Finding the Inv. of a rank deficient Matrix

[Shaddyab]

I have the following problem:

A * Phi = Ax' * Sx + Ay' * Sy

where,
A= Ax' * Ax + Ay' * Ay + Axy' * Axy

and I would like to solve for Phi.

Matrix A is:
1)symmetric
2) [89x89]
3) Rank(A)=88 ( I guess it means that there is no unique solution )
4) Det(A)~=0 ( I guess it means that A is not Singular )
5) A is a Sparse Matrix ( 673 (%8.5) non-zero elements out of 7921 )

How can I find the inverse of A and solve for Phi ?

Thank you

 

[HallsofIvy]

You don't. A "rank deficient" matrix which necessarily has determinant 0 does NOT have an inverse!

Now, what are "Ax' ", "Sx", etc.?

 

[Shaddyab]

Det(A) is NOT equal to zero.

I thought that I can solve it using SVD or pseudo-inverse, but I do not know how to implement this solution.

I am trying to find the least square phase ( Phi ) fit to a given slopes slopes.

Ax, Ay, and Axy are derivative matrix operators.

Can I Impose a B.C to solve the problem ? and how I do this ?

Thank you

 

[Office_Shredder]

If the determinant is not zero the matrix has to have full rank. You should double check that the problem is intended as written

 

[Shaddyab]

the determinant is not zero BUT it is almost -Inf.

 

[jasonRF]

This is a problem you are solving numerically on a computer, yes?

How are you estimating the rank?

How did you compute the determinant?

What does "the determinant is not zero BUT it is almost -Inf" mean?

It sounds like at the very least this is a poorly conditioned matrix. Yes, the pseudo-inverse can be used for these kinds of things to give a kind of solution (although I do not understand your notation at all so don't know the exact problem you are trying to solve ...). Wikipedia has a reasonable page on the Moore-Penrose pseudoinverse that may be worth your while.

jason

 

[Shaddyab]

I am running a Matlab code to solve the problem.
The rank and Determinant are estimated using Matlab commands 'rank' and 'det'

By saying that "the determinant is not zero BUT it is almost -Inf" I mean that the result of det(A) is around -1e24.

I am solving the following problem:

\vec{S} = \nabla Phi

Where
\vec{S} = Sx \hat{x} + Sy \hat{y}

Ax, Ay And Axy are matrix operators such as:
Ax=\partial / \partial x
Ay=\partial / \partial y
Axy=\partial / \partial x \partial y

Solving the least squares estimate I end out with the following equation:
A * Phi = Ax^{T} * Sx + Ay^{T}* Sy

where,
A=Ax^{T} * Ax + Ay^{T} * Ay + Axy^{T} * Axy

Hence I need the inverse of A to find Phi.
such that:
Phi = A^{-1} * Ax^{T} * Sx + A^{-1} * Ay^{T}* Sy

But Matrix A is rank deficient and I can not find the the inverse in the normal way.

 

[jasonRF]

I still don't understand your problem. From what you wrote Phi a scalar function? Thus \mathbf{S} is a vector (column vector?). Ax, Ay, Axy are all matrices, so Ax*sx is a vector? ...

So to me your equation looks like: matrix * scalar = vector.

Which seems crazy, since you state that the matrix A is square ...

What are the dimensions of each term in your equation? It still makes absolutely no sense to me.

jason

 

[jasonRF]

For some reason the fact that you are doing least squares just sunk in ...

I still do not understand your problem at all, but my hunch is you are doing least-squares and are showing us the "normal equations" that you derive that way. That approach is usually very poor numerically. If you have a linear least squares problem with a set of parameters \matbf{x} (n x 1)that you want to solve to minimize
| A \mathbf{x} - \mathbf{b} |^2,
where matrix A (m x n, with m>n) and the vector \mathbf{b} (m x 1) are both known, then in Matlab you solve it by using mldivide (the \ operator). Just like this:

x = A\b;

Inside matlab, type

doc mldivide

to see the algorithm it uses. For the m>n case (usual least-squares) it uses the QR factorization which is a reasonable way to do least squares.

My hunch is that you are trying to solve (using MATLAB notation)
A'*A*x = A'*b

Mathematically that is the exact way to solve the least squares problem, but it is usually a disaster numerically since A'*A often has a very poor condition number and can be unstable.

jason