Mixing
Approximation of eigenvalues of matrix
$begingroup$
We have the matrix begin{equation*}A=begin{pmatrix}1/2 & 1/5 & 1/10 & 1/17 \ 1/5 & 1/2 & 1/5 & 1/10 \ 1/10 & 1/5 & 1/2 & 1/5 \ 1/17 & 1/10 & 1/5 & 1/10end{pmatrix}end{equation*}
I want to find an approximation of the eigenvalues using 1 step of the QR method.
I have done the following:
First we find the QR decomposition of $A$, so we get the matrices begin{equation*}Q=begin{pmatrix}0.907651 & -0.415323 & -0.055244 & -0.0249738 \ 0.363061 & 0.846984 & -0.38676 & -0.0349633 \ 0.18153 & 0.302883 & 0.865764 & -0.354628 \ 0.106783 & 0.135599 & 0.312758 & 0.93402end{pmatrix}end{equation*} and begin{equation*}R=begin{pmatrix}0.550872 & 0.410045 & 0.275499 & 0.136682 \ 0 & 0.414564 & 0.306426 & 0.134404 \ 0 & 0 & 0.412557 & 0.162503 \ 0 & 0 & 0 & 0.017511end{pmatrix}end{equation*}
Then we get begin{equation*}A_1=RQ=begin{pmatrix}0.709622 & 0.212327 & 0.0966156 & 0.0014864 \ 0.219891 & 0.462078 & 0.1464 & 0.0016014 \ 0.091623 & 0.147077 & 0.408059 & 0.00481704 \ 0.001869 & 0.00238 & 0.005474 & 0.016275end{pmatrix}end{equation*}
Therefore an approximation of the eigenvalues are the elements of the diagonal because the result of the QR method is approximately a diagonal matrix or not?
numerical-methods eigenvalues-eigenvectors approximation matrix-decomposition
asked Feb 3 at 3:43
Mary StarMary Star
3,10982477
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$begingroup$
We have the matrix begin{equation*}A=begin{pmatrix}1/2 & 1/5 & 1/10 & 1/17 \ 1/5 & 1/2 & 1/5 & 1/10 \ 1/10 & 1/5 & 1/2 & 1/5 \ 1/17 & 1/10 & 1/5 & 1/10end{pmatrix}end{equation*}
I want to find an approximation of the eigenvalues using 1 step of the QR method.
I have done the following:
First we find the QR decomposition of $A$, so we get the matrices begin{equation*}Q=begin{pmatrix}0.907651 & -0.415323 & -0.055244 & -0.0249738 \ 0.363061 & 0.846984 & -0.38676 & -0.0349633 \ 0.18153 & 0.302883 & 0.865764 & -0.354628 \ 0.106783 & 0.135599 & 0.312758 & 0.93402end{pmatrix}end{equation*} and begin{equation*}R=begin{pmatrix}0.550872 & 0.410045 & 0.275499 & 0.136682 \ 0 & 0.414564 & 0.306426 & 0.134404 \ 0 & 0 & 0.412557 & 0.162503 \ 0 & 0 & 0 & 0.017511end{pmatrix}end{equation*}
Then we get begin{equation*}A_1=RQ=begin{pmatrix}0.709622 & 0.212327 & 0.0966156 & 0.0014864 \ 0.219891 & 0.462078 & 0.1464 & 0.0016014 \ 0.091623 & 0.147077 & 0.408059 & 0.00481704 \ 0.001869 & 0.00238 & 0.005474 & 0.016275end{pmatrix}end{equation*}
Therefore an approximation of the eigenvalues are the elements of the diagonal because the result of the QR method is approximately a diagonal matrix or not?
numerical-methods eigenvalues-eigenvectors approximation matrix-decomposition
asked Feb 3 at 3:43
Mary StarMary Star
3,10982477
$endgroup$
$begingroup$
So, are the eements of the diagonal not an approximation of the eigenvalues? @Moo
$endgroup$
– Mary Star
Feb 3 at 7:29
$begingroup$
Or is the matrix that I have calculated wrong? @Moo
$endgroup$
– Mary Star
Feb 3 at 7:50
$begingroup$
Ok! Thank you!! :-) @Moo
$endgroup$
– Mary Star
Feb 4 at 2:39
add a comment |
$begingroup$
We have the matrix begin{equation*}A=begin{pmatrix}1/2 & 1/5 & 1/10 & 1/17 \ 1/5 & 1/2 & 1/5 & 1/10 \ 1/10 & 1/5 & 1/2 & 1/5 \ 1/17 & 1/10 & 1/5 & 1/10end{pmatrix}end{equation*}
I want to find an approximation of the eigenvalues using 1 step of the QR method.
