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#1 2013-11-17 10:50:07

Bezoux
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Registered: 2013-11-17
Posts: 7

Polynomials and matrixes

Hi everyone,
I've encountered a problem while studying matrices.
A={{2,1},{3,-1}}, B={{4,-2},{3,-1}}
Prove that there does not exist a polynomial with real coefficients such that p(A)=B or p(B)=A.
I've read up on eigenvalues, eigenvectors, characteristic polynomials and diagonalization, but nothing seems to be making sense, as the whole thing gets way too complicated for a high school problem.
I think there's a way to do this without using any of the aforementioned.
Can you help me out, please?

Last edited by Bezoux (2013-11-17 10:51:04)

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#2 2013-11-18 01:48:05

bob bundy
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Registered: 2010-06-20
Posts: 6,269

Re: Polynomials and matrixes

hi Bezoux

Welcome to the forum.

I have to admit straight away that I cannot do this question.  I held off making a post in the hope someone else would, but it doesn't look like that is going to happen, so I'll jump in with what I have.  Maybe someone will notice who can put us both straight.  smile

Firstly, I'm assuming those are 2 by 2 matrices.  This is how to display them; click the matrix and you will see the underlying Latex code.

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

p(A)=B or p(B)=A.

If p(A) means 'the characteristic polynomial' then it cannot be equal to a matrix.  They just aren't the same thing.

So what does p(A) mean?  Do you have anything in your notes / textbook that tells us, because I don't recognise the notation.

The only thing I can think of is this:

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

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#3 2013-11-18 02:54:59

Nehushtan
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From: London
Registered: 2013-03-09
Posts: 611
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Re: Polynomials and matrixes

Bezoux is referring to a matrix polynomial:

http://en.wikipedia.org/wiki/Matrix_polynomial

(Not to be confused with “polynomial matrix”, which is a matrix whose entries are polynomials.)


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#4 2013-11-18 06:49:44

bob bundy
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Registered: 2010-06-20
Posts: 6,269

Re: Polynomials and matrixes

hi Nehushtan

Many thanks for that info.  I'd not met that before.  Any idea about how to do the problem?

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#5 2013-11-18 11:35:43

Nehushtan
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From: London
Registered: 2013-03-09
Posts: 611
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Re: Polynomials and matrixes

Not a clue, unfortunately. sad

The Wikipedia article is very badly written. Also, Wolfram MathWorld has a different definition of matrix polynomial (http://mathworld.wolfram.com/MatrixPolynomial.html) defining it as a polynomial with matrix coefficients rather than matrix variables – but I think the Wikipedia definition makes more sense. In other words, if p(x) is a polynomial, p(A) is the matrix polynomial obtained by replacing the variable x by the matrix A and the constant term by a[sub]0[/sub]I[sub]2[/sub] where I[sub]2[/sub] (= A[sup]0[/sup]) is the 2×2 identity matrix.

I’ve Google-searched but found very little in the way of help on tackling problems of this sort. hmm

Last edited by Nehushtan (2013-11-18 11:41:54)


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#6 2013-11-20 10:56:40

Bezoux
Member
Registered: 2013-11-17
Posts: 7

Re: Polynomials and matrixes

Essentially, 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).
Then,

, so if I find A^n for any natural number N (which can be done via diagonalization, if I'm not mistaken), I could get a contradiction out of it, but the process is a bit too complicated for such ugly numbers (the roots of the first quadratic equation have square roots of 21 in them).
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.

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#7 2013-11-20 11:08:28

anonimnystefy
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Registered: 2011-05-23
Posts: 14,883

Re: Polynomials and matrixes

Hi

The matrix B has nicer eigen mvalues.


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#8 2013-11-20 19:42:10

bob bundy
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Registered: 2010-06-20
Posts: 6,269

Re: Polynomials and matrixes

Bezoux wrote:

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 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.

I'm also trying to construct a series of geometric transformations equivalent to A and also for B, in the hope that something will show the required result. 

Stefy wrote:

The matrix B has nicer eigen values.

Yes, but how do eigenvalues enter into this problem anyway?  dizzy

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#9 2013-11-20 22:20:55

anonimnystefy
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Re: Polynomials and matrixes

You need eigenvalues to diagonalize a matrix.


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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#10 2013-11-21 03:01:22

bob bundy
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Registered: 2010-06-20
Posts: 6,269

Re: Polynomials and matrixes

But why would I want to do that to solve this problem?

B


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#11 2013-11-21 06:54:09

anonimnystefy
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Registered: 2011-05-23
Posts: 14,883

Re: Polynomials and matrixes

Because you can then get the general form of A^n or B^n.


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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