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Dec 26, 2021 at 2:33 history edited Michael Hardy CC BY-SA 4.0
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Dec 25, 2021 at 14:50 history edited dohmatob CC BY-SA 4.0
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Dec 25, 2021 at 14:20 history edited dohmatob CC BY-SA 4.0
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Dec 25, 2021 at 13:15 comment added dohmatob Yes, that's a typo that I couldn't correct due to time limit for editing comments. But ma point still remains, about simplifying by introducing an intermediate step.
Dec 25, 2021 at 12:48 comment added Dustin G. Mixon @dohmatob - In your notation, $\lambda$ is the vector of square roots of eigenvalues of $A$.
Dec 25, 2021 at 10:04 vote accept dohmatob
Dec 25, 2021 at 10:02 comment added dohmatob Also, in the final calculation, just before applying J. Tropp's (4.6), it might help to write $\|\Lambda^{1/2} G\|^2_{2 \to 2} = \|B \odot G\|^2_{2 \to 2}$, where $B = \lambda \otimes 1_k \in \mathbb R^{n \times k}$, where $\lambda \in \mathbb R^n$ is the vector of eigenvalues of $A$ (in any order).
Dec 25, 2021 at 3:01 history answered Dustin G. Mixon CC BY-SA 4.0