Z. L.
Reputation
503
Top tag
Next privilege 1,000 Rep.
Create new tags
 Apr 1 comment A matrix with trace $\leq n$ is conjugate to a matrix with all entries $0,1,-1$ As I said in other comment, I suspect that all matrices coming from the action on homology of (maybe periodic) automorphisms of manifolds satisfy that they are conjugate to matrices with only 0,1,-1. I asked to see if someone knew a characterization or something related. But I agree that you could ask the same question about other kind of matrices. For example, you can ask pretty much anything (without context) about finite groups and someone will answer. Mathematics should be self-explanatory. I don't know why telling you the reason for posing that question to myself would help to solve it. Mar 31 comment A matrix with trace $\leq n$ is conjugate to a matrix with all entries $0,1,-1$ I don't really suspect it should exist. I came up with this question by finding some families of matrices that did satisfy the property, and since it is a question that makes sense and I couldn't answer it, I posted it. Also, I am working on something unrelated at the moment so I'll rethink it in a few days and wait if someone knows something about the second question (or its difficulty) Mar 31 comment A matrix with trace $\leq n$ is conjugate to a matrix with all entries $0,1,-1$ Do you think that question applies for MO? Mar 31 comment A matrix with trace $\leq n$ is conjugate to a matrix with all entries $0,1,-1$ Ok, I upvoted because that answers my first question. Do you know anything about the second? Maybe some geometrical criterion, such the matrix coming from the action of a diffeomorphism on the integral homology of a manifold? Mar 31 comment A matrix with trace $\leq n$ is conjugate to a matrix with all entries $0,1,-1$ Thank you for the source, I will have a look at it. Mar 31 awarded Custodian Mar 31 reviewed Approve A matrix with trace $\leq n$ is conjugate to a matrix with all entries $0,1,-1$ Mar 31 revised A matrix with trace $\leq n$ is conjugate to a matrix with all entries $0,1,-1$ I changed the trace by the absolute value of the trace, doesn't change a lot the question... Mar 31 comment Show A is diagonalizable if and only if A is similar to a diagonal matrix. Also your condition for similarity is very strong. You should write that there exists a matrix $M$ such that $A=MBM^{-1}$ Mar 31 comment Show A is diagonalizable if and only if A is similar to a diagonal matrix. But that is not true, the square identity matrix of order n is diagonalizable and has not n distinct eigenvalues Mar 31 revised A matrix with trace $\leq n$ is conjugate to a matrix with all entries $0,1,-1$ I explained a little bit more notation Mar 31 comment Induced orientation on manifold There are several (two) conventions about orientation "induced". Also, outward has meaning in an ambience space and codimension $1$, otherwise you need to define it. Mar 31 revised A matrix with trace $\leq n$ is conjugate to a matrix with all entries $0,1,-1$ added 58 characters in body Mar 31 revised A matrix with trace $\leq n$ is conjugate to a matrix with all entries $0,1,-1$ added 22 characters in body; edited tags Mar 31 comment What would be the impact of a formula which explains the structure of primes? I don't agree that question 1. is not subjective. What does "exhaustively and precisely" mean? If that means that that formula would be enough for being able to solve any formal question about prime numbers, then probably that formula would end with what is known by "number theory" now. Mar 31 asked A matrix with trace $\leq n$ is conjugate to a matrix with all entries $0,1,-1$ Jan 14 awarded Nice Question Feb 21 comment To prove: $[K : \mathbb{Q}] = 2 \ \Longrightarrow \ \exists \zeta \text{ primitive root of unity}, \ \mathbb{Q}(\zeta) \ \supseteq \ K$ Do you mean $Q \subset L \subset K$? Jan 8 awarded Yearling Sep 24 awarded Autobiographer