---
name: "clatrs"
section: "3lapack"
platform: "linux"
package: "liblapack-doc"
version: "3.2.1-8"
release: "Debian 6.0 (Squeeze)"
platforms: ["linux"]
url: "https://new.carta.tech/clatrs/3lapack"
---

# clatrs(3lapack)

solves one of the triangular systems A \* x = s\*b, A\*\*T \* x = s\*b, or A\*\*H \* x = s\*b,

## NAME

CLATRS - solves one of the triangular systems   A \* x = s\*b, A\*\*T \* x = s\*b, or A\*\*H \* x = s\*b,

## SYNOPSIS

- SUBROUTINE CLATRS(

  UPLO, TRANS, DIAG, NORMIN, N, A, LDA, X, SCALE, CNORM, INFO )

- CHARACTER

  DIAG, NORMIN, TRANS, UPLO

- INTEGER

  INFO, LDA, N

- REAL

  SCALE

- REAL

  CNORM( \* )

- COMPLEX

  A( LDA, \* ), X( \* )

## PURPOSE

CLATRS solves one of the triangular systems with scaling to prevent overflow.  Here A is an upper or lower triangular matrix, A\*\*T denotes the transpose of A, A\*\*H denotes the conjugate transpose of A, x and b are n-element vectors, and s is a scaling factor, usually less than or equal to 1, chosen so that the components of x will be less than the overflow threshold.  If the unscaled problem will not cause overflow, the Level 2 BLAS routine CTRSV is called. If the matrix A is singular (A(j,j) = 0 for some j), then s is set to 0 and a non-trivial solution to A\*x = 0 is returned.

## ARGUMENTS

- UPLO (input) CHARACTER\*1

  Specifies whether the matrix A is upper or lower triangular. = 'U':  Upper triangular\
   \= 'L':  Lower triangular

- TRANS (input) CHARACTER\*1

  Specifies the operation applied to A. = 'N':  Solve A \* x = s\*b     (No transpose)\
   \= 'T':  Solve A\*\*T \* x = s\*b  (Transpose)\
   \= 'C':  Solve A\*\*H \* x = s\*b  (Conjugate transpose)

- DIAG (input) CHARACTER\*1

  Specifies whether or not the matrix A is unit triangular. = 'N':  Non-unit triangular\
   \= 'U':  Unit triangular

- NORMIN (input) CHARACTER\*1

  Specifies whether CNORM has been set or not. = 'Y':  CNORM contains the column norms on entry\
   \= 'N':  CNORM is not set on entry.  On exit, the norms will be computed and stored in CNORM.

- N (input) INTEGER

  The order of the matrix A.  N \>= 0.

- A (input) COMPLEX array, dimension (LDA,N)

  The triangular matrix A.  If UPLO = 'U', the leading n by n upper triangular part of the array A contains the upper triangular matrix, and the strictly lower triangular part of A is not referenced.  If UPLO = 'L', the leading n by n lower triangular part of the array A contains the lower triangular matrix, and the strictly upper triangular part of A is not referenced.  If DIAG = 'U', the diagonal elements of A are also not referenced and are assumed to be 1.

- LDA (input) INTEGER

  The leading dimension of the array A.  LDA \>= max (1,N).

- X (input/output) COMPLEX array, dimension (N)

  On entry, the right hand side b of the triangular system. On exit, X is overwritten by the solution vector x.

- SCALE (output) REAL

  The scaling factor s for the triangular system A \* x = s\*b,  A\*\*T \* x = s\*b,  or  A\*\*H \* x = s\*b. If SCALE = 0, the matrix A is singular or badly scaled, and the vector x is an exact or approximate solution to A\*x = 0.

- CNORM (input or output) REAL array, dimension (N)

  If NORMIN = 'Y', CNORM is an input argument and CNORM(j) contains the norm of the off-diagonal part of the j-th column of A.  If TRANS = 'N', CNORM(j) must be greater than or equal to the infinity-norm, and if TRANS = 'T' or 'C', CNORM(j) must be greater than or equal to the 1-norm. If NORMIN = 'N', CNORM is an output argument and CNORM(j) returns the 1-norm of the offdiagonal part of the j-th column of A.

- INFO (output) INTEGER

  \= 0:  successful exit\
   \< 0:  if INFO = -k, the k-th argument had an illegal value

## FURTHER DETAILS

A rough bound on x is computed; if that is less than overflow, CTRSV is called, otherwise, specific code is used which checks for possible overflow or divide-by-zero at every operation.\
 A columnwise scheme is used for solving A\*x = b.  The basic algorithm if A is lower triangular is\
\
      x\[1:n\] := b\[1:n\]\
\
      for j = 1, ..., n\
\
           x(j) := x(j) / A(j,j)\
\
           x\[j+1:n\] := x\[j+1:n\] - x(j) \* A\[j+1:n,j\]\
\
      end\
 Define bounds on the components of x after j iterations of the loop:\
    M(j) = bound on x\[1:j\]\
\
    G(j) = bound on x\[j+1:n\]\
 Initially, let M(0) = 0 and G(0) = max{x(i), i=1,...,n}.\
 Then for iteration j+1 we have\
\
    M(j+1) \<= G(j) / \| A(j+1,j+1) \|\
\
    G(j+1) \<= G(j) + M(j+1) \* \| A\[j+2:n,j+1\] \|\
\
           \<= G(j) ( 1 + CNORM(j+1) / \| A(j+1,j+1) \| )\
 where CNORM(j+1) is greater than or equal to the infinity-norm of column j+1 of A, not counting the diagonal.  Hence\
\
    G(j) \<= G(0) product ( 1 + CNORM(i) / \| A(i,i) \| )\
\
                 1\<=i\<=j\
 and\
\
    \|x(j)\| \<= ( G(0) / \|A(j,j)\| ) product ( 1 + CNORM(i) / \|A(i,i)\| )\
                                  1\<=i\< j\
 Since \|x(j)\| \<= M(j), we use the Level 2 BLAS routine CTRSV if the reciprocal of the largest M(j), j=1,..,n, is larger than\
 max(underflow, 1/overflow).\
 The bound on x(j) is also used to determine when a step in the columnwise method can be performed without fear of overflow.  If the computed bound is greater than a large constant, x is scaled to prevent overflow, but if the bound overflows, x is set to 0, x(j) to 1, and scale to 0, and a non-trivial solution to A\*x = 0 is found. Similarly, a row-wise scheme is used to solve A\*\*T \*x = b  or A\*\*H \*x = b.  The basic algorithm for A upper triangular is\
      for j = 1, ..., n\
\
           x(j) := ( b(j) - A\[1:j-1,j\]' \* x\[1:j-1\] ) / A(j,j)\
      end\
 We simultaneously compute two bounds\
\
      G(j) = bound on ( b(i) - A\[1:i-1,i\]' \* x\[1:i-1\] ), 1\<=i\<=j\
      M(j) = bound on x(i), 1\<=i\<=j\
 The initial values are G(0) = 0, M(0) = max{b(i), i=1,..,n}, and we add the constraint G(j) \>= G(j-1) and M(j) \>= M(j-1) for j \>= 1. Then the bound on x(j) is\
\
      M(j) \<= M(j-1) \* ( 1 + CNORM(j) ) / \| A(j,j) \|\
\
           \<= M(0) \* product ( ( 1 + CNORM(i) ) / \|A(i,i)\| )\
                     1\<=i\<=j\
 and we can safely call CTRSV if 1/M(n) and 1/G(n) are both greater than max(underflow, 1/overflow).\
