Partial Pivoting and Lower Triangularity
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In partial pivoting, I encounter something like the following: Let $P_1,...,P_n$ be permutation matrices, and $L$ is a lower, unit triangular matrix corresponding to the subtraction of rows underneath. In a text, the author simply said to $P_1...P_n LP_1^{-1}...P_n^{-1}$ is lower triangular. I cannot understand why it is the case, since if I consider the matrices
$L=begin{bmatrix}
1 & 0\
-1 & 1
end{bmatrix}$,
$P=begin{bmatrix}
0 & 1\
1 & 0
end{bmatrix}$, the matrix $PLP^{-1}$ turns out to be $begin{bmatrix}
1 & -1\
0 & 1
end{bmatrix}$ (if I did not calculate it wrong), which is clearly not lower triangular.
Remark: The following is the text I read. The formula (4.15) refers to the formula
$$U=L_{n-1}P_{n-1}...L_1P_1A$$
where $L_i,P_j$ are elementary matrices corresponding to substraction of rows and permutation of rows respectively. The formula comes from the process of partial pivoting.
linear-algebra numerical-methods
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up vote
0
down vote
favorite
In partial pivoting, I encounter something like the following: Let $P_1,...,P_n$ be permutation matrices, and $L$ is a lower, unit triangular matrix corresponding to the subtraction of rows underneath. In a text, the author simply said to $P_1...P_n LP_1^{-1}...P_n^{-1}$ is lower triangular. I cannot understand why it is the case, since if I consider the matrices
$L=begin{bmatrix}
1 & 0\
-1 & 1
end{bmatrix}$,
$P=begin{bmatrix}
0 & 1\
1 & 0
end{bmatrix}$, the matrix $PLP^{-1}$ turns out to be $begin{bmatrix}
1 & -1\
0 & 1
end{bmatrix}$ (if I did not calculate it wrong), which is clearly not lower triangular.
Remark: The following is the text I read. The formula (4.15) refers to the formula
$$U=L_{n-1}P_{n-1}...L_1P_1A$$
where $L_i,P_j$ are elementary matrices corresponding to substraction of rows and permutation of rows respectively. The formula comes from the process of partial pivoting.
linear-algebra numerical-methods
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In partial pivoting, I encounter something like the following: Let $P_1,...,P_n$ be permutation matrices, and $L$ is a lower, unit triangular matrix corresponding to the subtraction of rows underneath. In a text, the author simply said to $P_1...P_n LP_1^{-1}...P_n^{-1}$ is lower triangular. I cannot understand why it is the case, since if I consider the matrices
$L=begin{bmatrix}
1 & 0\
-1 & 1
end{bmatrix}$,
$P=begin{bmatrix}
0 & 1\
1 & 0
end{bmatrix}$, the matrix $PLP^{-1}$ turns out to be $begin{bmatrix}
1 & -1\
0 & 1
end{bmatrix}$ (if I did not calculate it wrong), which is clearly not lower triangular.
Remark: The following is the text I read. The formula (4.15) refers to the formula
$$U=L_{n-1}P_{n-1}...L_1P_1A$$
where $L_i,P_j$ are elementary matrices corresponding to substraction of rows and permutation of rows respectively. The formula comes from the process of partial pivoting.
linear-algebra numerical-methods
In partial pivoting, I encounter something like the following: Let $P_1,...,P_n$ be permutation matrices, and $L$ is a lower, unit triangular matrix corresponding to the subtraction of rows underneath. In a text, the author simply said to $P_1...P_n LP_1^{-1}...P_n^{-1}$ is lower triangular. I cannot understand why it is the case, since if I consider the matrices
$L=begin{bmatrix}
1 & 0\
-1 & 1
end{bmatrix}$,
$P=begin{bmatrix}
0 & 1\
1 & 0
end{bmatrix}$, the matrix $PLP^{-1}$ turns out to be $begin{bmatrix}
1 & -1\
0 & 1
end{bmatrix}$ (if I did not calculate it wrong), which is clearly not lower triangular.
Remark: The following is the text I read. The formula (4.15) refers to the formula
$$U=L_{n-1}P_{n-1}...L_1P_1A$$
where $L_i,P_j$ are elementary matrices corresponding to substraction of rows and permutation of rows respectively. The formula comes from the process of partial pivoting.
linear-algebra numerical-methods
linear-algebra numerical-methods
asked Nov 26 at 16:47
Jerry
512313
512313
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