Questions about left and right eigenvectors of symmetric complex matrix
$begingroup$
recently a question about right and left eigenvectors bother me a lot.
Here R is the matrix of the right eigenvectors of A, every column is one of A's eigenvectors; L is the matrix of the left eigenvectors of A, every row is one of A's eigenvectors; and "a" is a diagonalized matrix composed of the eigenvalues of A. A is a symmetric complex matrix
$AR=Ra$,
- we can take the transpose on both sides: $R^{T}A=aR^{T}$
- We can also multiply the inverse of R on the right left on both sides: $R^{-1}A=aR^{-1}$
- $LA=aL$
- So, according a,b we can get $R^{T}=R^{-1}=L$
- we can take the transpose on both sides: $R^{T}A=aR^{T}$
$LA=aL$
- we can take the conjugate transpose of both sides:$A^{dagger}L^{dagger}=L^{dagger}a*$
- So we can get the left eigenvalue matrix just by getting the right eigenvalues of $A^{dagger}$ and then do the conjugate transpose of $L^{dagger}$
- we can take the conjugate transpose of both sides:$A^{dagger}L^{dagger}=L^{dagger}a*$
These two methods both seem to be right and I should have consistency in the answers got from the two ways for the same eigenvalue(in the second method we can get the same a by doing the conjugate).
$$A=begin{pmatrix}1+i&1-i&1-3.5i\1-i&5+5.5i&-1.5+3i\1-3.5i&-1.5+3i&1.1+2iend{pmatrix}$$
However, the result I got is totally different.The Lone is got from the inverse of the right eigenvector matrix, and the Ltwo is got from the second method.
- method 1:
$$a={0.364039+7.3815i,4.52991+3.24693i,2.20606-2.12843i}$$
$$R=begin{pmatrix}-0.491206-0.0760069 i&0.466312 - 0.324685 i& 0.655799\0.403372 - 0.138801 i&0.800166 & -0.250746 + 0.338948i\0.716577& -0.101062 - 0.121558 i&0.668775 + 0.119451i
end{pmatrix}$$
$$Lone=begin{pmatrix}-0.555223 - 0.26064i& 0.543653 + 0.0197978i&
0.732705 - 0.143397 i\0.664945 - 0.222459i&
0.952185 + 0.36692 i& -0.127662 - 0.10407 i\ 0.776518 +
0.227821i& -0.45381 + 0.288362i& 0.707731 - 0.0116924i
end{pmatrix}$$
- method 2:
$$a^{*}={0.364039-7.3815i,4.52991-3.24693i,2.20606+2.12843i}$$
$$R=begin{pmatrix}-0.491206-0.0760069 i&0.466312 - 0.324685 i& 0.655799\0.403372 - 0.138801 i&0.800166 & -0.250746 + 0.338948i\0.716577& -0.101062 - 0.121558 i&0.668775 + 0.119451i
end{pmatrix}$$
$$Ltwo^{dagger}=begin{pmatrix}-0.491206 - 0.0760069i& 0.403372 - 0.138801i&
0.716577 \ 0.466312 - 0.324685i&
0.800166 &-0.101062 - 0.121558i\ 0.655799&-0.250746 + 0.338948i& 0.668775 + 0.119451iend{pmatrix}$$
Is there anybody can help me to solve the problem and point out my mistake or something? I really appreciate!
eigenvalues-eigenvectors
asked Jan 8 at 22:31
mqymqy
11
$endgroup$
|
show 4 more comments
$begingroup$
recently a question about right and left eigenvectors bother me a lot.
Here R is the matrix of the right eigenvectors of A, every column is one of A's eigenvectors; L is the matrix of the left eigenvectors of A, every row is one of A's eigenvectors; and "a" is a diagonalized matrix composed of the eigenvalues of A. A is a symmetric complex matrix
$AR=Ra$,
- we can take the transpose on both sides: $R^{T}A=aR^{T}$
- We can also multiply the inverse of R on the right left on both sides: $R^{-1}A=aR^{-1}$
- $LA=aL$
- So, according a,b we can get $R^{T}=R^{-1}=L$
- we can take the transpose on both sides: $R^{T}A=aR^{T}$
$LA=aL$
- we can take the conjugate transpose of both sides:$A^{dagger}L^{dagger}=L^{dagger}a*$
- So we can get the left eigenvalue matrix just by getting the right eigenvalues of $A^{dagger}$ and then do the conjugate transpose of $L^{dagger}$
- we can take the conjugate transpose of both sides:$A^{dagger}L^{dagger}=L^{dagger}a*$
These two methods both seem to be right and I should have consistency in the answers got from the two ways for the same eigenvalue(in the second method we can get the same a by doing the conjugate).
