Jan 26, 2026 · That is, there seems to be fairly strong symbolic evidence that for $n=4$, if $ABBA-BAAB = A-B$ and $A$ is nilpotent, then $B^4 = \lambda I$ for some $\lambda$.

Jun 23, 2022 · Are you required to make it wiht polar transformation? Because with the change $x=uv$ $y=u (1-v)$ it's easier

Mar 22, 2026 · Back in the day, CS was considered a branch of mathematics. Knuth and Dijkstra and the other giants of that era were all mathematicians first. Nothing much has changed besides the …

Jan 14, 2023 · According to this question, there is CW complex with one 0-cell,one 1-cell and one 2-cell. No such CW structure exists on the the Möbius strip. Moreover the linked question doesn't claim that, …

I get the trick. Use the fact that matrices "commute under determinants". +1

Apr 19, 2022 · Although both belong to a much broad combination of N=2 and n=4 (AAAA, ABBA, BBBB...), where order matters and repetition is allowed, both can be rearranged in different ways: …

Because abab is the same as aabb. I was how to solve these problems with the blank slot method, i.e. _ _ _ _. If I do this manually, it's clear to me the answer is 6, aabb abab abba baba bbaa baab Which is …

Hint: Here you can limit the size of the possible solutions reasonably easily - suppose $a\ge b \gt 3$ then $a^b\gt a^3 \gt 3ab$

Jun 8, 2018 · Assume that matrix $A$ and $B$ are square with same dimensions. prove by mathematical induction that $(ABA^{-1})^k=AB^kA^{-1}$ if I let $k=2$ then I think I can do ...

Feb 28, 2018 · Hint: in digits the number is $abba$ with $2 (a+b)$ divisible by $3$.