Let $C $ be a non-zero real matrix and let $A$ and $B$ be such that $C = AB.$

For any arbitrary matrix $X$ let $X^{\dagger}$ denote the Moore-Penrose inverse of $X$ and let $\|X\|$ denote the largest singular value of $X$.

Is it true that $\|C^{\dagger}\| \leq \| A^{\dagger}\| \| B^{\dagger} \|$?

The above is obviously true for $A$,$B$ for which $(AB)^{\dagger} = B^{\dagger}A^{\dagger}$, there are many such matrices for this is true ( see https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_pseudoinverse ) but there are $A$,$B$ for which this is not true see https://math.stackexchange.com/q/457408 (but even in the case discussed in the previous link the above inequality is true).

In case the inequality is false, are there any good bounds for $\|A^{\dagger}\| \|B^{\dagger}\| - \|(AB)^{\dagger}\|$?

Added later: Mark Yasuda has constructed a counterexample here: https://mathoverflow.net/questions/248367/norm-of-moore-penrose-inverse-of-a-product/248445#248445

  • $\begingroup$ Note: I constructed and posted a small counterexample to the question of whether $\|(AB)^{\dagger}\| \leq \| A^{\dagger}\| \| B^{\dagger} \|$ always holds for the spectral norm at this question's MathOverflow repost. $\endgroup$ – Mark Yasuda Aug 28 '16 at 0:08

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