$$A=begin{pmatrix}1+i&1-i&1-3.5i\1-i&5+5.5i&-1.5+3i\1-3.5i&-1.5+3i&1.1+2iend{pmatrix}$$
However, the result I got is totally different.The Lone is got from the inverse of the right eigenvector matrix, and the Ltwo is got from the second method.
- method 1:
$$a={0.364039+7.3815i,4.52991+3.24693i,2.20606-2.12843i}$$
$$R=begin{pmatrix}-0.491206-0.0760069 i&0.466312 - 0.324685 i& 0.655799\0.403372 - 0.138801 i&0.800166 & -0.250746 + 0.338948i\0.716577& -0.101062 - 0.121558 i&0.668775 + 0.119451i
end{pmatrix}$$
$$Lone=begin{pmatrix}-0.555223 - 0.26064i& 0.543653 + 0.0197978i&
0.732705 - 0.143397 i\0.664945 - 0.222459i&
0.952185 + 0.36692 i& -0.127662 - 0.10407 i\ 0.776518 +
0.227821i& -0.45381 + 0.288362i& 0.707731 - 0.0116924i
end{pmatrix}$$
- method 2:
$$a^{*}={0.364039-7.3815i,4.52991-3.24693i,2.20606+2.12843i}$$
$$R=begin{pmatrix}-0.491206-0.0760069 i&0.466312 - 0.324685 i& 0.655799\0.403372 - 0.138801 i&0.800166 & -0.250746 + 0.338948i\0.716577& -0.101062 - 0.121558 i&0.668775 + 0.119451i
end{pmatrix}$$
$$Ltwo^{dagger}=begin{pmatrix}-0.491206 - 0.0760069i& 0.403372 - 0.138801i&
0.716577 \ 0.466312 - 0.324685i&
0.800166 &-0.101062 - 0.121558i\ 0.655799&-0.250746 + 0.338948i& 0.668775 + 0.119451iend{pmatrix}$$
Is there anybody can help me to solve the problem and point out my mistake or something? I really appreciate!
eigenvalues-eigenvectors
asked Jan 8 at 22:31
mqymqy
11
$endgroup$
$begingroup$
Please take the time to enter your work as text instead of linking to a picture of it. Images are neither searchable nor accessible to screen readers, nor do they show up in question summaries.
$endgroup$
– amd
Jan 9 at 5:31
$begingroup$
Have you actually tested your two results to see whether or not the rows are in fact independent eigenvectors of $A$? You do remember that eigenvectors are not unique, don’t you? There’s no particular reason to expect that the left eigenvectors produced by these two different methods will be identical.
$endgroup$
– amd
Jan 9 at 7:53
$begingroup$
@amd Thank you for your help!!! Since they have the same eigenvalues (L^dagger can be transformed to L), the L1 and L2 (the left eigenvectors in 2 methods) should be linear dependent? I thought the results are all normalized so they should have the same value.
$endgroup$
– mqy
Jan 14 at 22:16
$begingroup$
This is really hard to understand. Maybe you can put a sentence or two in the beginning to help us have some context on what assumptions about $A$, $R$, $L$, and $a$ you are making?
$endgroup$
– Morgan Rodgers
Jan 14 at 22:16
$begingroup$
@MorganRodgers I'm sorry about the unclear express. I edited it again. Thank you very much!
$endgroup$
– mqy
Jan 14 at 22:22
|
show 4 more comments
$begingroup$
recently a question about right and left eigenvectors bother me a lot.
