Finding U in SVD with eigenvectors

Question

So while working with SVD. I'm having some trouble consistently finding $U$. Sometimes the dimensions of $U$ are not correct using the method where we want $C=UEV^\top$ and obtain this by using $C^\top C=VE^\top EV^\top$ and $CV=UE$.

I was wondering if instead of extending the orthonormal basis you could just take the unit eigenvectors of $CC^\top$ to obtain $U$. In the way similar to how $V$ gets obtained?

I tried this for one problem and it worked, but I tried it for some other questions and ran into some problems. Its possible my calculations are all wrong, I've been working on this all day.

Any help trying to obtain $U$ would be appreciated.

Answer 1

You can obtain $\mathbf{U}$ as the matrix of eigenvectors of $\mathbf{C}\mathbf{C}^T = \mathbf{U} \mathbf{\Sigma}^2\mathbf{U}^T$ where $\mathbf{C}=\mathbf{U}\mathbf{\Sigma}\mathbf{V}^T$.

Answer 2

When $A$ is singular then indeed you can find the (normalized) eigenvectors of the symmetric matrix $A A ^ T$. Where $A$ is non-singular you can also use the fact that $$ A = U \Sigma V ^ T \to AV \Sigma ^{-1}=U. $$ The second approach is useful e.g., when $A$ is unitary, then both $ A ^ T A$ and $A A ^ T$ will give you $I_n$ which is not helpful in determining $U$.