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You are not logged in. #1 20131118 09:50:07
Polynomials and matrixesHi everyone, Last edited by Bezoux (20131118 09:51:04) #2 20131119 00:48:05
Re: Polynomials and matrixeshi Bezoux and Now, what makes you think that this is a question about eigenvalues? Here's how to get a characteristic polynomial for a square matrix: Multiplying and equating the first and second entries: Solving for lambda: And by a similar method for B These are called the characteristic polynomials for A and for B. But what has that got to do with
If p(A) means 'the characteristic polynomial' then it cannot be equal to a matrix. They just aren't the same thing. where the small letters are the coefficients of a normal polynomial and there are a number of powers of A. [note: There can be no 'constant' matrix at the end, because it would be easy to make any sum of matrices equal to B by a suitable choice of constant term.] If that is correct, then we have to show that no combination of powers of A, multiplied by coefficients, can ever sum to give B (and similarly the other way round). At the moment I cannot think how to do that; but I'm working on it. RSVP Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #3 20131119 01:54:59
Re: Polynomials and matrixesBezoux is referring to a matrix polynomial: 134 books currently added on Goodreads #5 20131119 10:35:43
Re: Polynomials and matrixesNot a clue, unfortunately. Last edited by Nehushtan (20131119 10:41:54) 134 books currently added on Goodreads #6 20131121 09:56:40
Re: Polynomials and matrixesEssentially, my first idea was to assume the opposite, i.e. that there is a polynomial such that p(A)=B, for example (I'd do the p(B)=A one separately). I found a very useful link on finding the nth degree of a matrix (I can't link it unfortunately, but it's the first result that comes up when you google "finding the nth power of a matrix"). Still, something makes me think there's a much more elegant solution that I'm not seeing. #7 20131121 10:08:28
Re: Polynomials and matrixesHi The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #8 20131121 18:42:10
Re: Polynomials and matrixes
I wondered if that would work too. I was hoping that the powers of A would have some common feature that would be incompatible with the equivalent in B, but nothing obvious occurs.
Yes, but how do eigenvalues enter into this problem anyway? You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #9 20131121 21:20:55
Re: Polynomials and matrixesYou need eigenvalues to diagonalize a matrix. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #11 20131122 05:54:09
Re: Polynomials and matrixesBecause you can then get the general form of A^n or B^n. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment 