Consider fast matrix multiplication of XDX^T for X an n by m matrix, and D an m by m diagonal matrix. Here m>>n (suppose n around 1000, m around 100000). In my application, X is a fixed matrix and values of D can change at every iteration.
What would be a fast way to calculate this? At the moment I am just doing simple multiplication in C++.
EDIT: I should clarify my current procedure, it is not "simple multiplication". In particular, I am columnise multiplying the X by the square root of diagonal entries of D to get A:=XD^{1/2}. Then I am directly calculating A*t(A) (which is the multiplication of an n by m matrix with its transpose).
Thank you.
If you know that D is diagonal, the you can just do simple multiplication. Hopefully, you are not multiplying the zeros.
Related
I'm trying to estimate a 3D rotation matrix between two sets of points, and I want to do that by computing the SVD of the covariance matrix, say C, as follows:
U,S,V = svd(C)
R = V * U^T
C in my case is 3x3 . I am using the Eigen's JacobiSVD module for this and I only recently found out that it stores matrices in column-major format. So that has had me confused.
So, when using Eigen, should I do:
V*U.transpose() or V.transpose()*U ?
Additionally, the rotation is accurate upto changing the sign of the column of U corresponding to the smallest singular value,such that determinant of R is positive. Let's say the index of the smallest singular value is minIndex .
So when the determinant is negative, because of the column major confusion, should I do:
U.col(minIndex) *= -1 or U.row(minIndex) *= -1
Thanks!
This has nothing to do with matrices being stored row-major or column major. svd(C) gives you:
U * S.asDiagonal() * V.transpose() == C
so the closest rotation R to C is:
R = U * V.transpose();
If you want to apply R to a point p (stored as column-vector), then you do:
q = R * p;
Now whether you are interested R or its inverse R.transpose()==V.transpose()*U is up to you.
The singular values scale the columns of U, so you should invert the columns to get det(U)=1. Again, nothing to do with storage layout.
How can I multiply an NxN matrix A in Fortran x times to get its power without amplifying rounding errors?
If A can be diagonalized as
A P = P D,
where P is some NxN matrix (each column is called 'eigenvector'), and D is an NxN diagonal matrix (the diagonal elements are called 'eigenvalues'), then
A = P D P^{-1},
where P^{-1} is the inverse matrix of P. Therefore the second power of A results in
A A= P D P^{-1} P D P^{-1} = P D D P^{-1}.
Repeating multiplication of A for x times yields
A^x = P D^x P^{-1}.
Note here that D^x is still a diagonal matrix. Let the i-th diagonal element of D be D_{ii}. Then, the i-th diagonal element of D^x is
[D^x]_{ii} = (D_{ii})^x.
That is, the elements of D^x is simply x-th power of the elements of D and can be computed without much rounding error, I guess. Now, you multiply P and P^{-1} from left and right, respectively, to this D^x to obtain A^x. The error in A^x depends on the error of P and P^{-1}, which can be calculated by some subroutines in numerical packages such as LAPACK.
as alluded to in the answer by norio, one can employ in general the Jordan (or alternatively Schur) decomposition and proceed in a similar fashion - for details (including brief error analysis) see, e.g., Chapter 11 of Matrix computations by Golub and Loan.
I have 2 matrices of dimension 1*280000.
I wanted to multiply one matrix with transposed second matrix using opencv.
I tried to multiply them using Multiplication operator(*).
But it is giving me error :'The total size matrix does not fit to size_t type'
As after multiplication the size will be 280000*28000 of matrix.
So,I am thinking multiplication should 32 bit.
Is there any method to do the 32bit multiplication?
Why do you want to multiply them like that? But because this is an answer, I would like to help you thinking more than just do it:
supposing that you have the two matrix: A and B (A.size() == B.size() == [1x280000]).
and A * B.t() = AB (AB is the result)
then AB = [A[0][0]*B A[0][1]*B ... A[0][279999]*B] (each column is the transposed matrix multiplied by the corresponding element of the other matrix)
AB may also be written as:
[ B[0][0]*A
B[0][1]*A
...
B[0][279999]*A]
(each row of the result will be the row matrix multiplied by the corresponding element of the column (transposed) matrix)
Hope that this will help you in what you are doing... Using a for loop you can print, or store, or what you need with the result
I am little confused of definition of normalized laplacian matrix edfined as:
D^{-1/2}*A*D^{-1/2}
The sentence of wiki: (isnt too much clear for me}::
the positive square roots of the corresponding positive diagonal entries of D
THe problem is if my D matrix have over 99 diagonal entries less then zero, so the square root of negative numbers can be done only with complex numbers...
does it mean that this sum have to be with abs() value? or input A matrix have to be rewrite with abs() values,....?
This is how I have it now:
for ( int i=0; i < data.cols(); i++) {
//Deg.sqrt();
Deg(i,i)=1/(sqrt((data.cwiseAbs().row(i).sum())) );
}
Lapla=Deg*data*Deg;
The normalized laplacian matrix is usually defined for networks, where A is the adjacency matrix and D is the diagonal matrix containing the degrees of the nodes. As an adjacency matrix, A has a one when the corresponding nodes are connected. Thus, as you say, the entries in D can be computing by taking row sums in A. But these sums are always positive, because you are adding zeros and ones. So I do not understand why your D matrix has negative numbers.
Maybe the edges in the network are weighted and some of the weights are negative? In that case, the definition of normalized laplacian matrix as D^{-1/2}*A*D^{-1/2} does not make sense, but I do not know what the correct definition is. I guess it depends on what you are using it for.
I have performed block SVD decomposition over image and I stored results.
Now, I need to make reconstruction from this results. I found few examples all written in Matlab, which is a mystery for me.
I only need formula from which I can reconstruct my picture, or example written in C language.
Matrix A is equal U*S*V'. How will look formula, e.g. for calculating first five singular values (product of which rows and columns)? Please provide formula with indexes in C like style. U and V' are matrices and S is vector (not matrix).
Not sure if I get your question right, but if you just need to know singular values, they are the diagonal values of the middle matrix S. S in general is a diagonal matrix, which is stored here as a vector. I mean, only the diagonal is stored, you should imagine it as a matrix if you're thinking in matrix calculations.
Those diagonal values are your singular values, if you need the first biggest singular values, just take the 5 biggest values of the vector S.
Quoting from Wikipedia:
The diagonal entries Σi,i of Σ are known as the singular values of M.
The m columns of U and the n columns of V are called the left-singular
vectors and right-singular vectors of M, respectively.
In the above quote, sigma is your S, and M is the original matrix.
You have asked for C code, yet my hope is that pseudocode will suffice (it's late, I'm tired). The target matrix A has m rows, c columns and rank rho. The variable p = min(m,n).
One strategy is to first form the the intermediate matrix product B = US. This is trivial due to the diagonal-like nature of the matrix of singular values. Assume you have rho ( = 5 ) singular values. You must enforce rho <= p.
Replace column vector u1 with s1u1.
Replace column vector u2 with s2u2.
...
Replace column vector urho with srhourho.
Replace column vector urho+1 with a zero vector of length m.
Replace column vector urho+2 with a zero vector of length m.
...
Replace column vector up with a zero vector of length m.
Next form the new image matrix A = BVT. The matrix element in row r and column c is the dot product of the rth row vector (length rho) of B with the cth column vector (length rho) of VT.
Another strategy is to jump to the form where the matrix elements of A in row r and column c are
ar,c = sum ( skur,kvc,k, { k, 1, rho } )
The row counter r runs from 1 to m; the column counter c runs from 1 to n.