0
$\begingroup$

Consider $\| B(AB)^\dagger \|_2$ where $A$ is a real matrix, $B$ is a real, square and symmetric matrix, and $(AB)^\dagger$ is the Moore-Penrose pseudoinverse of $AB$. Is $\| B(AB)^\dagger \|_2$ uniformly bounded over all non-singular symmetric matrices $B$? If not, what about over all positive diagonal matrices $B$?

For example, when $A$ and $B$ are both non-singular, $\| B(AB)^\dagger \|_2 = \| A^{-1} \|_2 $, and the norm of interest is uniformly bounded for all $B$. My question is about whether a similar result exists for when $A$ is not invertible. Some relevant references would be very helpful.

$\endgroup$
0
$\begingroup$

The answer to your first question is "no". Consider e.g. $$ A=\pmatrix{0&1\\ 0&0},\ B=\pmatrix{1&t\\ t&t} $$ where $0<t<1$. Then $$ B(AB)^+ =\pmatrix{1&t\\ t&t}\pmatrix{t&t\\ 0&0}^+ =\pmatrix{1&t\\ t&t}\pmatrix{\frac12t^{-1}&0\\ \frac12t^{-1}&0} =\pmatrix{\frac12(t^{-1}+1)&0\\ 1&0}, $$ whose norm is not bounded above.

$\endgroup$
1
  • $\begingroup$ Good example. I do think it is uniformly bounded over all positive diagonal matrices. Do you have any suggestions here? $\endgroup$
    – user123456
    May 23 '21 at 5:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.