Here R is the matrix of the right eigenvectors of A, every column is one of A's eigenvectors; L is the matrix of the left eigenvectors of A, every row is one of A's eigenvectors; and "a" is a diagonalized matrix composed of the eigenvalues of A. A is a symmetric complex matrix
$AR=Ra$,
- we can take the transpose on both sides: $R^{T}A=aR^{T}$
- We can also multiply the inverse of R on the right left on both sides: $R^{-1}A=aR^{-1}$
- $LA=aL$
- So, according a,b we can get $R^{T}=R^{-1}=L$
- we can take the transpose on both sides: $R^{T}A=aR^{T}$
$LA=aL$
- we can take the conjugate transpose of both sides:$A^{dagger}L^{dagger}=L^{dagger}a*$
- So we can get the left eigenvalue matrix just by getting the right eigenvalues of $A^{dagger}$ and then do the conjugate transpose of $L^{dagger}$
- we can take the conjugate transpose of both sides:$A^{dagger}L^{dagger}=L^{dagger}a*$
These two methods both seem to be right and I should have consistency in the answers got from the two ways for the same eigenvalue(in the second method we can get the same a by doing the conjugate).
$$A=begin{pmatrix}1+i&1-i&1-3.5i\1-i&5+5.5i&-1.5+3i\1-3.5i&-1.5+3i&1.1+2iend{pmatrix}$$
However, the result I got is totally different.The Lone is got from the inverse of the right eigenvector matrix, and the Ltwo is got from the second method.
- method 1:
$$a={0.364039+7.3815i,4.52991+3.24693i,2.20606-2.12843i}$$
$$R=begin{pmatrix}-0.491206-0.0760069 i&0.466312 - 0.324685 i& 0.655799\0.403372 - 0.138801 i&0.800166 & -0.250746 + 0.338948i\0.716577& -0.101062 - 0.121558 i&0.668775 + 0.119451i
end{pmatrix}$$
$$Lone=begin{pmatrix}-0.555223 - 0.26064i& 0.543653 + 0.0197978i&
0.732705 - 0.143397 i\0.664945 - 0.222459i&
0.952185 + 0.36692 i& -0.127662 - 0.10407 i\ 0.776518 +
0.227821i& -0.45381 + 0.288362i& 0.707731 - 0.0116924i
end{pmatrix}$$
- method 2:
$$a^{*}={0.364039-7.3815i,4.52991-3.24693i,2.20606+2.12843i}$$
$$R=begin{pmatrix}-0.491206-0.0760069 i&0.466312 - 0.324685 i& 0.655799\0.403372 - 0.138801 i&0.800166 & -0.250746 + 0.338948i\0.716577& -0.101062 - 0.121558 i&0.668775 + 0.119451i
end{pmatrix}$$
$$Ltwo^{dagger}=begin{pmatrix}-0.491206 - 0.0760069i& 0.403372 - 0.138801i&
0.716577 \ 0.466312 - 0.324685i&
0.800166 &-0.101062 - 0.121558i\ 0.655799&-0.250746 + 0.338948i& 0.668775 + 0.119451iend{pmatrix}$$
Is there anybody can help me to solve the problem and point out my mistake or something? I really appreciate!
eigenvalues-eigenvectors
asked Jan 8 at 22:31
mqymqy
11
$endgroup$
recently a question about right and left eigenvectors bother me a lot.
Here R is the matrix of the right eigenvectors of A, every column is one of A's eigenvectors; L is the matrix of the left eigenvectors of A, every row is one of A's eigenvectors; and "a" is a diagonalized matrix composed of the eigenvalues of A. A is a symmetric complex matrix
$AR=Ra$,
- we can take the transpose on both sides: $R^{T}A=aR^{T}$
- We can also multiply the inverse of R on the right left on both sides: $R^{-1}A=aR^{-1}$
- $LA=aL$
- So, according a,b we can get $R^{T}=R^{-1}=L$
- we can take the transpose on both sides: $R^{T}A=aR^{T}$
$LA=aL$
- we can take the conjugate transpose of both sides:$A^{dagger}L^{dagger}=L^{dagger}a*$
- So we can get the left eigenvalue matrix just by getting the right eigenvalues of $A^{dagger}$ and then do the conjugate transpose of $L^{dagger}$
- we can take the conjugate transpose of both sides:$A^{dagger}L^{dagger}=L^{dagger}a*$
These two methods both seem to be right and I should have consistency in the answers got from the two ways for the same eigenvalue(in the second method we can get the same a by doing the conjugate).
$$A=begin{pmatrix}1+i&1-i&1-3.5i\1-i&5+5.5i&-1.5+3i\1-3.5i&-1.5+3i&1.1+2iend{pmatrix}$$
However, the result I got is totally different.The Lone is got from the inverse of the right eigenvector matrix, and the Ltwo is got from the second method.
