Timeline for Can a perturbation of a matrix product always be represented as product of perturbations of its factor matrices?
Current License: CC BY-SA 3.0
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| when toggle format | what | by | license | comment | |
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| Mar 15, 2017 at 12:37 | vote | accept | jayki | ||
| Mar 15, 2017 at 12:35 | comment | added | jayki | I think this has been almost the answer. Since it turns out that there always exist a sufficiently small perturbation $A'$ of $A$ such that the corresponding matrices $B'$ and $C'$ can always be chosen within any arbitrary small neighborhood of $B$ and $C.$ But this does not mean that any perturbation of $A$ would directly imply an upper-bound on perturbation of $B$ and $C.$ Thanks everyone!! | |
| Mar 15, 2017 at 10:33 | comment | added | Noah Stein | Given that the OP only asked for two perturbations and in the comments only asked to bound their distance from $B$ and $C$ in terms of $\epsilon$ without specifying any particular desired dependence, it seemed reasonable to interpret the question topologically. But I'll let the OP clarify if he wants something stronger. It's also possible that a quantitative bound could be extracted from the paper, but I have not read it in detail. | |
| Mar 15, 2017 at 6:58 | comment | added | domenico fiorenza | so i guess the real question is: can an a priori bound $\delta_{B,C,|\tilde{A}-BC|}$ in terms of $B,C$ and $|\tilde{A}-BC|$ given such that there exist $\tilde{B}$ and $\tilde{C}$ with $\tilde{B}\tilde{C}=\tilde{A}$ and $|\tilde{B}-B|,|\tilde{C}-C|<\delta_{B,C,|\tilde{A}-BC|}$ | |
| Mar 15, 2017 at 6:53 | comment | added | domenico fiorenza | My impression is that the question asks for more than a topological information. For instance, when $n=m=k=1$ consider $B=C=0$. Then the image of the neighborhood $|\tilde{B}|,|\tilde{C}|<\epsilon$ of $(0,0)$ via the multiplication contains an open neighborhood of $0$. But how big? This clearly is $|\tilde{A}|<\epsilon^2$. That is, if we choose an arbitary $\tilde{A}$ with $|\tilde{A}|<\epsilon<1$, we can surely write $\tilde{A}=\tilde{B}\tilde{C}$, with $|\tilde{B}|,|\tilde{C}|<\delta_\epsilon$ for some $\delta_\epsilon$, but we need $\delta_\epsilon\geq\sqrt{\epsilon}>\epsilon$. | |
| Mar 15, 2017 at 3:56 | history | answered | Noah Stein | CC BY-SA 3.0 |