Negative Value Matrix: Finding the Singular Value Decomposition

[asif zaidi]

I did some problems from the example and the questions at end of chapter. I got all of them right except this one.

Problem Statement:

Consider the matrix [3 0; 0 -2]. Find its singular value decompositions

Problem Solution

Goal is to find A = U*S*V as below

Step1: Find AA', A'A. In this case they both are equal and are [9 0; 0 4];

Step2: Find U = eig vector (AA'). Doing so gives [1 0; 0 1];

Step 3: Find S = [3 0; 0 2] (I am not showing the steps)

Step 4: Find V = eig vector (A'A). Doing so gives [1 0; 0 1];

Verify: Multiply U*S*V and it should give back A.

My problem is it gives [3 0; 0 2] which is different than A = [3 0; 0 -2].

I know that if I change V to [1 0 ; 0 -1] I will get A back. But why do my computations not show this. What am I missing?

Like I said, I did the above procedure for a lot of other numbers and I get it right. Only when I have a negative value in the matrix then it seems I am missing a -1 factor which I cannot get from my procedure.

Thanks

Asif

 

[Dick]

I'm not really sure exactly what recipe you are following. Can you tell us where you found the algorithm? U could also be [[1,0],[0,-1]]. I'm not sure what 'eig vector(AA')' means, for example.

 

[asif zaidi]

Hi:

I am following the recipe from Linear Algebra with Applications (Steven J Leon)

Basically the text says on pg 346

- V matrix is the eigenvector of A'A
- U matrix is the eigenvector of AA'

I agree with you that U could also be as you said... But why didn't my calculations come up with that.

Thanks

Asif

 

[Dick]

Well, I asked for that. I don't have that reference so I can't really tell you. Maybe somebody else does. But I still don't get exactly what "V matrix is the eigenvector of A'A" means. [0,1] is an eigenvector of A'A. But so is [0,-1]. I don't know how you are supposed to put the signs in. Maybe just by hand?

 

[asif zaidi]

I am not typing the whole thing, so perhaps it is creating confusion

- The goal is to find SVD of a matrix A. Once we do this it will be comprised of 3 different matrices, U, S, V

- The text explains how to get S. It says compute A*A' and take the square root of the eigenvalues

- To get u: compute A*A' and find the eigen vector.

The example which I have, one can easily see that (1,1) or (1,-1) is the eigenvector. But what if I am given a problem in an exam which I cannot easily see. This is my main concern. I would have thought the math computation would have given this result.

Hope this clarifiesThanks

Asif