Timeline for A question about the generalized Lidskii-Wielandt inequality for matrices proved by Thompson and Freede
Current License: CC BY-SA 2.5
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| Dec 29, 2010 at 7:32 | comment | added | user11870 | Yes, such the inequality doesn't hold for eigenvalues. You can consider a well-know result, that is $\vert \sigma(X)-\sigma(Y) \vert$ is weakly majorized by $\sigma(X-Y)$, wherer $\sigma(X)$ deontes the vector of singular values of $X$ arranged in the decreasing order. This inequality just means that $\sigma_i(X)$ and $\simga_i(Y)$ can be exchanged in the inequality due to the absolute value. This is just a special case of the Likskii-Wielandt inequality for singular values. My question is about whether the generalized Likskii-Widlandt still have such "exchange" property. | |
| Dec 29, 2010 at 3:39 | comment | added | Mark Meckes | @Denis: $\alpha_j,\beta_j,\gamma_j$ are singular values, not eigenvalues, so they are all nonnegative. | |
| Dec 28, 2010 at 20:28 | comment | added | Denis Serre | The inequality that you call Lidskii-Wielandt is incorrect. It contains the inequality $|\gamma_i-\alpha_j|\le\beta_1$. There is no reason why $B$ be non-negative. Take $B$ such that $\beta_1$ is negative and you have a contradiction. | |
| Dec 28, 2010 at 20:25 | history | edited | Denis Serre | CC BY-SA 2.5 |
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| Dec 28, 2010 at 17:05 | history | edited | user11870 | CC BY-SA 2.5 |
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| Dec 28, 2010 at 16:59 | history | edited | user11870 | CC BY-SA 2.5 |
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| Dec 28, 2010 at 16:53 | history | asked | user11870 | CC BY-SA 2.5 |