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Nov 26, 2022 at 21:04 vote accept Andrew
Nov 26, 2022 at 19:20 comment added Andrew Jason DeVito Thanks, you really helped me out
Nov 26, 2022 at 19:19 comment added Jason DeVito - on hiatus @Andrew: The matrix $-B$ of my answer works for $odd\times odd$.
Nov 26, 2022 at 18:15 comment added Jason DeVito - on hiatus My guess is that by playing around with various Jordan blocks, you can find examples of $n\times n$ matrices having precisely $2k$ square roots for any given $k\leq n$. I didn't think too deeply about this, though.
Nov 26, 2022 at 18:14 history answered Jason DeVito - on hiatus CC BY-SA 4.0