we see that $(\sqrt{D}A)^\dagger\sqrt{D}A=(DA)^\dagger DA$. Right-multiply both sides by the right inverse of $A$, we obtain $(\sqrt{D}A)^\dagger\sqrt{D}=(DA)^\dagger D$.


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