Consider the following statements: ● a) Let (L,U) with L,UER dxd be an LU factorisation of an invertible matrix AER dxd, and let beR d. Then we can solve the linear system Ax=b for x by first solving Ly=b for y, and then solving Ux=y for x. ● ● b) The matrix possesses an LU factorisation. c) The matrix 2 ^= (²³) A 3 3 A = (²₂³) 3 possesses a PALU factorisation. (Or, in other words, our PALU factorisation algorithm from Theorem 2.43 can be safely applied to this matrix A.) dxd ● d) Assume that AER possesses an LU factorisation A=LU. If we compute a PALU factorisation of this A as in Theorem 2.43, then P=I.
Consider the following statements: ● a) Let (L,U) with L,UER dxd be an LU factorisation of an invertible matrix AER dxd, and let beR d. Then we can solve the linear system Ax=b for x by first solving Ly=b for y, and then solving Ux=y for x. ● ● b) The matrix possesses an LU factorisation. c) The matrix 2 ^= (²³) A 3 3 A = (²₂³) 3 possesses a PALU factorisation. (Or, in other words, our PALU factorisation algorithm from Theorem 2.43 can be safely applied to this matrix A.) dxd ● d) Assume that AER possesses an LU factorisation A=LU. If we compute a PALU factorisation of this A as in Theorem 2.43, then P=I.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Need help deciding which of the statements are true (Theorem 2.43 also provided for c) and d)). Thank you :)
![Theorem 2.43: LU-factorisation with column pivoting
Let A € Rdxd be invertible. Then the sequence given by A₁ := A and
Tk € argmaxs{k,...,.d} |(Ak)s,k|
k‡ Tk,
Ãk :=
otherwise,
[P[k, Tk]Ak,
Ak,
Ak+1 := Lk(Ãk)Ãk
:
for all k € {1,..., d-1} is well-defined. The matrix
U := Ld-1(Ãd–1)P[d — 1, rd–1] ··· L2(Ã₂) P[2, r2]L1(Ã₁)P[1, r₁]A1
is upper triangular and can be rearranged as
U = Îd-1 · · · Î₂Î1 · P[d-1, rd-1]... P[2, r2] P[1,₁]A
with lower triangular matrices Î₁,..., Îd-1 satisfying (Îk)ü = 1 for all k and
i, and the matrix
L := Ρ¹'Îz¹ ….. ‚ο¹'ı
is lower triangular. The pair (L,U) is an LU-factorisation of the matrix PA,
where P = P[d-1, Td-1]... P[2, r2] P[1, T₁].](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6fe1fc35-672a-49fd-831b-9642c77888ed%2F0308de75-a092-4c0e-b8ba-71c7f796217c%2Fbq5x1cq_processed.png&w=3840&q=75)
Transcribed Image Text:Theorem 2.43: LU-factorisation with column pivoting
Let A € Rdxd be invertible. Then the sequence given by A₁ := A and
Tk € argmaxs{k,...,.d} |(Ak)s,k|
k‡ Tk,
Ãk :=
otherwise,
[P[k, Tk]Ak,
Ak,
Ak+1 := Lk(Ãk)Ãk
:
for all k € {1,..., d-1} is well-defined. The matrix
U := Ld-1(Ãd–1)P[d — 1, rd–1] ··· L2(Ã₂) P[2, r2]L1(Ã₁)P[1, r₁]A1
is upper triangular and can be rearranged as
U = Îd-1 · · · Î₂Î1 · P[d-1, rd-1]... P[2, r2] P[1,₁]A
with lower triangular matrices Î₁,..., Îd-1 satisfying (Îk)ü = 1 for all k and
i, and the matrix
L := Ρ¹'Îz¹ ….. ‚ο¹'ı
is lower triangular. The pair (L,U) is an LU-factorisation of the matrix PA,
where P = P[d-1, Td-1]... P[2, r2] P[1, T₁].
Transcribed Image Text:Consider the following statements:
●
a) Let (L,U) with L,UER dxd be an LU factorisation of an invertible matrix AER dxd, and let
berd. Then we can solve the linear system Ax=b for x by first solving Ly=b for y, and then
solving Ux=y for x.
b) The matrix
●
●
possesses an LU factorisation.
c) The matrix
2
5
A:
¹-(3²)
A
¹ = (33)
possesses a PALU factorisation. (Or, in other words, our PALU factorisation algorithm
from Theorem 2.43 can be safely applied to this matrix A.)
dxd
●
d) Assume that AER possesses an LU factorisation A=LU. If we compute a PALU
factorisation of this A as in Theorem 2.43, then P=I.
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