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- Thread starter jostpuur
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HallsofIvy

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No, it is not the same thing. It is simply taking A to be the inverse of *Df*_{a}.

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All right

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I haven't made much progress with this proof. The problem is not anymore about diagonalizability (well it wasn't in the beginning either...), but since I started talking about this proof here, I might as well continue it here.

First it says that there exists such r, that for [itex]||x||<2r[/itex], [itex]||Dg_x|| < 1/2[/itex] holds. A small question: When a norm of a matrix is written without explanations, does it usually mean the operator norm [itex]||A||:=\textrm{sup}_{||x||<1} ||Ax||[/itex]? Anyway, then it says that "It follows from the mean value theorem that [itex]||g(x)|| < ||x||/2[/itex]". I encountered some problems in this step.

Doesn't the mean value theorem in this case say, that there exists such [itex]0\leq\lambda\leq 1[/itex], that

[tex]

||g(x)|| = \big(\frac{d}{d\lambda'}||g(\lambda' x)||\big) \Big|_{\lambda'=\lambda} + ||g(0)||

[/tex]

I computed

[tex]

\frac{d}{d\lambda'}||g(\lambda' x)|| = \sum_{i=1}^n \frac{g_i(\lambda'x) (x\cdot\nabla g_i(\lambda'x))}{||g(\lambda'x)||} = \frac{g^T(\lambda'x) (Dg_{\lambda'x}) x}{||g(\lambda'x)||}

[/tex]

after which I could estimate

[tex]

||g(x)|| \leq ||Dg_{\lambda x}||\; ||x|| + ||g(0)|| \leq \frac{1}{2}||x|| + ||g(0)||

[/tex]

A big difference is that the proof in the book didn't have [itex]||g(0)||[/itex] term. Perhaps that is not a big problem, we can get rig of it by redefining the original function with some translation of the image. Although it is strange that it was not mentioned in the proof, so is there some thing that I'm already getting wrong here?

Another matter is, that the mapping [itex]\lambda\mapsto ||g(\lambda x)||[/itex] is not nessecarily differentiable, if g reaches zero with some lambda, and I cannot see how to justify that g would remain nonzero here. So the use of the mean value theorem doesn't seem fully justified.

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At least my problem is not, that I could not think complicatedly enough.

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