I am aware that both are Matrix decomposition techniques.

  • SVD decomposes the Matrix as $\mathbf C = \mathbf U_1 \mathbf L \mathbf V_1^\top$.

  • UV decomposes the Matrix as $\mathbf C = \mathbf U_2V_2^\top$.

Obviously UV can be seen as a special case of SVD if when $\mathbf L$ is equal to the identity matrix.

Nevertheless can't we also say that UV is another way of looking at SVD when $\mathbf U_2 = \mathbf U_1 L$? Am I missing something?


I believe you are missing out on the fact that in UV decomposition there are no orthogonality constraints on $U$.

When you perform the SVD decomposition of a matrix $A$ such that $A = U S V^T$, you impose the restriction $U^T U = I$. Even if the matrix holding the singular values was equal to the identity matrix $I$ so you could write $A = U V^T$ that would not equate the results of the original UV decomposition as you are arguably solving a different optimisation problem. This is one of the reason the UV decomposition has multiple local optima too; it does not restrict its solution space enough, it only try to reduce the relevant RMSE.

There a some well-documented "cheap SVD" variants you could consider instead of UV decomposition if you are in need of a cheap decomposition. These should get you much further both in terms of statistical coherence as well as existing algorithmic implementations.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.