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May 29, 2018 at 20:35 comment added Mahdi - Free Palestine @DenisSerre: I think I fixed it.
May 29, 2018 at 20:31 history edited Mahdi - Free Palestine CC BY-SA 4.0
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May 29, 2018 at 6:30 comment added Denis Serre Von Neumann's trace Inequality is in terms of singular values, not eigenvalues. Since $C^{1/2}A^{1/2}$ is not symmetric in general, its singular values do not equal its eigenvalues (they tend to be larger). Your second inequality is therefore false.
May 29, 2018 at 6:28 history edited Denis Serre CC BY-SA 4.0
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May 28, 2018 at 19:19 vote accept Yair Carmon
May 28, 2018 at 13:26 history edited Mahdi - Free Palestine CC BY-SA 4.0
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May 28, 2018 at 13:24 comment added Denis Serre How do you justify the second inequality ?
May 28, 2018 at 13:13 history edited Mahdi - Free Palestine CC BY-SA 4.0
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May 28, 2018 at 13:05 history answered Mahdi - Free Palestine CC BY-SA 4.0