Norm of Moore-Penrose inverse of a product

Question

I had asked this question in math.stackexchange (link: https://math.stackexchange.com/questions/1902276/bounds-on-the-moore-penrose-inverse-of-a-product ) but I did not get any response so I am trying my luck here.

Let $A^{\dagger}$ denote the Moore-Penrose inverse of a real matrix and let $\|A\|$ denote the usual matrix norm given by the largest singular value of $A.$

Then is it true that $\|(AB)^{\dagger}\| \leq \|A^{\dagger}\| \|B^{\dagger}\|?$

This is trivially true whenever $(AB)^{\dagger} = B^{\dagger}A^{\dagger}$, which happens, for example, when $A$ is full column rank and $B$ is full row rank. But what about the general case?

Answer 1

Here’s a small counterexample involving two rank-deficient matrices $A$ and $B$: $$A = \left( {\begin{array}{*{20}c} 0 & { - 1} & 0 \\ 0 & 0 & 0 \\ 0 & { - 1} & 0 \\ \end{array}} \right), B = \left( {\begin{array}{*{20}c} 4 & 0 & 2 \\ 2 & 0 & 1 \\ 4 & 0 & 2 \\ \end{array}} \right)$$

The corresponding Moore–Penrose pseudoinverses of interest are: $$ A^{\dagger} = \left( {\begin{array}{*{20}c} 0 & 0 & 0 \\ { -\frac{1}{2}} & 0 & { -\frac{1}{2}} \\ 0 & 0 & 0 \\ \end{array}} \right), B^{\dagger} = \left( {\begin{array}{*{20}c} {\frac{4}{{45}}} & {\frac{2}{{45}}} & {\frac{4}{{45}}} \\ 0 & 0 & 0 \\ {\frac{2}{{45}}} & {\frac{1}{{45}}} & {\frac{2}{{45}}} \\ \end{array}} \right), (AB)^{\dagger} = \left( {\begin{array}{*{20}c} {-\frac{1}{{5}}} & 0 & {-\frac{1}{{5}}} \\ 0 & 0 & 0 \\ {-\frac{1}{{10}}} & 0 & {-\frac{1}{{10}}} \\ \end{array}} \right)$$

From this, we have that $\sqrt {\frac{1}{{10}}} = \left\| {\left( {AB} \right)^{\dagger} } \right\|_2 > \left\| {B^{\dagger}} \right\|_2 \left\| {A^{\dagger}} \right\|_2 = \sqrt {\frac{1}{{45}}} \sqrt {\frac{1}{2}} = \frac{1}{3}\sqrt {\frac{1}{{10}}}$.