- method 1:
$$a={0.364039+7.3815i,4.52991+3.24693i,2.20606-2.12843i}$$
$$R=begin{pmatrix}-0.491206-0.0760069 i&0.466312 - 0.324685 i& 0.655799\0.403372 - 0.138801 i&0.800166 & -0.250746 + 0.338948i\0.716577& -0.101062 - 0.121558 i&0.668775 + 0.119451i
end{pmatrix}$$
$$Lone=begin{pmatrix}-0.555223 - 0.26064i& 0.543653 + 0.0197978i&
0.732705 - 0.143397 i\0.664945 - 0.222459i&
0.952185 + 0.36692 i& -0.127662 - 0.10407 i\ 0.776518 +
0.227821i& -0.45381 + 0.288362i& 0.707731 - 0.0116924i
end{pmatrix}$$
- method 2:
$$a^{*}={0.364039-7.3815i,4.52991-3.24693i,2.20606+2.12843i}$$
$$R=begin{pmatrix}-0.491206-0.0760069 i&0.466312 - 0.324685 i& 0.655799\0.403372 - 0.138801 i&0.800166 & -0.250746 + 0.338948i\0.716577& -0.101062 - 0.121558 i&0.668775 + 0.119451i
end{pmatrix}$$
$$Ltwo^{dagger}=begin{pmatrix}-0.491206 - 0.0760069i& 0.403372 - 0.138801i&
0.716577 \ 0.466312 - 0.324685i&
0.800166 &-0.101062 - 0.121558i\ 0.655799&-0.250746 + 0.338948i& 0.668775 + 0.119451iend{pmatrix}$$
Is there anybody can help me to solve the problem and point out my mistake or something? I really appreciate!
eigenvalues-eigenvectors
eigenvalues-eigenvectors
asked Jan 8 at 22:31
mqymqy
11
asked Jan 8 at 22:31
mqymqy
11
edited Jan 15 at 16:34
mqy
asked Jan 8 at 22:31
mqymqy
11
asked Jan 8 at 22:31
mqymqy
11
asked Jan 8 at 22:31
mqymqy
11
11
$begingroup$
Please take the time to enter your work as text instead of linking to a picture of it. Images are neither searchable nor accessible to screen readers, nor do they show up in question summaries.
$endgroup$
– amd
Jan 9 at 5:31
$begingroup$
Have you actually tested your two results to see whether or not the rows are in fact independent eigenvectors of $A$? You do remember that eigenvectors are not unique, don’t you? There’s no particular reason to expect that the left eigenvectors produced by these two different methods will be identical.
$endgroup$
– amd
Jan 9 at 7:53
$begingroup$
@amd Thank you for your help!!! Since they have the same eigenvalues (L^dagger can be transformed to L), the L1 and L2 (the left eigenvectors in 2 methods) should be linear dependent? I thought the results are all normalized so they should have the same value.
$endgroup$
– mqy
Jan 14 at 22:16
$begingroup$
This is really hard to understand. Maybe you can put a sentence or two in the beginning to help us have some context on what assumptions about $A$, $R$, $L$, and $a$ you are making?
$endgroup$
– Morgan Rodgers
Jan 14 at 22:16
$begingroup$
@MorganRodgers I'm sorry about the unclear express. I edited it again. Thank you very much!
$endgroup$
– mqy
Jan 14 at 22:22
|
show 4 more comments
$begingroup$
Please take the time to enter your work as text instead of linking to a picture of it. Images are neither searchable nor accessible to screen readers, nor do they show up in question summaries.
$endgroup$
– amd
Jan 9 at 5:31
$begingroup$
Have you actually tested your two results to see whether or not the rows are in fact independent eigenvectors of $A$? You do remember that eigenvectors are not unique, don’t you? There’s no particular reason to expect that the left eigenvectors produced by these two different methods will be identical.
$endgroup$
– amd
Jan 9 at 7:53
$begingroup$
@amd Thank you for your help!!! Since they have the same eigenvalues (L^dagger can be transformed to L), the L1 and L2 (the left eigenvectors in 2 methods) should be linear dependent? I thought the results are all normalized so they should have the same value.
