Given a matrix $A \in \mathbb{R}^{n\times m}$, and its perturbation $$ A_p = A + \Delta $$ is there a way to represent $$ (A_p)^{\star}= (A)^{\star} + f(\Delta) $$ where $(A_p)^{\star}$ ($(A)^{\star}$) is the pseudo-inverse of $A_p$ ($A$)? What can be said about the spectral norm of $f(\Delta)$?
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substituting $A_p=A+\Delta$ into the definition $A_p^\star=\lim_{\epsilon\rightarrow 0}(A_p^\ast A_p+\epsilon I)^{-1}A_p^\ast$, and linearizing in the perturbation $\Delta$, gives $A_p^\star=A^\star+f(\Delta)$ with
$$f(\Delta)= -A^\star\Delta A^\star+ \lim_{\epsilon\rightarrow 0}(A^\ast A+\epsilon I)^{-1}\Delta^*P+{\rm order}(\Delta^2),$$
with $P=I-AA^\star$ the orthogonal projector onto the range of $A$. If $\Delta^\ast P=0$ the simple result
$$A_p^\star=A^\star-A^\star\Delta A^\star+{\rm order}(\Delta^2)$$
is obtained.
answered May 9, 2016 at 10:24
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$\begingroup$ Thanks! If $A$ is full rank than $\Delta^\ast P=0$, correct? Do you have a reference for this? $\endgroup$ Commented May 9, 2016 at 13:59
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$\begingroup$ if $A$ is square and full rank, then $P=0$, but not so if $A$ is full row or column rank and non-square. $\endgroup$ Commented May 9, 2016 at 14:48
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$\begingroup$ I also found here sciencedirect.com/science/article/pii/S0024379509005230 this bound on the norms that is nice: $$ ||f(\Delta)|| \leq \mu \max({{||A||_2}^2,{||A_p||_2}^2})||\Delta ||_2 $$ where $\mu$ is a constant $\endgroup$ Commented May 10, 2016 at 20:10
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$\begingroup$ Isn't $P$ the orthogonal projector onto $\mathcal{R}(A)^\perp$? $\endgroup$ Commented Feb 10, 2023 at 0:48