B
]]>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.
The matrix B has nicer eigen values.
Yes, but how do eigenvalues enter into this problem anyway?
Bob
]]>The matrix B has nicer eigen mvalues.
]]>Still, something makes me think there's a much more elegant solution that I'm not seeing.
]]>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.
]]>Many thanks for that info. I'd not met that before. Any idea about how to do the problem?
Bob
]]>http://en.wikipedia.org/wiki/Matrix_polynomial
(Not to be confused with polynomial matrix, which is a matrix whose entries are polynomials.)
]]>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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