Mixing
Calculating the real triagonal form from a complex triagonal matrix
$begingroup$
I am writing a custom 3x3 Matrix Exponentiator in C for specific complex Hermitian matrices of the form
$$
left(begin{matrix}
q+z & x-iy & 0 \
x+iy & 0 & x-iy \
0 & x+iy & q-z \
end{matrix}right)
$$
I am basing my algorithm off Kopp's hybrid algorithm (https://arxiv.org/pdf/physics/0610206.pdf, https://www.mpi-hd.mpg.de/personalhomes/globes/3x3/index.html). His algorithm works fine for my purposes but I am looking for additional speed ups.
This uses the QL algorithm as a method for calculating eigenvectors when the analytical method fails.
The first step of the QL algorithm is to convert the matrix to real tridiagonal form.
The matrix is already in a complex tridiagonal form. Is there a simple analytical way to convert from a complex tridiagonal matrix to a real tridiagonal? Specifically what is the real tridiagonal matrix of this matrix?
linear-algebra numerical-linear-algebra
asked Jan 10 at 21:52
mathsifunmathsifun
12
$endgroup$
add a comment |
$begingroup$
I am writing a custom 3x3 Matrix Exponentiator in C for specific complex Hermitian matrices of the form
$$
left(begin{matrix}
q+z & x-iy & 0 \
x+iy & 0 & x-iy \
0 & x+iy & q-z \
end{matrix}right)
$$
I am basing my algorithm off Kopp's hybrid algorithm (https://arxiv.org/pdf/physics/0610206.pdf, https://www.mpi-hd.mpg.de/personalhomes/globes/3x3/index.html). His algorithm works fine for my purposes but I am looking for additional speed ups.
This uses the QL algorithm as a method for calculating eigenvectors when the analytical method fails.
The first step of the QL algorithm is to convert the matrix to real tridiagonal form.
The matrix is already in a complex tridiagonal form. Is there a simple analytical way to convert from a complex tridiagonal matrix to a real tridiagonal? Specifically what is the real tridiagonal matrix of this matrix?
linear-algebra numerical-linear-algebra
asked Jan 10 at 21:52
mathsifunmathsifun
12
$endgroup$
add a comment |
$begingroup$
I am writing a custom 3x3 Matrix Exponentiator in C for specific complex Hermitian matrices of the form
$$
left(begin{matrix}
q+z & x-iy & 0 \
x+iy & 0 & x-iy \
0 & x+iy & q-z \
end{matrix}right)
$$
I am basing my algorithm off Kopp's hybrid algorithm (https://arxiv.org/pdf/physics/0610206.pdf, https://www.mpi-hd.mpg.de/personalhomes/globes/3x3/index.html). His algorithm works fine for my purposes but I am looking for additional speed ups.
This uses the QL algorithm as a method for calculating eigenvectors when the analytical method fails.
The first step of the QL algorithm is to convert the matrix to real tridiagonal form.
The matrix is already in a complex tridiagonal form. Is there a simple analytical way to convert from a complex tridiagonal matrix to a real tridiagonal? Specifically what is the real tridiagonal matrix of this matrix?
linear-algebra numerical-linear-algebra
asked Jan 10 at 21:52
mathsifunmathsifun
12
$endgroup$
I am writing a custom 3x3 Matrix Exponentiator in C for specific complex Hermitian matrices of the form
$$
left(begin{matrix}
q+z & x-iy & 0 \
x+iy & 0 & x-iy \
0 & x+iy & q-z \
end{matrix}right)
$$
I am basing my algorithm off Kopp's hybrid algorithm (https://arxiv.org/pdf/physics/0610206.pdf, https://www.mpi-hd.mpg.de/personalhomes/globes/3x3/index.html). His algorithm works fine for my purposes but I am looking for additional speed ups.
This uses the QL algorithm as a method for calculating eigenvectors when the analytical method fails.
The first step of the QL algorithm is to convert the matrix to real tridiagonal form.
The matrix is already in a complex tridiagonal form. Is there a simple analytical way to convert from a complex tridiagonal matrix to a real tridiagonal? Specifically what is the real tridiagonal matrix of this matrix?
linear-algebra numerical-linear-algebra
linear-algebra numerical-linear-algebra
asked Jan 10 at 21:52
mathsifunmathsifun
12
asked Jan 10 at 21:52
mathsifunmathsifun
12
asked Jan 10 at 21:52
mathsifunmathsifun
12
asked Jan 10 at 21:52
mathsifunmathsifun
12
asked Jan 10 at 21:52
mathsifunmathsifun
12
12
add a comment |
add a comment |
1 Answer
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The unitary transformation required was
$$
left(begin{matrix}
e^{iphi} & 0 & 0 \
0 & 1 & 0 \
0 & 0 & e^{-iphi} \
end{matrix}right)
$$
where $phi=Arg(x+iy)$
answered Jan 14 at 0:56
mathsifunmathsifun
12
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
The unitary transformation required was
$$
left(begin{matrix}
e^{iphi} & 0 & 0 \
0 & 1 & 0 \
0 & 0 & e^{-iphi} \
end{matrix}right)
$$
where $phi=Arg(x+iy)$
answered Jan 14 at 0:56
mathsifunmathsifun
12
$endgroup$
add a comment |
$begingroup$
The unitary transformation required was
$$
left(begin{matrix}
e^{iphi} & 0 & 0 \
0 & 1 & 0 \
0 & 0 & e^{-iphi} \
end{matrix}right)
$$
where $phi=Arg(x+iy)$
answered Jan 14 at 0:56
mathsifunmathsifun
12
$endgroup$
add a comment |
$begingroup$
The unitary transformation required was
$$
left(begin{matrix}
e^{iphi} & 0 & 0 \
0 & 1 & 0 \
0 & 0 & e^{-iphi} \
end{matrix}right)
$$
where $phi=Arg(x+iy)$
answered Jan 14 at 0:56
mathsifunmathsifun
12
$endgroup$
The unitary transformation required was
$$
left(begin{matrix}
e^{iphi} & 0 & 0 \
0 & 1 & 0 \
0 & 0 & e^{-iphi} \
end{matrix}right)
$$
where $phi=Arg(x+iy)$
answered Jan 14 at 0:56
mathsifunmathsifun
12
answered Jan 14 at 0:56
mathsifunmathsifun
12
answered Jan 14 at 0:56
mathsifunmathsifun
12
answered Jan 14 at 0:56
mathsifunmathsifun
12
12
add a comment |
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