# Bound the pseudo-inverse matrix of a product

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.

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

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.