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Timeline for On minimal eigenvalue

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Sep 11 at 7:25 comment added Jasmine Then how would the method work for $M = \mathbf{x}\mathbf{x}^\top + \mathbf{y}\mathbf{y}^\top + \mathbf{z}\mathbf{z}^\top$ for $\mathbf{x}, \mathbf{y}, \mathbf{z} \in \mathbb{R}^5$ with $\|\mathbf{x}\| = \|\mathbf{y}\| = \|\mathbf{z}\| = 1$? Thanks.
Sep 11 at 6:41 history edited Ilya Bogdanov CC BY-SA 4.0
A wrong proof og Gen.1 is removed.
Sep 11 at 6:38 comment added Ilya Bogdanov Oh, you are right! So *this * generalization is not proved yet, I will edit the answer. (Anyway, there is no need in commenting at both posts...)
Sep 11 at 2:00 comment added Jasmine Why did you obtain $$ d_i\geq \lambda_i\cdot \left(\frac{t_i-\lambda_i}{d-2}\right)^{d-2} $$ rather than the opposite direction using the AM-GM inequality? Thanks.
Sep 10 at 20:20 comment added Ilya Bogdanov That was a typo; now corrected, thanks.
Sep 10 at 20:19 history edited Ilya Bogdanov CC BY-SA 4.0
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Sep 10 at 14:35 comment added Jasmine How did you obtain $\det\Gamma=\det XX^\top$? Thanks.
Sep 9 at 7:57 comment added Ilya Bogdanov Sorry, I think now you can think a bit by yourself. Here, pne post is usually for one Q+A, not for such a large thread.
Sep 9 at 1:08 comment added Jasmine What if $ M = \mathbf{x} \mathbf{x}^{\top} + \mathbf{y} \mathbf{y}^{\top} + \mathbf{z} \mathbf{z}^{\top} $ for all $\mathbf{x}, \mathbf{y}, \mathbf{z} \in \mathbb{R}^5$ with $\|\mathbf{x}\| = \|\mathbf{y}\| = \|\mathbf{z}\| = 1$? Thanks.
Sep 8 at 7:02 comment added Jasmine Thank you for your answer. What if $M = x_1 x_1^\top + x_2 x_2^\top$, for all $x_1, x_2 \in \mathbb{R}^3$ with $\|x_1\| = \|x_2\| = 1$ and $\langle x_1, x_2 \rangle = 0$? Thanks!
Sep 6 at 8:45 comment added Ilya Bogdanov Upper; sorry, edited.
Sep 6 at 8:44 history edited Ilya Bogdanov CC BY-SA 4.0
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Sep 4 at 10:30 history edited Ilya Bogdanov CC BY-SA 4.0
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Sep 4 at 7:44 comment added Ilya Bogdanov I realized that the ikey estimate becomes more transparent by means of the Cauchy--Binet formula, see edits. I have also added an estimate for $\mathbf x_1,\mathbf x_2\in\mathbb R^4$, but I do not know whether it is sharp.
Sep 4 at 7:43 history edited Ilya Bogdanov CC BY-SA 4.0
The initial estimate is rewritten by means of Cauchy--The secind generalization is added.
Sep 3 at 13:16 comment added Ilya Bogdanov @Jasmine I am unsure of what result would you expect from the method in the general setup…
Aug 31 at 14:26 comment added Ilya Bogdanov I have added a proof of the $d$-dimensional generalization.
Aug 31 at 14:23 history edited Ilya Bogdanov CC BY-SA 4.0
A proof of the generalization is added.
Aug 31 at 13:07 history answered Ilya Bogdanov CC BY-SA 4.0