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Mar 5, 2019 at 10:40 comment added Mateusz Kwaśnicki If $\alpha_j$ are the coefficients in the eigenvector expansion of $x$, then $\sum |\alpha_j|^2 = 1$ and $\sum |\alpha_j|^2 |\lambda_j - \lambda|^2 \leqslant \varepsilon^2$. That's all one can say, and of course this implies that, for example, $\sum_{j : |\lambda_j - \lambda| > k \varepsilon} |\alpha_j|^2 \leqslant 1/k^2$.
Mar 4, 2019 at 21:45 review First posts
Mar 4, 2019 at 22:10
Mar 4, 2019 at 21:40 history asked user136577 CC BY-SA 4.0