I have done the following:
First we find the QR decomposition of $A$, so we get the matrices begin{equation*}Q=begin{pmatrix}0.907651 & -0.415323 & -0.055244 & -0.0249738 \ 0.363061 & 0.846984 & -0.38676 & -0.0349633 \ 0.18153 & 0.302883 & 0.865764 & -0.354628 \ 0.106783 & 0.135599 & 0.312758 & 0.93402end{pmatrix}end{equation*} and begin{equation*}R=begin{pmatrix}0.550872 & 0.410045 & 0.275499 & 0.136682 \ 0 & 0.414564 & 0.306426 & 0.134404 \ 0 & 0 & 0.412557 & 0.162503 \ 0 & 0 & 0 & 0.017511end{pmatrix}end{equation*}
Then we get begin{equation*}A_1=RQ=begin{pmatrix}0.709622 & 0.212327 & 0.0966156 & 0.0014864 \ 0.219891 & 0.462078 & 0.1464 & 0.0016014 \ 0.091623 & 0.147077 & 0.408059 & 0.00481704 \ 0.001869 & 0.00238 & 0.005474 & 0.016275end{pmatrix}end{equation*}
Therefore an approximation of the eigenvalues are the elements of the diagonal because the result of the QR method is approximately a diagonal matrix or not?
numerical-methods eigenvalues-eigenvectors approximation matrix-decomposition
asked Feb 3 at 3:43
Mary StarMary Star
3,10982477
$endgroup$
We have the matrix begin{equation*}A=begin{pmatrix}1/2 & 1/5 & 1/10 & 1/17 \ 1/5 & 1/2 & 1/5 & 1/10 \ 1/10 & 1/5 & 1/2 & 1/5 \ 1/17 & 1/10 & 1/5 & 1/10end{pmatrix}end{equation*}
I want to find an approximation of the eigenvalues using 1 step of the QR method.
I have done the following:
First we find the QR decomposition of $A$, so we get the matrices begin{equation*}Q=begin{pmatrix}0.907651 & -0.415323 & -0.055244 & -0.0249738 \ 0.363061 & 0.846984 & -0.38676 & -0.0349633 \ 0.18153 & 0.302883 & 0.865764 & -0.354628 \ 0.106783 & 0.135599 & 0.312758 & 0.93402end{pmatrix}end{equation*} and begin{equation*}R=begin{pmatrix}0.550872 & 0.410045 & 0.275499 & 0.136682 \ 0 & 0.414564 & 0.306426 & 0.134404 \ 0 & 0 & 0.412557 & 0.162503 \ 0 & 0 & 0 & 0.017511end{pmatrix}end{equation*}
Then we get begin{equation*}A_1=RQ=begin{pmatrix}0.709622 & 0.212327 & 0.0966156 & 0.0014864 \ 0.219891 & 0.462078 & 0.1464 & 0.0016014 \ 0.091623 & 0.147077 & 0.408059 & 0.00481704 \ 0.001869 & 0.00238 & 0.005474 & 0.016275end{pmatrix}end{equation*}
Therefore an approximation of the eigenvalues are the elements of the diagonal because the result of the QR method is approximately a diagonal matrix or not?