$endgroup$
– mqy
Jan 14 at 22:16
$begingroup$
This is really hard to understand. Maybe you can put a sentence or two in the beginning to help us have some context on what assumptions about $A$, $R$, $L$, and $a$ you are making?
$endgroup$
– Morgan Rodgers
Jan 14 at 22:16
$begingroup$
@MorganRodgers I'm sorry about the unclear express. I edited it again. Thank you very much!
$endgroup$
– mqy
Jan 14 at 22:22
$begingroup$
Please take the time to enter your work as text instead of linking to a picture of it. Images are neither searchable nor accessible to screen readers, nor do they show up in question summaries.
$endgroup$
– amd
Jan 9 at 5:31
$begingroup$
Please take the time to enter your work as text instead of linking to a picture of it. Images are neither searchable nor accessible to screen readers, nor do they show up in question summaries.
$endgroup$
– amd
Jan 9 at 5:31
$begingroup$
Have you actually tested your two results to see whether or not the rows are in fact independent eigenvectors of $A$? You do remember that eigenvectors are not unique, don’t you? There’s no particular reason to expect that the left eigenvectors produced by these two different methods will be identical.
$endgroup$
– amd
Jan 9 at 7:53
$begingroup$
Have you actually tested your two results to see whether or not the rows are in fact independent eigenvectors of $A$? You do remember that eigenvectors are not unique, don’t you? There’s no particular reason to expect that the left eigenvectors produced by these two different methods will be identical.
$endgroup$
– amd
Jan 9 at 7:53
$begingroup$
@amd Thank you for your help!!! Since they have the same eigenvalues (L^dagger can be transformed to L), the L1 and L2 (the left eigenvectors in 2 methods) should be linear dependent? I thought the results are all normalized so they should have the same value.
$endgroup$
– mqy
Jan 14 at 22:16
$begingroup$
@amd Thank you for your help!!! Since they have the same eigenvalues (L^dagger can be transformed to L), the L1 and L2 (the left eigenvectors in 2 methods) should be linear dependent? I thought the results are all normalized so they should have the same value.
$endgroup$
– mqy
Jan 14 at 22:16
$begingroup$
This is really hard to understand. Maybe you can put a sentence or two in the beginning to help us have some context on what assumptions about $A$, $R$, $L$, and $a$ you are making?
$endgroup$
– Morgan Rodgers
Jan 14 at 22:16
$begingroup$
This is really hard to understand. Maybe you can put a sentence or two in the beginning to help us have some context on what assumptions about $A$, $R$, $L$, and $a$ you are making?
$endgroup$
– Morgan Rodgers
Jan 14 at 22:16
$begingroup$
@MorganRodgers I'm sorry about the unclear express. I edited it again. Thank you very much!
$endgroup$
– mqy
Jan 14 at 22:22
$begingroup$
@MorganRodgers I'm sorry about the unclear express. I edited it again. Thank you very much!
$endgroup$
– mqy
Jan 14 at 22:22
|
show 4 more comments
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$begingroup$
Please take the time to enter your work as text instead of linking to a picture of it. Images are neither searchable nor accessible to screen readers, nor do they show up in question summaries.
$endgroup$
– amd
Jan 9 at 5:31
$begingroup$
Have you actually tested your two results to see whether or not the rows are in fact independent eigenvectors of $A$? You do remember that eigenvectors are not unique, don’t you? There’s no particular reason to expect that the left eigenvectors produced by these two different methods will be identical.
$endgroup$
– amd
Jan 9 at 7:53
$begingroup$
@amd Thank you for your help!!! Since they have the same eigenvalues (L^dagger can be transformed to L), the L1 and L2 (the left eigenvectors in 2 methods) should be linear dependent? I thought the results are all normalized so they should have the same value.
$endgroup$
– mqy
Jan 14 at 22:16
$begingroup$
This is really hard to understand. Maybe you can put a sentence or two in the beginning to help us have some context on what assumptions about $A$, $R$, $L$, and $a$ you are making?
$endgroup$
– Morgan Rodgers
Jan 14 at 22:16
$begingroup$
@MorganRodgers I'm sorry about the unclear express. I edited it again. Thank you very much!
$endgroup$
– mqy
Jan 14 at 22:22