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

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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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,469

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.

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

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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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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,469

hi Nehushtan

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

Bob

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Not a clue, unfortunately.

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.

Ive Google-searched but found very little in the way of help on tackling problems of this sort.

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

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

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,

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,606

Hi

The matrix B has nicer eigen mvalues.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,469

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?

Bob

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,606

You need eigenvalues to diagonalize a matrix.

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**bob bundy****Moderator**- Registered: 2010-06-20
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But why would I want to do that to solve this problem?

B

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Because you can then get the general form of A^n or B^n.

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