numerical-methods eigenvalues-eigenvectors approximation matrix-decomposition
numerical-methods eigenvalues-eigenvectors approximation matrix-decomposition
asked Feb 3 at 3:43
Mary StarMary Star
3,10982477
asked Feb 3 at 3:43
Mary StarMary Star
3,10982477
asked Feb 3 at 3:43
Mary StarMary Star
3,10982477
asked Feb 3 at 3:43
Mary StarMary Star
3,10982477
asked Feb 3 at 3:43
Mary StarMary Star
3,10982477
3,10982477
$begingroup$
So, are the eements of the diagonal not an approximation of the eigenvalues? @Moo
$endgroup$
– Mary Star
Feb 3 at 7:29
$begingroup$
Or is the matrix that I have calculated wrong? @Moo
$endgroup$
– Mary Star
Feb 3 at 7:50
$begingroup$
Ok! Thank you!! :-) @Moo
$endgroup$
– Mary Star
Feb 4 at 2:39
add a comment |
$begingroup$
So, are the eements of the diagonal not an approximation of the eigenvalues? @Moo
$endgroup$
– Mary Star
Feb 3 at 7:29
$begingroup$
Or is the matrix that I have calculated wrong? @Moo
$endgroup$
– Mary Star
Feb 3 at 7:50
$begingroup$
Ok! Thank you!! :-) @Moo
$endgroup$
– Mary Star
Feb 4 at 2:39
$begingroup$
So, are the eements of the diagonal not an approximation of the eigenvalues? @Moo
$endgroup$
– Mary Star
Feb 3 at 7:29
$begingroup$
So, are the eements of the diagonal not an approximation of the eigenvalues? @Moo
$endgroup$
– Mary Star
Feb 3 at 7:29
$begingroup$
Or is the matrix that I have calculated wrong? @Moo
$endgroup$
– Mary Star
Feb 3 at 7:50
$begingroup$
Or is the matrix that I have calculated wrong? @Moo
$endgroup$
– Mary Star
Feb 3 at 7:50
$begingroup$
Ok! Thank you!! :-) @Moo
$endgroup$
– Mary Star
Feb 4 at 2:39
$begingroup$
Ok! Thank you!! :-) @Moo
$endgroup$
– Mary Star
Feb 4 at 2:39
add a comment |
1 Answer
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If the method converges, the matrices $A_k$ converge to a triangular matrix with the same eigenvalues as $A$. So the answer is yes, if you just perform one step of the $QR$ method, your estimate for the eigenvalues would be the diagonal elements of $A_1$.
answered Feb 4 at 11:07
PierreCarrePierreCarre
2,158215
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add a comment |
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$begingroup$
If the method converges, the matrices $A_k$ converge to a triangular matrix with the same eigenvalues as $A$. So the answer is yes, if you just perform one step of the $QR$ method, your estimate for the eigenvalues would be the diagonal elements of $A_1$.
answered Feb 4 at 11:07
PierreCarrePierreCarre
2,158215
$endgroup$
add a comment |
$begingroup$
If the method converges, the matrices $A_k$ converge to a triangular matrix with the same eigenvalues as $A$. So the answer is yes, if you just perform one step of the $QR$ method, your estimate for the eigenvalues would be the diagonal elements of $A_1$.
answered Feb 4 at 11:07
PierreCarrePierreCarre
2,158215
$endgroup$
add a comment |
$begingroup$
If the method converges, the matrices $A_k$ converge to a triangular matrix with the same eigenvalues as $A$. So the answer is yes, if you just perform one step of the $QR$ method, your estimate for the eigenvalues would be the diagonal elements of $A_1$.
answered Feb 4 at 11:07
PierreCarrePierreCarre
2,158215
$endgroup$
If the method converges, the matrices $A_k$ converge to a triangular matrix with the same eigenvalues as $A$. So the answer is yes, if you just perform one step of the $QR$ method, your estimate for the eigenvalues would be the diagonal elements of $A_1$.
answered Feb 4 at 11:07
PierreCarrePierreCarre
2,158215
answered Feb 4 at 11:07
PierreCarrePierreCarre
2,158215
answered Feb 4 at 11:07
PierreCarrePierreCarre
2,158215
answered Feb 4 at 11:07
PierreCarrePierreCarre
2,158215
2,158215
add a comment |
add a comment |
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$begingroup$
So, are the eements of the diagonal not an approximation of the eigenvalues? @Moo
$endgroup$
– Mary Star
Feb 3 at 7:29
$begingroup$
Or is the matrix that I have calculated wrong? @Moo
$endgroup$
– Mary Star
Feb 3 at 7:50
$begingroup$
Ok! Thank you!! :-) @Moo
$endgroup$
– Mary Star
Feb 4 at 2:39