SRC/dspsv.f(3)

πŸ’– SRC/dspsv.f πŸ’–

okay so this is a LAPACK manual page (the Library Functions Manual kind, iykyk) and honestly bestie, it's a whole vibe. Buckle up.


✨ SYNOPSIS ✨

🌸 Functions/Subroutines 🌸

  • subroutine dspsv (uplo, n, nrhs, ap, ipiv, b, ldb, info) β€” DSPSV computes the solution to system of linear equations A * X = B for OTHER matrices

πŸŽ€ Function/Subroutine Documentation πŸŽ€

subroutine dspsv (character uplo, integer n, integer nrhs, double precision,
    dimension( * ) ap, integer, dimension( * ) ipiv, double precision,
    dimension( ldb, * ) b, integer ldb, integer info)

DSPSV computes the solution to system of linear equations A * X = B for OTHER matrices β€” like, that's the whole assignment and she ate πŸ’…

πŸ¦‹ Purpose πŸ¦‹

DSPSV computes the solution to a real system of linear equations A * X = B, where A is an N-by-N symmetric matrix stored in packed format and X and B are N-by-NRHS matrices. The diagonal pivoting method is used to factor A as A = U * D * UT, if UPLO = 'U', or A = L * D * LT, if UPLO = 'L', where U (or L) is a product of permutation and unit upper (lower) triangular matrices, D is symmetric and block diagonal with 1-by-1 and 2-by-2 diagonal blocks. The factored form of A is then used to solve the system of equations A * X = B.

ngl, that's the entire plot. She factors A with the diagonal pivoting method, gets that D looking all block-diagonal with her 1-by-1 and 2-by-2 blocks, and then uses the factored form to actually solve A * X = B. main character energy fr fr πŸ”₯


πŸ’œ Parameters πŸ’œ

  • UPLO β€” UPLO is CHARACTER*1. Which triangle are we even working with, bestie:
    • = 'U': Upper triangle of A is stored;
    • = 'L': Lower triangle of A is stored.
  • N β€” N is INTEGER. The number of linear equations, i.e., the order of the matrix A. N >= 0. (Yes, she has standards. Zero is allowed, negative is not. πŸ™ˆ)
  • NRHS β€” NRHS is INTEGER. The number of right hand sides, i.e., the number of columns of the matrix B. NRHS >= 0.
  • AP β€” AP is DOUBLE PRECISION array, dimension (N*(N+1)/2). On entry: the upper or lower triangle of the symmetric matrix A, packed columnwise in a linear array. The j-th column of A is stored in the array AP as follows:
    • if UPLO = 'U', AP(i + (j-1)*j/2) = A(i,j) for 1<=i<=j;
    • if UPLO = 'L', AP(i + (j-1)*(2n-j)/2) = A(i,j) for j<=i<=n.

See below for further details (she's got a whole diagram for you, don't worry πŸ’–).

On exit: the block diagonal matrix D and the multipliers used to obtain the factor U or L from the factorization A = U*D*U**T or A = L*D*L**T as computed by DSPTRF, stored as a packed triangular matrix in the same storage format as A. Same storage format, no cap.

  • IPIV β€” IPIV is INTEGER array, dimension (N). Details of the interchanges and the block structure of D, as determined by DSPTRF. So here's the situation:
    • If IPIV(k) > 0, then rows and columns k and IPIV(k) were interchanged, and D(k,k) is a 1-by-1 diagonal block.
    • If UPLO = 'U' and IPIV(k) = IPIV(k-1) < 0, then rows and columns k-1 and -IPIV(k) were interchanged and D(k-1:k,k-1:k) is a 2-by-2 diagonal block.
    • If UPLO = 'L' and IPIV(k) = IPIV(k+1) < 0, then rows and columns k+1 and -IPIV(k) were interchanged and D(k:k+1,k:k+1) is a 2-by-2 diagonal block.

Negative IPIV is giving "we swapped as a pair" and honestly iconic of her 🫢

  • B β€” B is DOUBLE PRECISION array, dimension (LDB,NRHS). On entry: the N-by-NRHS right hand side matrix B. On exit: if INFO = 0, the N-by-NRHS solution matrix X. Glow-up moment: B walks in as the problem and leaves as the answer. ✨
  • LDB β€” LDB is INTEGER. The leading dimension of the array B. LDB >= max(1,N). Respect the boundary, bestie.
  • INFO β€” INFO is INTEGER:
    • = 0: successful exit πŸŽ‰
    • < 0: if INFO = -i, the i-th argument had an illegal value (someone did NOT read the spec)
    • > 0: if INFO = i, D(i,i) is exactly zero. The factorization has been completed, but the block diagonal matrix D is exactly singular, so the solution could not be computed. πŸ’€ Like, the factorization finished but the answer is simply not computable. Not her fault. Truly.

πŸ‘‘ Author πŸ‘‘

  • Univ. of Tennessee
  • Univ. of California Berkeley
  • Univ. of Colorado Denver
  • NAG Ltd.

(the whole squad showed up for this one 🌈)


⭐ Further Details ⭐

The packed storage scheme is illustrated by the following example when N = 4, UPLO = 'U': Two-dimensional storage of the symmetric matrix A: text a11 a12 a13 a14 a22 a23 a24 a33 a34 (aij = aji) a44 Packed storage of the upper triangle of A: text AP = [ a11, a12, a22, a13, a23, a33, a14, a24, a34, a44 ]

Notice how the lower-left is just... not there? Because aij = aji, she doesn't need to store the mirror. Efficient queen. πŸŽ€


Definition at line 161 of file dspsv.f.

🌸 Author 🌸

Generated automatically by Doxygen for LAPACK from the source code. (so don't come for her formatting, she was literally generated πŸ™ˆ)


LAPACK β€” Version 3.12.0 β€” SRC/dspsv.f(3) πŸ’– mwah, go solve those symmetric packed systems bestie, you got this πŸ¦‹πŸŒΈπŸ’…

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