Timeline for Inequality on matrix trace
Current License: CC BY-SA 4.0
17 events
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| Feb 2, 2023 at 8:20 | comment | added | Fawen90 | @NarutakaOZAWA More precisely, what is the definition of $\|\gamma-\gamma_1\|_{tr}$? In particular, how can we dominate the difference $|tr((\gamma-\gamma_1) L)|$ in terms of $\|\gamma-\gamma_1\|_{tr}\le 2(1-tr(\gamma B))$? | |
| Feb 1, 2023 at 22:30 | comment | added | Fawen90 | @NarutakaOZAWA For example, how to estimate the difference $tr\big(L(\gamma-\gamma_1)LB\big)$? Why this estimation allows to derive the desired inequality? | |
| Feb 1, 2023 at 22:29 | comment | added | Fawen90 | @NarutakaOZAWA Thanks a lot for the precision, while I still don't know how to estimate the difference for the terms $tr(L\gamma LB), tr(B(L\gamma + \gamma L))tr(BL + \gamma L), tr(B\gamma)tr(BL)tr(\gamma L), tr\big((Id-\gamma)B\big)$ when replacing $\gamma$ by $\gamma_1$. Do you mind provide more details on the calculus? Many thanks | |
| Feb 1, 2023 at 6:21 | comment | added | Narutaka OZAWA | @Fawen90: You misunderstood what I say. Just replace $\gamma=\mathrm{diag}(\lambda_1,\lambda_2,\ldots)$ everywhere with the rank-one projection $\gamma_1 = \mathrm{diag}(1,0,\ldots)$ by paying the cost $\| \gamma-\gamma_1 \|_{\mathrm{tr}}\le 2(1-\mathrm{tr}(\gamma B))$ at each time. | |
| Feb 1, 2023 at 5:38 | comment | added | Fawen90 | @NarutakaOZAWA Let $I^{i,j}$ be the matrix whose only non zero element is $I^{i,j}_{i,j}=1$. Then for $\gamma=\sum_i \gamma_i I^{i,i}$, how can we obtain the estimation $f(\gamma) \le C \sum_i f(\gamma_i I^{i,i})$ for some constant $C$? | |
| Jan 31, 2023 at 6:35 | comment | added | Fawen90 | @NarutakaOZAWA Thank you Narutaka for your explanation. The difficulty for me is that the r.h.s. of the desired inequality is affine in $\gamma$ while the l.h.s. is not (I don't know how to show it is convex in $\gamma$), see PS2 above. Could you please let me know how to proceed to the case where $\gamma$ has rank one? A detailed solution is highly appreciated and thanks in advance for your help! | |
| Jan 31, 2023 at 6:32 | history | edited | Fawen90 | CC BY-SA 4.0 |
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| Jan 31, 2023 at 0:10 | comment | added | Narutaka OZAWA | @Fawen90: We may assume $\gamma=\mathrm{diag}(\lambda_1,\lambda_2,\ldots)$ with $\lambda_1\geq\lambda_2\geq\cdots$. Then the only way $\mathrm{tr}(\gamma B)=\sum_i \lambda_i B_{ii}$ is close to $\mathrm{tr}(B)=1$ is $\lambda_1\approx1$. | |
| Jan 30, 2023 at 14:03 | comment | added | Fawen90 | @NarutakaOZAWA More precisely, "The tr-norm distance from $\gamma$ (or $B$) to a rank-one projection is dominated by a constant (perhaps 2) multiple of $tr((1-\gamma)B)$" is not clear to me. Could you please write the corresponding inequality? | |
| Jan 30, 2023 at 13:34 | comment | added | Fawen90 | @NarutakaOZAWA Thanks so much for pointing out this. Could you please detail how to prove what you have mentioned? I'm not very familiar with the tricks in matrix analysis and I'm interested in this inequality. Thank you very kindly for your consideration | |
| Jan 30, 2023 at 6:58 | comment | added | Narutaka OZAWA | This is true, but for an uninteresting reason. The $\mathrm{tr}$-norm distance from $\gamma$ (or $B$) to a rank-one projection is dominated by a constant (perhaps $2$) multiple of $\mathrm{tr}((1-\gamma)B)$. Hence the problem reduces to the case where $\gamma$ has rank one (modulo changing the constant in front of $\mathrm{tr}((1-\gamma)B)$ as you suggested). | |
| Jan 27, 2023 at 17:06 | comment | added | Joseph Van Name | Using computer experiments with gradient descent (I used the Flux library in the language Julia). I conjecture the inequality $$|\text{tr}(L\gamma LB)−0.5\cdot \text{tr}(B(L\gamma+\gamma L))\text{tr}(BL+\gamma L)+\text{tr}(B\gamma)\text{tr}(BL)\text{tr}(\gamma L)|\leq \|L\|_p^2\cdot\text{tr}((Id−\gamma)B)⋅4^{1/p}$$ for $1\leq p\leq\infty$ where $\|\cdot\|_p$ denotes the Schatten $p$-norm. | |
| Jan 27, 2023 at 17:04 | comment | added | Joseph Van Name | What you had is right. I was getting the inequality in the wrong direction. | |
| Jan 27, 2023 at 9:47 | history | edited | Fawen90 | CC BY-SA 4.0 |
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| Jan 27, 2023 at 7:53 | comment | added | Fawen90 | @JosephVanName That's a really unexpected observation. Could you please give more details how computation yields the inverse inequality? I know very few on computer experiments. Many thanks | |
| Jan 26, 2023 at 16:54 | history | edited | Fawen90 | CC BY-SA 4.0 |
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| Jan 26, 2023 at 15:10 | history | asked | Fawen90 | CC BY-SA 4.0 |