Desperate for Matrix Algebra help!

[Math_Struggles]

I really need some help with this proof:

 

 

If A and B are nxn symmetric matrices, and x is a column vector with x^t as it's transpose:

 

Is it true that x^tAxBx = x^tBxAx?

 

I arrived at this because I am trying to prove that the Generalized Rayleigh Quotient (Ax dot x)/(Bx dot x) has at least one critical point.

 

 

 

If the first statement is not true, please help me with the second part.

 

 

 

Thanks!

 

There will be more proofs to follow after this if I run into more trouble, because the problem is multiple steps of proofs.

[DrRocket]

I really need some help with this proof:

 

 

If A and B are nxn symmetric matrices, and x is a column vector with x^t as it's transpose:

 

Is it true that x^tAxBx = x^tBxAx?

 

I arrived at this because I am trying to prove that the Generalized Rayleigh Quotient (Ax dot x)/(Bx dot x) has at least one critical point.

 

 

 

If the first statement is not true, please help me with the second part.

 

 

 

Thanks!

 

There will be more proofs to follow after this if I run into more trouble, because the problem is multiple steps of proofs.

 

You have a problem with the term, AxBx. It doesn't make sense since Ax and Bx are both column vectors.

 

 

Did you mean to write x^tABx = x^tBAx ?

 

If so, think about the transpose of each side and about the dimension of the indicated product.

[Math_Struggles]

I forgot some parentheses:

 

It should be (x^tAx)Bx = (x^tBx)Ax

 

If that's not right can you please help with proving that the Generalized Rayleigh Quotient has at least one critical point?

[DrRocket]

I forgot some parentheses:

 

It should be (x^tAx)Bx = (x^tBx)Ax

 

If that's not right can you please help with proving that the Generalized Rayleigh Quotient has at least one critical point?

 

It is not correct that in general (x^tAx)Bx = (x^tBx)Ax for symmetric A and B (even if they are positive-definite).

 

For an ordinary Rayleigh quotient R(M,x) where M is a symmetric matrix, the critical points are eigenvectors of M. This problem is well beyond just matrix algebra. The generalized Rayleigh quotient requires symmetric positive-definite matrices. Are your A and B positive-definite ?

 

What are you really doing ?

 

You might want to read this Wiki article http://en.wikipedia.org/wiki/Rayleigh_quotient

[Math_Struggles]

We only know that B is positive definite. I have no idea where to start. If I had some idea where to start I think I could figure much of it out, but I really have on clue!

 

 

 

Any help is so much appreciated.

 

Thanks!

[DrRocket]

We only know that B is positive definite. I have no idea where to start. If I had some idea where to start I think I could figure much of it out, but I really have on clue!

 

 

 

Any help is so much appreciated.

 

Thanks!

 

 

You still have not explained what it really is that you are trying to do, or the source of the question. It would also help to know something of your background in mathematics.

 

The question of critical points of the rayleigh quotient ties back to your thread on maxima and minima. But this question is a bit advanced for either a calculus class or a class on matrices and linear algebra.

 

Moreover, the Wiki article ought to give you a couple of ideas where to start.

 

This is not your basic undergraduate homework problem.

[Math_Struggles]

I thought I said that we are trying to prove that Q(x) has at least one critical point, where Q(x) is the Generalize Rayleigh Quotient (Ax dot x)/(Bx dot x). And it is an undergraduate homework problem, it's just the weekend so the teacher doesn't have office hours... The homework is for a multivariable calculus class. This is only the first part of the problem too.

 

The given information is that A and B are real symmetric n x n matrices, and B is positive definite.

 

(a) I first need to prove Q has at least one critical point. (b) Then I need to prove that x is a CP of Q iff Ax = (lambda)Bx where lambda is a real eigenvalue. © Next I need to prove that lambda from b satisfies lambda = Q(x). (D) Finally, I need to prove that in part, det(A-(lambda)B) = 0.

 

 

I have no idea where to start, our teacher has a habit of giving us extremely difficult homework problems, but she usually has office hours so we may get help.

 

 

 

Thank you in advance for any help you can give me!

[DrRocket]

I thought I said that we are trying to prove that Q(x) has at least one critical point, where Q(x) is the Generalize Rayleigh Quotient (Ax dot x)/(Bx dot x). And it is an undergraduate homework problem, it's just the weekend so the teacher doesn't have office hours... The homework is for a multivariable calculus class. This is only the first part of the problem too.

 

The given information is that A and B are real symmetric n x n matrices, and B is positive definite.

 

(a) I first need to prove Q has at least one critical point. (b) Then I need to prove that x is a CP of Q iff Ax = (lambda)Bx where lambda is a real eigenvalue. © Next I need to prove that lambda from b satisfies lambda = Q(x). (D) Finally, I need to prove that in part, det(A-(lambda)B) = 0.

 

 

I have no idea where to start, our teacher has a habit of giving us extremely difficult homework problems, but she usually has office hours so we may get help.

 

 

 

Thank you in advance for any help you can give me!

 

Choose a basis in which B is diagonal.

 

Read the Wiki article and vview this as a constrained optimization problem. Use Lagrange multipliers.

[Math_Struggles]

How would I prove that if x is a critical point, Ax = (lambda)Bx?

[A Tripolation]

How would I prove that if x is a critical point, Ax = (lambda)Bx?

 

For an ordinary Rayleigh quotient R(M,x) where M is a symmetric matrix, the critical points are eigenvectors of M.

 

 

Math_Struggles, you do realize that [math]Ax = \lambda Bx[/math] is the formula for an eigenvalue, from which you can then obtain an eigenvector, correct?

[Math_Struggles]

Yes but we don't know that B is necessarily a diagonal matrix. Does it matter?

[A Tripolation]

Yes but we don't know that B is necessarily a diagonal matrix.

 

Oh, ok. I must've skipped that bit.

[DrRocket]

Yes but we don't know that B is necessarily a diagonal matrix. Does it matter?

 

B is symmetric and positive definite, therefore diagonalizable with positive entries on the diagonal.

http://en.wikipedia....alizable_matrix

 

You don't have to diagonalize B, but I think it will make life easier. It certainly doesn't hurt and there is nothing lost.

 

Note that you cannot necessarily diagonalize your matrices A and B simultaneously. They need not commute.

Edited November 22, 2011 by DrRocket