Timeline for Is the following map a diffeomorphism?
Current License: CC BY-SA 3.0
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| Jul 28, 2017 at 9:18 | comment | added | Matt Geleta | @SimingTu. I think you're right, thanks. One concern is that I don't know the dimension of the domain of $R$, because of the restriction on the rank of $U$. Since $x_0,\dots, x_k, x_{k+1}$ are assumed to be linearly independent, we know that $x_{k+1}$ is in a space of dimension $n - (k+1)$. So $R$ could only possibly be a diffeomorphism if $n = 2(k+1)$. Thanks for the answer! | |
| Jul 27, 2017 at 12:29 | comment | added | Siming Tu | The image of $R$ is contained in the linear space spanned by $x_0, x_1,ldots,x_k$ (In fact just R(x) is a convex combination of the $k+1$ vectors) . Then the dimension of the domain $n$ is large than the dimension of the image $k+1$ so they can not be homeomorphic to each other so $R$ can not be a diffeomorphism I think. | |
| Jul 26, 2017 at 13:15 | history | edited | Matt Geleta | CC BY-SA 3.0 |
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| Jul 26, 2017 at 13:11 | comment | added | Matt Geleta | @SimingTu: Thanks, good point. I mean that I restrict the map to the $x$ such that $U$ has rank $k+1$. If it so happens that the iterative algorithm produces $x = x_{k+1}$ that results in a $U$ with lower rank, then the mapping $R$ will not be applied. We can assume that $x$ is in an open neighbourhood such that $U$ has rank $k+1$ for all $x_{k+1}$ in that neighbourhood. | |
| Jul 25, 2017 at 15:56 | comment | added | Siming Tu | I am sorry I do not quite understand that why $U$ has rank $k+1$ or you mean that you restrict your map to the $x$ such that $U$ has rank $k+1$? | |
| Jul 21, 2017 at 8:54 | review | First posts | |||
| Jul 21, 2017 at 9:40 | |||||
| Jul 21, 2017 at 8:52 | history | asked | Matt Geleta | CC BY-SA 3.0 |