Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given rectangular matrices $A$ and $B$ so that $A \cdot B$ is well-defined, can we bound $\| (AB)^\dagger \|$ in terms of $\| A^\dagger \|$ and $\| B^\dagger \|$? Here, $A^\dagger$ and $B^\dagger$ are the pseudo-inverse matrices of $A$ and $B$, respectively.

share|improve this question
    
my guess $\| (AB)^\dagger \| \leq \| A^\dagger \| \| B^\dagger \|$ –  chaohuang Jul 18 '12 at 15:39

1 Answer 1

There is no such bound. Let $A=\begin{pmatrix}1&0\end{pmatrix}$ be orthogonal projection from plane to line. The norm of $A^\dagger$ is $1$. Next, let $B=\begin{pmatrix}\cos t\\ \sin t\end{pmatrix} $ be isometric embedding of line into plane. The norm of $B^\dagger$ is also $1$, no matter what $t$ is. But $AB=\begin{pmatrix} \cos t\end{pmatrix}$ has pseudoinverse of norm $1/|\cos t|$ as long as $\cos t$ is not zero.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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