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I'm trying to resolve the following:
given a function
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hi onako,
I'm sorry but this isn't making sense to me.
(i) If x, A, B, R, D are matrices then f(x) is, too.
So how do you expect to prove f(x zero) > f(x one)
I am not aware that one matrix can be greater than another.
(ii)
with the being the diagonal matrix with diagonal entries of A (and R containing the off-diagonal entries of A),
Don't understand what this means. Can you post an example?
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
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Sorry, I thought the notation would suffice.
x and b are vectors;
A, R, D are matrices;
as for the second question: suppose a square matrix A is given.
Now split A = D + R, such that D is a square diagonal matrix with diagonal entries of A
(hence, the matrix R entries would be the off-diagonal entries of A)
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Hi onako;
Are x0 and b column vectors or row vectors?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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The function is f(x), resulting in a value, it is clear that
x_0, x_1, and b are column vectors, and their transposes are row vectors.
(vectors are oriented to allow the operations)
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Hi onako;
You say the fuction outputs a scalar. Have you made up a 2 X 2 example and tried it?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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OK, I reduced the problem to the following: Suppose a square matrix A is given, with D being a diagonal matrix containing only its diagonal entries
which are positive.
It is known that the eigenvalues of
Last edited by onako (2012-03-01 03:07:27)
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Given that all the eigenvalues of I-B, where I is the identity matrix, satisfy
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