Search results
Results From The WOW.Com Content Network
The rank of A is equal to r if and only if there exists an invertible m × m matrix X and an invertible n × n matrix Y such that = [], where I r denotes the r × r identity matrix. Sylvester’s rank inequality: if A is an m × n matrix and B is n × k, then [ii] + ().
The th column of an identity matrix is the unit vector, a vector whose th entry is 1 and 0 elsewhere. The determinant of the identity matrix is 1, and its trace is . The identity matrix is the only idempotent matrix with non-zero determinant. That is, it is the only matrix such that:
Every finite-dimensional matrix has a rank decomposition: Let be an matrix whose column rank is . Therefore, there are r {\textstyle r} linearly independent columns in A {\textstyle A} ; equivalently, the dimension of the column space of A {\textstyle A} is r {\textstyle r} .
Applicable to: m-by-n matrix A of rank r Decomposition: A = C F {\displaystyle A=CF} where C is an m -by- r full column rank matrix and F is an r -by- n full row rank matrix Comment: The rank factorization can be used to compute the Moore–Penrose pseudoinverse of A , [ 2 ] which one can apply to obtain all solutions of the linear system A x ...
The matrix determinant lemma performs a rank-1 update to a determinant. Woodbury matrix identity; Quasi-Newton method; Binomial inverse theorem; Bunch–Nielsen–Sorensen formula; Maxwell stress tensor contains an application of the Sherman–Morrison formula.
In mathematics, specifically linear algebra, the Woodbury matrix identity – named after Max A. Woodbury [1] [2] – says that the inverse of a rank-k correction of some matrix can be computed by doing a rank-k correction to the inverse of the original matrix.
It is the group of complex orthogonal matrices, complex matrices whose product with their transpose is the identity matrix. ... = rank(I − f) modulo 2, ...
In mathematics, the general linear group of degree n is the set of n×n invertible matrices, together with the operation of ordinary matrix multiplication.This forms a group, because the product of two invertible matrices is again invertible, and the inverse of an invertible matrix is invertible, with the identity matrix as the identity element of the group.