Requirements for SVD to work

  • Thread starter spookyfw
  • Start date
  • #1
spookyfw
25
0
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 [itex]\neq[/itex] 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
 

Answers and Replies

  • #2
chiro
Science Advisor
4,815
134
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
spookyfw
25
0
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
 

Attachments

  • equ.PNG
    equ.PNG
    47.3 KB · Views: 448
  • #4
chiro
Science Advisor
4,815
134
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
spookyfw
25
0
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?
 
  • #7
spookyfw
25
0
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
PhilDSP
643
15
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
PhilDSP
643
15
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.
 

Suggested for: Requirements for SVD to work

Replies
14
Views
765
  • Last Post
Replies
7
Views
815
Replies
9
Views
469
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
657
Replies
0
Views
127
Replies
8
Views
639
Replies
11
Views
671
Top