MATLAB documentation of SVD states that the diagonal matrix returned has singular values in decreasing order. Is there a way to find out what the natural ordering of singular values would be?
The reason I ask is because the singular values correspond to dimensions associated with rows of the input matrix.
No, the very definition of SVD does not introduce an ordering. Restricting the discussion to square X matrices and adopting the same notation of the cited matlab documentation, if X = U*S*V' is a SVD of X, then for every permutation matrix P, we can form a valid SVD as X = (U*P)*(P'*S*P)*(V*P)'. Presenting matrix S with descending values is just a matter of convenience: every permutation P'*S*P would do the same job.
As a side note: P*X = P*U*S*V' showing that a row permutation of matrix X does not change the singular values S, which can be considered independent from any row (or column) permutation of X.
I was hoping to get some idea of what is being asked here before responding. For example, the eigenshuffle tool I've posted on the file exchange allows you to reorder the eigenvalues and eigenvectors of a sequence of eigen-problems, so they are maximally consistent with each other in sequence. Perhaps your problem is similar, thus you might think of the singular values as functions that vary along with some parameter that drives a system.
But really, there is no natural ordering of the singular values that comes from the method used to compute the SVD. In fact, the only ordering that makes sense is the one that comes out - decreasing order. The order of the singular values is not dependent on the sequence of the rows of your matrix, as the question seems to vaguely imply, so I'm not sure what is meant there.
Feel free to modify the question in case you can make your needs clearer.
Related
I have 50x49 matrix A that has 49 linearly independent columns. However, my software (octave) tells me its rank is 44:
Is it due to some computational error? If so, then how to prevent such errors?
If the software was able to correctly calculate rref(A), then why did it fail with rank(A)? Does it mean that calculating rank(A) is more error prone than calculating rref(A), or vice versa? I mean rref(A) actually tells you the rank, but here's a contradiction.
P.S. I've checked, Python makes the same error.
EDIT 1: Here is the matrix A itself. The first 9 columns were given. The rest was obtained with polynomial features.
EDIT 2: I was able to found a similar issue. Here is 10x10 matrix B of rank 10 (and octave calculates its rank correctly). However, octave says that rank(B * B) = 9 which is impossible.
The distinction between an invertible matrix (i.e. full rank) and a non-invertible one is clear-cut in theory, but not so in practice. A matrix B with large condition number (as in your example) can be inverted, but computing the inverse is numerically unstable. It roughly corresponds to B having a determinant that is "small" (using an appropriate, relative measure of "small"), so the matrix is almost singular. As a result, the inverse matrix will be computed with bad accuracy. In your example B, the condition number (computed with cond) is 2.069e9.
Another way to look at this is: when the condition number is large, it well could be that B is "really" singular, but small numerical errors from previous computations make it look barely non-singular. So you can't be sure.
The rank and rref functions use different algorithms (singular-value decomposition for rank, Gauss-Jordan elimination with partial pivoting for rref). For well-behaved matrices the numerical errors will be small in both cases, and the results will be consistent. But for a bad-conditioned matrix the numerical errors will be large and potentially different in each case, giving inconsistent results.
This is a well known issue with numerical algebra. In general, avoid inverting matrices with large condition number.
I am wondering if there is technical or theoretical reason on why Matlab on rank function considers as zero the value max(size(A))*eps(norm(A)). Can you please provide some intuition?
Thank you!
The following answer is not based on proper mathematical reasoning, it is just some speculations (as you were asking for intuition):
norm(A) is the order of magnitude of the matrix entries.
eps(norm(A)) is thus the accuracy that the floating point representation of the matrix entries typically has.
Now, consider you add N numbers that should theoretically add up to zero, but each of them has an error of eps to it ... I think we would expect an error in the order of sqrt(N) * eps for the result.
Then, given that the algorithm that computes the rank performs N^2 operations on the matrix entries (where N is its size) to result in a number that is checked against zero, the error that we would then expect is what you stated in your question.
What I don't know, is the algorithm that Matlab uses really of complexity N^2?
I use the function below to generate the betas for a given set of guess lambdas from my optimiser.
When running I often get the following warning message:
Warning: Matrix is singular to working precision.
In NSS_betas at 9
In DElambda at 19
In Individual_Lambdas at 36
I'd like to be able to exclude any betas that form a singular matrix form the solution set, however I don't know how to test for it?
I've been trying to use rcond() but I don't know where to make the cut off between singular and non singular?
Surely if Matlab is generating the warning message it already knows if the matrix is singular or not so if I could just find where that variable was stored I could use that?
function betas=NSS_betas(lambda,data)
mats=data.mats2';
lambda=lambda;
yM=data.y2';
nObs=size(yM,1);
G= [ones(nObs,1) (1-exp(-mats./lambda(1)))./(mats./lambda(1)) ((1-exp(-mats./lambda(1)))./(mats./lambda(1))-exp(-mats./lambda(1))) ((1-exp(-mats./lambda(2)))./(mats./lambda(2))-exp(-mats./lambda(2)))];
betas=G\yM;
r=rcond(G);
end
Thanks for the advice:
I tested all three examples below after setting the lambda values to be equal so guiving a singular matrix
if (~isinf(G))
r=rank(G);
r2=rcond(G);
r3=min(svd(G));
end
r=3, r2 =2.602085213965190e-16; r3= 1.075949299504113e-15;
So in this test rank() and rcond () worked assuming I take the benchmark values as given below.
However what happens when I have two values that are close but not exactly equal?
How can I decide what is too close?
rcond is the right way to go here. If it nears the machine precision of zero, your matrix is singular. I usually go with:
if( rcond(A) < 1e-12 )
% This matrix doesn't look good
end
You can experiment with a value that suites your needs, but taking the inverse of a matrix that is even close to singular with MATLAB can produce garbage results.
