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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