Requirements for SVD to work

1. Aug 17, 2012

spookyfw

Dear fellows,

during my internship I've stumbled over a problem of analysis. To cut things short some pseudoinverses have to be calculated. For one of them it does not work, s.t. A'*A $\neq$ I.

I just wondered about the requirements to find a pseudoinverse. One of the eigenvalues is zero, but as far as I understand that eigenvalues different than zero are not a requirement.

The matrix under consideration is that one:

A = [1 -1 0; 1 0 -1; 0 -1 1; 0 -1 1];

If anyone could help me that would be really great :).

Cheers and thank you very much in advance,
spookyfw

2. Aug 17, 2012

chiro

Hey spookyfw and welcome to the forums.

I just took a look at the Moore-Penrose pseudo-inverse from a grad linear algebra book and for that pseudo-inverse, the relation ship is that BAB = A and ABA = B.

Are you using this one or some other one?

3. Aug 17, 2012

spookyfw

Hi chiro,

thank you very much for your reply. Yes I am using the Moore-Penrose pseudo-inverse. Actually the original problem can be seen in the attached picture. After using the pseudo-inverses it is stated that only the first three systems have a solution, but that there is too little information to solve for A00, B00 and C00. When I am multiplying with the pseudo-inverses I get the unitary matrix on the left hand side, but for the last system. So I wondered how that is related to the last one not having a solution and what the reasoning is. I thought about independence, but there is also row-degeneracy in the matrix of the first system.

I hope that this still makes sense. Any idea?
spookyfw

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4. Aug 17, 2012

chiro

I'm wondering just as a curiosity, whether you have tried using a least squares approach to get some kind of reference for your solutions?

5. Aug 20, 2012

spookyfw

hmm..what do you mean by that least square approach? I just wanted to check a solution, I just don't understand the mathematical reason why the last system cannot be solved for A00, B00 and C00.
How would you go about the least square approach though?

6. Aug 20, 2012

chiro

7. Aug 22, 2012

spookyfw

Thank you very much. I think I spotted the problem with the matrix, as for square-matrices the Moore-Penrose inverse needs independent vectors, but unfortunately there is two dependent column vectors. At first sight I just spotted degenerate row vectors, but of course there has to be dependent ones with a 4x3.

8. Aug 22, 2012

PhilDSP

Hi spookyfw,

I'm a bit rusty on this now, but I believe SVD requires the matrix to be positive definite. That's probably true for any procedure based on Spectral Decomposition.

9. Aug 23, 2012

PhilDSP

Sorry, it's Cholesky Decomposition that requires the matrix to be positive definite. In that case, Cholesky Decomposition is far quicker to arrive at a result. The SVD is especially sensitive to a large variation in the range of cells values. Very small numbers can wreak havoc with the results.