You could compare the result of rank(G) with the number of columns of G. If the rank is less than the column dimension, you will have a singular matrix.
you can also check this by:
min(svd(A))>eps
and verifying that the smallest singular value is larger than eps, or any other numerical tolerance that is relevant to your needs. (the code will return 1 or 0)
Here's more info about it...
Condition number (Maximal singular value/Minimal singular value) is another good method:
cond(A)
It uses svd. It should be as close to 1 as possible. Very large values mean that the matrix is almost singular. Inf means that it is precisely singular.
Note that almost all of the methods mentioned in other answers use somehow svd :
There are special tools designed for this problem, appropriately called "rank revealing matrix factorizations". To my best (albeit a little old) knowledge, a good enough way to decide whether a n x n matrix A is nonsingular is to go with
det(A) <> 0 <=> rank(A) = n
and use a rank-revealing QR factorization of A:
AP = QR
where Q is orthogonal, P is a permutation matrix and R is an upper triangular matrix with the property that the magnitude of the diagonal elements is decreased along the diagonal.
I am using LAPACK's ssteqr function to calculate eigenvalues/eigenvectors. The documentation for ssteqr says that the eigenvalues are sorted "in ascending order". Is it reasonable to assume that the list of eigenvectors is also sorted in ascending order?
Yes, it is reasonable to assume that the eigenvectors are ordered so that the i-th eigenvector corresponds to the i-th eigenvalue.
Nevertheless, if I were you, I would check for each eigenvalue the result of the multiplication of the eigenvector by the matrix. This way you are sure that you interpret the output right, and you see explicitly the accuracy of the calculations.
An old question, this, but I struggled with this recently, so am adding this for current and future readers.
The basic answer is that, yes, the eigenvectors are sorted such that the ith eigenvector corresponds to the ith eigenvalue. However, note that the eigenvectors thus obtained may not be the actual eigenvectors you want. This is so because of the following.
Since the ?steqr functions work only on tridiagonal matrices, one typically uses LAPACK's ?sytrd functions to first transform one's original symmetric matrix, call it M, to a tridiagonal form, call it T, such that M = QTQT where Q is an orthogonal matrix (and QT denotes its transpose). One then applies the ?steqr function on this tridiagonal matrix T to find its eigenvalues and eigenvectors. Now the eigenvalues thus obtained (of T) are exactly the same as the eigenvalues of M, so if one only wants the eigenvalues one can stop here. But if one is interested in the eigenvectors, like the OP, then one needs to bear in mind that the eigenvectors of T and M are different. To find the eigenvectors of the original matrix M, one needs to left-multiply the obtained eigenvectors of T by Q. This is very easily done by using the LAPACK functions orgtr or ormtr. See here for a clear explanation: https://software.intel.com/en-us/mkl-developer-reference-fortran-sytrd.
from a simulation problem, I want to calculate complex square matrices on the order of 1000x1000 in MATLAB. Since the values refer to those of Bessel functions, the matrices are not at all sparse.
Since I am interested in the change of the determinant with respect to some parameter (the energy of a searched eigenfunction in my case), I overcome the problem at the moment by first searching a rescaling factor for the studied range and then calculate the determinants,
result(k) = det(pre_factor*Matrix{k});
Now this is a very awkward solution and only works for matrix dimensions of, say, maximum 500x500.
Does anybody know a nice solution to the problem? Interfacing to Mathematica might work in principle but I have my doubts concerning feasibility.
Thank you in advance
Robert
Edit: I did not find a convient solution to the calculation problem since this would require changing to a higher precision. Instead, I used that
ln det M = trace ln M
which is, when I derive it with respect to k
A = trace(inv(M(k))*dM/dk)
So I at least had the change of the logarithm of the determinant with respect to k. From the physical background of the problem I could derive constraints on A which in the end gave me a workaround valid for my problem. Unfortunately I do not know if such a workaround could be generalized.
You should realize that when you multiply a matrix by a constant k, then you scale the determinant of the matrix by k^n, where n is the dimension of the matrix. So for n = 1000, and k = 2, you scale the determinant by
>> 2^1000
ans =
1.07150860718627e+301
This is of course a huge number, so you might expect that it should fail, since in double precision, MATLAB will only represent floating point numbers as large as realmax.
>> realmax
ans =
1.79769313486232e+308
There is no need to do all the work of recomputing that determinant, not that computing the determinant of a huge matrix like that is a terribly well-posed problem anyway.
If speed is not a concern, you may want to use det(e^A) = e^(tr A) and take as A some scaling constant times your matrix (so that A - I has spectral radius less than one).
EDIT: In MatLab, the log of a matrix (logm) is calculated via trigonalization. So it is better for you to compute the eigenvalues of your matrix and multiply them (or better, add their logarithm). You did not specify whether your matrix was symmetric or not: if it is, finding eigenvalues are easier than if it is not.
You said the current value of the determinant is about 10^-300.
Are you trying to get the determinant at a certain value, say 1? If so, rescaling is awkward: the matrix you are considering is ill-conditioned, and, considering the precision of the machine, you should consider the output determinant to be zero. It is impossible to get a reliable inverse in other words.
I would suggest to modify the columns or lines of the matrix rather than rescale it.
I used R to make a small test with a random matrix (random normal values), it seems the determinant should be clearly non-zero.
> n=100
> M=matrix(rnorm(n**2),n,n)
> det(M)
[1] -1.977380e+77
> kappa(M)
[1] 2318.188
This is not strictly a matlab solution, but you might want to consider using Mahout. It's specifically designed for large-scale linear algebra. (1000x1000 is no problem for the scales it's used to.)
You would call into java to pass data to/from Mahout.