# What is the relationship between SVD and UV decomposition?

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.