Without knowing more about the data, I cannot explain why the singular values appear the way they do here. However, generally in mathematics, larger singular values imply more "importance" to that data. I'm not sure why we are looking at the normalized cumulative sum; however, from these results we can...

[Uh,Sh,Vh]=svd(Ih); [Uw,Sw,Vw]=svd(double(watermark)); When you run svd, the resultant matrix Sh has the same dimensions as Ih, and resultant Sw has the dimensions as watermark. http://www.mathworks.com/help/matlab/ref/svd.html Now, Shw=Sh+a*Sw; you are adding 2 matrices together. Matrix addition requires that the matrices you are adding together have the same dimensions (same number of...

matlab,matrix,linear-algebra,svd

The answer is simply diag(S) Why? There's a theorem1 that says that the error between a matrix A and its rank-k approximation Ak has a (spectral) norm2 given by the k+1-th singular value of A. That is, the error is given by the first not used singular value. Isn't it...

At this time the cuSolver gesvd function only supports jobu = 'A' and jobvt = 'A' So the error when you specify other combinations is expected. From the documentation: Remark 2: gesvd only supports jobu='A' and jobvt='A' and returns matrix U and VH ...

The easiest way (if you are restricted to that library) would be to pad your matrix with rows of zeros to N x N, which you can then pass to your function. The padded matrix will have the same null space.

r,svd,matrix-decomposition,function

Cannot you just wrap your matrix arithmetic in a small function of your own? recover_matrix_from_svd <- function(svd) { score <- 0 for(i in 1:ncol(svd$u)) { score <- score + svd$u[,i] %*% t(svd$v[,i]) * svd$d[i] } score } alternatively, the diag function is very useful for this. Using it results in...

The sparse SVD can be computed indirectly. For example, first calculate X'*X or X*X', and then pass the resultant matrix to eigs_sym(). Another way is to first construct a sparse matrix like [zeros(m,m) X; X' zeros(n,n)], where m and n indicate the number of rows and columns in X. You...

cv::Mat can handle only continously stored data, so there are no suitable conversion from std::list. But you can implement it by yourself, as follows: std::list<cv::Point3d> points; cv::Mat matPoints(points.size(), 1, CV_64FC3); int i = 0; for (auto &p : points) { matPoints.at<cv::Vec3d>(i++) = p; } matPoints = matPoints.reshape(1); ...

machine-learning,recommendation-engine,svd,collaborative-filtering

In case of recommender systems one usually splits each user's history into train and test. More detailed: For each user we write out items he interacted with. Preferably, we order them by (incresing) time to overcome "time-traveling issue" (user can revisit already known items, so you don't want to test...

Just play with small numbers to debug your problem. Start with A=np.random.randn(3,2) instead of your much larger matrix with size (50,20) In my random case, I find that v1 = array([[-0.33872745, 0.94088454], [-0.94088454, -0.33872745]]) and for v2: v2 = array([[ 0.33872745, -0.94088454], [ 0.94088454, 0.33872745]]) they only differ for a...

size(Sh) returns the dimensions of the matrix Sh. eye(size(Sh)) creates an identity matrix with the same dimensions as Sh. logical(eye(size(Sh))) converts the elements of the identity matrix to logical values. Sh(...) is selecting a submatrix of Sh using logical indexing. Here it looks like it's just getting the diagonal elements...

Assuming MATLAB (or Octave): A = [1 3 1 2; 0 2 1 4; 6 5 2 1]; [U,S,V] = svd(A); S(3,3) = 0; A_hat = U*S*V'; This gives: A_hat = 1.37047 2.50649 1.03003 2.30320 -0.20009 2.26654 0.98378 3.83625 5.90727 5.12352 1.99248 0.92411 ...

You don't need a loop if you use reshape: cols = [1:3]; z1 = reshape(U(:,cols), numel(U(:,cols)), 1); You can also use this for non-consecutive columns, for example: cols = [1 2 4 7]; Example: A = [1 2 3; 4 5 6; 7 8 9] cols = [1:2]; B =...

In short, the SVD decomposition is not unique. The singular vectors of M are the eigenvectors of M`M. Eigenvectors are not unique. Even when the matrix is full rank, the eigenvectors are only defined up to a sign: If v is an eigenvector of the matrix A for eigenvalue lambda,...