If $ABBA-BAAB=A-B$, show $\operatorname {tr} …
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 $\lambda$.
How many ways can we get 2 a's and 2 b's from aabb?
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 …
calculus - Is convolution kind of like “carry” in decimal addition ...
Apr 5, 2026 · I’ve been trying to build some intuition around convolution in signal processing and I thought of an analogy with carry in decimal arithmetic, but I’m not sure if it’s valid. When we add …
$A^2=AB+BA$. Prove that $\det (AB-BA)=0$ [duplicate]
I get the trick. Use the fact that matrices "commute under determinants". +1
If $ab+ba=1$ and $a^3=a$ in a ring, show that $a^2=1$.
Sep 19, 2023 · The first condition holds for all $b$ or for some particular $b$?
How to calculate total combinations for AABB and ABBB sets?
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: …
elementary number theory - Divisibility Tests for Palindromes ...
I've found and proven the following extensions to palindromes of the usual divisibility rules for 3 and 9: A palindrome is divisible by 27 if and only if its digit sum is. A palindrome is divisible by 81 if and only if …
Matrices - Conditions for $AB+BA=0$ - Mathematics Stack Exchange
There must be something missing since taking $B$ to be the zero matrix will work for any $A$.
Prove $\frac 12 (AB+BA)$ is hermitian when $A$ and $B$ are hermitian
Aug 31, 2019 · $A$ and $B$ are hermitian, so I know they must commute. So $AB-BA = 0$. But I don't think I can get very far with that. I just totally don't know how to start. Some ...
elementary number theory - Common factors for all palindromes ...
For example a palindrome of length $4$ is always divisible by $11$ because palindromes of length $4$ are in the form of: $$\\overline{abba}$$ so it is equal to $$1001a+110b$$ and $1001$ and $110$ are