# Why can SVD handle rank-deficient matrices?

I am currently reading a book on data analysis (Nathan Kutz, Data-Driven Modeling & Scientific Computation) and a bit stuck in the chapter about SVD.

It states that SVDs can be used to handle rank deficient matrices, but I do not understood why that actually is true. The book explains that "silent" columns are added to the Û matrix, these are orthonormal to the existing set in Û. Also, "silent" rows of all zero elements are added to the E matrix. This procedure would make it obvious that SVD can handle rank deficient matrices, the following illustration is provided: (copied from p. 379 of the book). Would someone be so kind and explain why these two steps are necessary to let SVD handle rank deficient matrices? Thank you!

David

The definition of SVD wants $U$ to be square (and unitary); if the $U$ you get from the first part of the procedure is not square you need to add some columns to make it square.
There's nothing inherently wrong with writing $A=U\Sigma V^*$ with a non-square $U$; it is just not what the SVD promises to give as a result.