## svd(A) vs. eig(A'*A) and eig(A*A')

on 20 Feb 2013

### Jan Simon (view profile)

When I use [U,S,V]=svd(A), I can reproduce A by U*S*V'. However, if I generate U and V by solving two eigenvalue problems [V,D]=eig(A'*A) and [U,D]=eig(A*A'), respectively, the resulting U and V may or may NOT satisfy U*S*V'=A, due possibly to the sign difference of some of the columns (eigenvectors). While the columns of the U's and V's produced by these two different methods have the same absolute values, they may have different signs, simply because if u_i is an eigenvector of A*A', so is -u_i.

But if U*S*V'~=A for some U and V, isn't the SVD theorem violated?

Lalit Patil

on 20 Feb 2013

## Products

### Jan Simon (view profile)

on 20 Feb 2013

I do not understand, where you see a violation. When you create U and V by another method, and consider, that they are not uniquely defined, it can be expected, that you get incompatible U and V matrices. If you want the orientation of the eigenvectors to satisfy U*S*V'=A, calculating them by solving the two separate eigenvalue problems eig(A'*A) and eig(A*A') is not sufficient.

Ruye Wang

on 7 May 2013