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S Jan 22 at 5:19 history suggested 5th decile CC BY-SA 4.0
Corrected my username
Jan 21 at 19:21 review Suggested edits
S Jan 22 at 5:19
Jan 20 at 23:06 comment added loup blanc @LSpice, yes it is.
Jan 20 at 19:04 history edited LSpice CC BY-SA 4.0
`\operatorname`
Jan 20 at 18:59 comment added LSpice Is the reference to a bound obtained/used by Thibaut Demaerel to the first answer by the user currently called Vergilius?
May 9, 2020 at 2:30 comment added Nathaniel Johnston Can we simplify this proof by using the fact that every $A \in M_n(K)$ can be written in the form $A = XSY$, where $X$ and $Y$ are invertible and $S = \mathrm{diag}(1,...,1,0,...,0)$ (with $\mathrm{rank}(A)$ ones on the diagonal)? I think that works over any field ($X$ and $Y$ just come from doing row and column operations on $A$, respectively), and it lets you assume that $A = \mathrm{diag}(1,...,1,0,...,0)$ instead of the more complicated block matrix form, right?
May 4, 2020 at 12:21 history answered loup blanc CC BY-SA 4.0