Determinant of a block-triangular matrix A block-upper-triangular matrix is a matrix of the form where and are square matrices. PropositionLet be a block-upper-triangular matrix, as defined above. Then, Proof A
A matrix with a row of zeros has det A = 0. If A is triangular then detA=a11a22...anndetA=a11a22...ann=product of diagnonal entries. If A is singular then det A = 0. If A is invertible then detA≠0detA≠0. The determinant of AB is det A times det B : |AB|=|A||B||...
Determinant of a triangular matrixThe first result concerns the determinant of a triangular matrix. Proposition Let be a triangular matrix (either upper or lower). Then, the determinant of is equal to the product of its diagonal entries: Proof...
Hence by Theorem 1, the determinant of every upper-left n×n section of this infinite matrix is 1. The same will be true if “squares” is replaced by “cubes” or any higher power, or for that matter by any increasing sequence {0, 1, . . .} at all! 2. Take f = 1 +x ...
The Woodbury matrix identity and the matrix determinant lemma are the fundamental relations respectively for the inverse and determinant of the sum of two matrices. Consider the matrix(1)N=A+XBY*,where A and B are invertible matrices and X and Y are matrices of conformable size. We assume ...
product of thediagonalentries. It turns out this is true for a slightly larger class of matrices called triangular. <definition> <idx><h>Matrix</h><h>diagonal entries of</h></idx> <idx><h>Matrix</h><h>upper-triangular</h</idx> <idx><h>Matrix/h><h>lower-triangular...
LEMMA. Let Vm denote the Vandermonde matrix Then there is a linear combination of the first m rows of Vm which, when subtracted from the last row, reduces the last row to Proof of Lemma. Since the upper-left m-by-m minor of Vm is also Vandermonde, it is nonsingular. Thus a linear...