svd(X,"econ") does not work as expected if X is square nxn matrix with rank<n

If X=[1 1;1 1] then with [U,S,V]=svd(X,"econ") I had expected that U=V=[1;1]/sqrt(2) and S=[2].

But "econ" does not work, S=[2 0;0 0] and U=V=[1 1;1 -1]/sqrt(2) (actually matlab gives opposite signs in U, but that does not matter).

What is the trick to make "econ" work in this case, so U is 2x1 and not 2x2 and S is 1x1 and not 2x2?

Answer 1

[Steven Lord]

If you want a number of singular values fewer than the maximum number, consider using svds. Though for a tiny problem like this if you're expecting it to be faster than just calling svd the difference is likely to be negligible.

X=[1 1;1 1];
[U, S, V] = svds(X, 1) % Ask for the 1 largest singular value

U = 2×1

-0.7071 -0.7071

S = 2

V = 2×1

-0.7071 -0.7071

[Bruno Luong]

The use case of svds is mainly for sparse matrix or linear operator where one does not have explicit matrix representation, where direct method cannot be used pratically.

sdvs like eigs is iterative (based on Krylov subspaced), it is less robust and accurate than direct method svd. Applied on full matrix it likely also slower.

[Eric]

Yes, this works. So in general I first give rank 1, then rank 2 etc. until a 0 appears on the diagonal of S. But still my question stays: why does svd(X,"econ") does as usual not work when X is squared? What is the reason why svd(X,"econ")=svd(X) when X is squared? There must be a reason.

[Steven Lord]

[U, S, V] = svd(X, 'econ') returns a square S matrix with a number of rows and columns each equal to the minimum of the number of rows in X and the number of columns in X.

If X is "tall" (more rows than columns) the size of S is based on the number of columns.

If X is "wide" (more columns than rows) the size of S is based on the number of rows.

If X is square (the same number of rows and columns) the size of S is based on that common value. In this case, S is exactly the same size as X.

[U, S, V] = svd(X) makes S the exact same size as X regardless of whether X is "tall", "wide", or square.

So svd(X, 'econ') is behaving as it is documented to behave. If you have a use case for a different syntax for svd, one that is rank-based, you can file that as an enhancement request via Technical Support. When or if you file that enhancement request, please describe how you would use such functionality were it to be implemented. Enhancement requests with real-world use cases tend to carry more weight with the developers when considering what enhancement requests to action.

[Eric]

Thank you very much!

Answer 2

[Chunru]

X=[1 1;1 1]

X = 2×2

1 1 1 1

[U,S,V]=svd(X,"econ") 

U = 2×2

-0.7071 -0.7071 -0.7071 0.7071

S = 2×2

2.0000 0 0 0.0000

V = 2×2

-0.7071 0.7071 -0.7071 -0.7071

%I had expected that U=V=[1;1]/sqrt(2) and S=[2].
%But "econ" does not work, S=[2 0;0 0] and U=V=[1 1;1 -1]/sqrt(2), except for signs. 

You can see svd with "econ" actually works. You expect U=V=[1;1]/sqrt(2) but another solution U=V=-[1;1]/sqrt(2) is actually the same solution (with opposite sign). It is obvious that X=U*S*V'=(-U)*S*(-V)'.

[Eric]

What I meant was that Matlab gives S a 2x2 matrix, which is not "econ". S should be 1x1 and U=V 2x1.

Answer 3

[Bruno Luong]

You should read the doc again:

"[___] = svd(A,"econ") produces an economy-size decomposition of A using either of the previous output argument combinations. If A is an m-by-n matrix, then:

• m > n — Only the first n columns of U are computed, and S is n-by-n.
• m = n — svd(A,"econ") is equivalent to svd(A).
• m < n — Only the first m columns of V are computed, and S is m-by-m.

"

The "econ" option truncates the output when the matrix is not square based on matrix size. It does not do any truncation based on rank.

If you want to truncate on rank you can do it as post ptocessing step

r = rank(X); % r = find(diag(S)>=S(1,1)*eps,1,'last')
U = U(:,r);
S = S(1:r,1:r);
V = V(:,1:r);

[Eric]

I had read that. But I am still wondering why Matlab does not acts "econ" in the case of a square matrix. Why not, and why do we have to do workarounds?

[Bruno Luong]

You said "does not work as expected". You expect something that the doc NEVER claim to return.

So you did not read careful the doc.

Now back the question "why": Because the notion of numerical "rank" can vary, and possibly there is no universal truncation suitable for all application downstream. But I agree that it would be nice to have an adjustable option to truncate less than min(size(A)).