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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:
Let A be an m × n matrix. Let the column rank of A be r, and let c 1, ..., c r be any basis for the column space of A. Place these as the columns of an m × r matrix C. Every column of A can be expressed as a linear combination of the r columns in C. This means that there is an r × n matrix R such that A = CR.
In numerical analysis, interpolative decomposition (ID) factors a matrix as the product of two matrices, one of which contains selected columns from the original matrix, and the other of which has a subset of columns consisting of the identity matrix and all its values are no greater than 2 in absolute value.
A Hankel matrix. Identity matrix: A square diagonal matrix, with all entries on the main diagonal equal to 1, and the rest 0. a ij = δ ij: Lehmer matrix: a ij = min(i, j) ÷ max(i, j). A positive symmetric matrix. Matrix of ones: A matrix with all entries equal to one. a ij = 1. Pascal matrix: A matrix containing the entries of Pascal's ...
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 ...
For the case of column vector c and row vector r, each with m components, the formula allows quick calculation of the determinant of a matrix that differs from the identity matrix by a matrix of rank 1: (+) = +. More generally, [14] for any invertible m × m matrix X,
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} .
The matrices R 1, ..., R k give conjugate pairs of eigenvalues lying on the unit circle in the complex plane; so this decomposition confirms that all eigenvalues have absolute value 1. If n is odd, there is at least one real eigenvalue, +1 or −1; for a 3 × 3 rotation, the eigenvector associated with +1 is the rotation axis.