0
$\begingroup$

In SVD we have $A = U \Sigma V^T$. When applying it for ML, e.g. to calculate Moore-Penrose pseudoinverse for linear regression, I have seen that we take columns of $A$ as vectors. Typically in ML I have seen rows as vectors, i.e. matrix $A$ is a collection of $n$ measurements, which form rows, and columns are $d$ dimensions.

  1. In case of SVD, why do we assume column vectors instead?
  2. If I wanted to use row vectors instead, should I just do $A = V \Sigma U^T$?
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.