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Jul 7, 2011 at 18:11 comment added Robert Israel More general in what way? Perhaps subtracting $d_i$ from $M_{i,i}$ for each $i$? Then if $\min_i d_i \ge \max_j \lambda_j$, the new matrix will be negative semidefinite.
Jul 6, 2011 at 19:17 comment added sbos Ok, let M be symmetrical matrix
Jul 6, 2011 at 19:02 comment added sbos That's nice, but what would be in more general case?
Jul 6, 2011 at 16:21 history answered Robert Israel CC BY-SA 3.0