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The orthogonal Procrustes problem [1] is a matrix approximation problem in linear algebra.In its classical form, one is given two matrices and and asked to find an orthogonal matrix which most closely maps to .
Rule of Sarrus: The determinant of the three columns on the left is the sum of the products along the down-right diagonals minus the sum of the products along the up-right diagonals.
M is a Q-matrix if there exists d > 0 such that LCP(M,0) and LCP(M,d) have a unique solution. [1] [2] Any P-matrix is a Q-matrix. Conversely, if a matrix is a Z-matrix and a Q-matrix, then it is also a P-matrix. [3]
The matrix X is subjected to an orthogonal decomposition, e.g., the QR decomposition as follows. = , where Q is an m×m orthogonal matrix (Q T Q=I) and R is an n×n upper triangular matrix with >. The residual vector is left-multiplied by Q T.
Consider the system of equations x + y + 2z = 3, x + y + z = 1, 2x + 2y + 2z = 2.. The coefficient matrix is = [], and the augmented matrix is (|) = [].Since both of these have the same rank, namely 2, there exists at least one solution; and since their rank is less than the number of unknowns, the latter being 3, there are infinitely many solutions.
A common use of the pseudoinverse is to compute a "best fit" (least squares) approximate solution to a system of linear equations that lacks an exact solution (see below under § Applications). Another use is to find the minimum norm solution to a system of linear equations with multiple solutions. The pseudoinverse facilitates the statement ...
Sparse approximation (also known as sparse representation) theory deals with sparse solutions for systems of linear equations.Techniques for finding these solutions and exploiting them in applications have found wide use in image processing, signal processing, machine learning, medical imaging, and more.
Now () is an matrix solution of ′ =. This fundamental matrix will provide the homogeneous solution, and if added to a particular solution will give the general solution to the inhomogeneous equation. Let = be the general solution. Now,