2
$\begingroup$

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: SVD

(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

$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Many thanks for this! So does this mean that SVD could also handle rank deficient matrices with only a non-square U matrix? My question might sound a bit silly, but I am just trying to figure out what enables SVD to handle linearly dependent (rank deficient) matrices, and how that is linked to the U matrix? (I can see why ordinary least square methods fail as it is not possible to compute the inverse of a rank deficient matrix. Thus, I guess SVD is a method used in such situation to compute the pseudo inverse? Sorry if my question is a bit imprecise, just trying to get my head around this). $\endgroup$
    – David Mehler
    Commented Dec 4, 2014 at 15:12
  • $\begingroup$ @DavidMehler: I think in order to get an answer that is useful to you, you need to explain more about why one might expect that the procedure wouldn't handle such matrices. (Don't do it as a comment here; edit the question). $\endgroup$
    – hmakholm left over Monica
    Commented Dec 4, 2014 at 15:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .