Generalized Rayleigh Quotient

Given the Generalized Rayleigh Quotient

Q(x) = (Ax dot x)/(Bx dot x)

and that A and B are both real n x n matrices, and B is positive definite.

How would I go about proving that Q(x) has at least one critical point. In class we restricted ||x|| to 1 so that ||x||^2 also is 1, but we never actually see ||x||^2 in this equation, so it doesn't seem like that would be helpful. If someone can please help me with where to start that'd be awesome! I know that taking the gradient is the first step, but after that I'm utterly confused.

Re: Generalized Rayleigh Quotient

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**tubetess123** Given the Generalized Rayleigh Quotient Q(x) = (Ax dot x)/(Bx dot x) and that A and B are both real n x n matrices, and B is positive definite

Surely you forgot to say that $\displaystyle A,B$ are symmetric. __Hint__: Use the substitution $\displaystyle x=Py$ where $\displaystyle P\in \mathbb{R}^{n\times n}$ is a non singular matrix such that $\displaystyle P^tAP=D=\textrm{diag}\;(\lambda_1,\ldots,\lambda_n )$ and $\displaystyle P^tBP=I$ being $\displaystyle \lambda_j$ the generalized eigenvalues associated to $\displaystyle A$ and $\displaystyle B$ .

Re: Generalized Rayleigh Quotient

We aren't given that A is positive definite. Also, why does substituting that particular matrix into the equation prove there is a critical point?

Re: Generalized Rayleigh Quotient

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**tubetess123** We aren't given that A is positive definite.

I meant symmetric.

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Also, why does substituting that particular matrix into the equation prove there is a critical point?

Suppose $\displaystyle \lambda_1\leq\lambda_2\leq\ldots\leq\lambda_n$ , then

$\displaystyle Q(x)=\dfrac{x^tAx}{x^tBx}=\frac{(Py)^tA(Py)}{(Py)^ tB(Py)}=\dfrac{y^tP^tAPy}{y^tP^tBPy}=\dfrac{y^tDy} {y^tIy}=$

$\displaystyle \frac{\lambda_1y_1^2+\ldots+\lambda_ny_n^2}{y_1^2+ \ldots+y_n^2} \geq \lambda_1\quad (\forall y=(y_1,\ldots,y_n)^t\in\mathbb{R}^n)$

If $\displaystyle y_0=(1,0,\ldots,0)^t$ and $\displaystyle x_0=Py_0$ then, $\displaystyle Q(x_0)=\lambda_1\geq Q(x)$ for all $\displaystyle x\in\mathbb{R}^n$ . This implies that $\displaystyle x_0$ is a point of minimum for $\displaystyle Q$ and as a consequence, a critical point of $\displaystyle Q$ .

Re: Generalized Rayleigh Quotient

Thank you for that explanation. Also, how do you make the nice equations show up instead of what I had shown above?

And is matrix Py a column vector? Because to do the dot product we need two column vectors. And why can we go from (Py)^tB(Py) to y^Py? Why can we change the order of y^t and P^t and why does the B change to A?

Re: Generalized Rayleigh Quotient

The next thing I need to do is prove that x is a CP of Q if and only if Ax = (lambda)Bx for some real lambda. I see that you've already proved one portion of this, but I don't know how I'd go about proving the other part?

Thanks!

Re: Generalized Rayleigh Quotient

So in your hint above, do you mean to say that $\displaystyle P^{t}BP = I$ and that $\displaystyle \lambda _{j}$ are the eigenvalues of A? It doesn't make sense to me if this is not the case.

Thanks!

Re: Generalized Rayleigh Quotient

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**tubetess123** Also, how do you make the nice equations show up instead of what I had shown above?

$\displaystyle (Ax)\cdot x=(Ax)^tx=x^tA^tx=x^tAx$ . The same for $\displaystyle B$ .

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And is matrix Py a column vector? Because to do the dot product we need two column vectors.

Yes, it is, $\displaystyle y=(y_1,\ldots,y_n)^t$ is a column vector, so $\displaystyle Py$ is also a column vector.

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And why can we go from (Py)^tB(Py) to y^Py?

We can't. It is $\displaystyle (Py)^tB(Py)=y^tP^tBPy=y^tIy=y^ty=y_1^2+y_2^2+...+y _n^2$

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Why can we change the order of y^t and P^t

Well known property: $\displaystyle (MN)^t=N^tM^t$ .

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and why does the B change to A?

A typo on my part. I've just corrected it.

Re: Generalized Rayleigh Quotient

Thanks! Those clarifications are great help!

Last thing, could you help me prove that if all the above given information is true, and that at a critical point x it's true that $\displaystyle Ax = \lambda Bx$, that the determinant of $\displaystyle A-\lambda B = 0$. I don't know how I'd go about showing this. Not sure if I said this, but x can't be 0.

Re: Generalized Rayleigh Quotient

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**tubetess123** Thanks! Those clarifications are great help!

Last thing, could you help me prove that if all the above given information is true, and that at a critical point x it's true that $\displaystyle Ax = \lambda Bx$, that the determinant of $\displaystyle A-\lambda B = 0$. I don't know how I'd go about showing this. Not sure if I said this, but x can't be 0.

$\displaystyle \exists x\neq 0:Ax=\lambda Bx\Leftrightarrow \exists x\neq 0: (A-\lambda B)x=0$

$\displaystyle \Leftrightarrow \textrm{rank}\;(A-\lambda B)<n \Leftrightarrow \det (A-\lambda B)=0$