91Ó°ÊÓ

In QR factorization of a matrix A, the matrix is decomposed into a product A = QR, where Q is an orthonormal matrix and R is an upper triangular matrix.

Let A be an invertible n x n upper triangular matrix.

Suppose, $A=\left[\begin{array}{ccccc}{a}_{11}& a_{12}& a_{13}& …& …\\ 0& a_{22}& a_{23}& …& …\\ 0& 0& a_{33}& …& …\\ â‹\circledR & â‹\circledR & â‹\pm & ⋯& ⋯\\ 0& 0& ⋯& ⋯& a_{nn}\end{array}\right]$

Here, A is an invertible matrix, so, $a_{ii}≠0$.

Now, for first case all the diagonal entry of the matrix A are positive, $a_{ii}>0$, then the matrix Q is the identity matrix such that $Q=\left[\begin{array}{cccc}{a}_{11}& 0& ⋯& 0\\ 0& a_{22}& â‹\circledR & 0\\ â‹\circledR & â‹\circledR & a_{33}& â‹\circledR \\ 0& 0& …& a_{33}\end{array}\right]$and the upper triangular matrix R will be R = A.

In the second case, let the diagonal entries of the matrix is less than zero, a_ii < 0, then the matrix Q will have entries $q_{ii}=-1$and the upper triangular matrix R has matrix $r_{ij}=-a_{ij}$.

Thus, if $a_{ii}>0$, then we have$q_{ii}=1$ and for any$j=1,2,...,n$ we have $r_{ij}=a_{ij}$. If $a_{ii}<0$, then we have$q_{ii}=-1$ and for any$j=1,2,...,n$ we have $r_{ij}=-a_{ij}$.