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 July 26th, 2010, 09:27 PM #1 Newbie   Joined: Jul 2010 From: Shanghai,China.P.R Posts: 3 Thanks: 0 How can I solve this equation? P'AP=0 A is a diag Metrix. To solve P of a particular solution or general solution. How to do that? Is there any Matlab function? Thanks.
 July 27th, 2010, 09:15 PM #2 Senior Member   Joined: Dec 2009 Posts: 150 Thanks: 0 Re: How can I solve this equation? Is P' P inverse or P transpose? If it's P inverse then P(P' A P) = AP = P * 0 = 0. The ith row of AP is the the ith diagonal element of A (a_{i}) times the ith row of P. So that either a_{i} = 0 or row i of P is zero.
July 27th, 2010, 10:08 PM   #3
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Re: How can I solve this equation?

Quote:
 Originally Posted by forcesofodin Is P' P inverse or P transpose? If it's P inverse then P(P' A P) = AP = P * 0 = 0. The ith row of AP is the the ith diagonal element of A (a_{i}) times the ith row of P. So that either a_{i} = 0 or row i of P is zero.
A is a symmetric matrix. P is a vector.

I solved by means of this:
[T D]=eig(A)
so P'TDT'P=0
(T'P)'D(T'P)=0
so D(T'P)=0
let x=T'P
then Dx=0, it a equation with several unknowns
x=fsolve(D,a), a is initial value;
then T'P=x
P=T'\x

But, x is affected by a.

 July 29th, 2010, 07:11 PM #4 Senior Member   Joined: Dec 2009 Posts: 150 Thanks: 0 Re: How can I solve this equation? Seems like there are a many possibilities. Firstly, if the kernel of A is nontrivial then any p in the kernel of A is a solution, but this seems less interesting. As you seem to know, the symmetric matrix A has an orthogonal eigenbasis $v_1 , v_2 , \, ... \, , v_n$ and can be diagonalized by $A= TDT^{-1}$ where $T= [v_1 \, v_2 \, ... \, v_n ]$. So y has a basis representation using the eigenvectors, that is y = Tx for some x. We have $y^{t}Ay= (T^{-1}y)^{t} D (T^{-1}y) = x^{t}Dx = \lambda_{1}x_{1}^{2} + \lambda_{2}x_{2}^{2} + \, ... \, + \lambda_{n}x_{n}^{2} \, = \, 0$. Since $x \in \mathbb{R}^{n}$ (the eigenvalues of A are also real because A is symmetric), we must have at least one eigenvalue whose sign differs from the others (since the squared terms are positive and the eigenvalues were assumed to be nonzero so that kernel(A) = {0} ). This amounts to saying: is there a vector $[x_{1}^{2} \, , \, x_{2}^{2} \, ,\, ... \, , \, x_{n}^{2}]$ which is orthogonal to the vector $[\lambda_{1} \, , \, ... \, , \, \lambda_{n}]$ so that their dot product is zero? In general in $\mathbb{R}^{n}$ there could be many such solutions unless we can somehow use theory to add additional constraints. Your problem I assume, is how do you find all them. If you're just guessing, your initial guess vector X_0 may not be orthogonal to the eigenvalue vector, but you can use the two to form an orthogonal solution. But there are infinitely many orthogonal vectors, so maybe it's best to look for orthonormal ones, and in addition to the positivity constraint... Now onto fsolve, I got this from the matlab website page dealing with fsolve (http://www.mathworks.com/access/helpdes ... solve.html) The Gauss-Newton, Levenberg-Marquardt, and trust-region-reflective methods are based on the nonlinear least-squares algorithms also used in lsqnonlin. Use one of these methods if the system may not have a zero. The algorithm still returns a point where the residual is small. However, if the Jacobian of the system is singular, the algorithm might converge to a point that is not a solution of the system of equations (see Limitations and Diagnostics following). The fsolve algorithm (at least the Gauss-Newton method) requires that the number of functions be greater than the number of variables, and since you're trying to compute the zeros of a single function in n variables, you may wish to try another method. I like this problem, and want to think about it more. Sorry for the jumble of remarks and no answer yet.

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