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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****Administrator**- Registered: 2010-06-20
- Posts: 8,447

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

Children are not defined by school ...........The Fonz

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

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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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****Administrator**- Registered: 2010-06-20
- Posts: 8,447

hi Nehushtan

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

Bob

Children are not defined by school ...........The Fonz

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

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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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: Harlan's World
- Registered: 2011-05-23
- Posts: 16,039

Hi

The matrix B has nicer eigen mvalues.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,447

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

Children are not defined by school ...........The Fonz

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

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,039

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

The knowledge of some things as a function of age is a delta function.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,447

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

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,039

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

The knowledge of some things as a function of age is a delta function.

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