Writing a script for finding the largest and second largest eigenvectors of a symmetric matrix.
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0
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For a final project in my linear algebra intro, I have been tasked with writing a script that finds the largest and the second largest eigenvectors of a symmetric matrix in Matlab. For the best possible grade, it must include a function as well. So far, I have been able to get my script to verify that a matrix is symmetric, and am feeling a little bit stuck. I need some guidance for finishing this assignment, as my Matlab experience is extremely limited.
Here is what I have so far:
prompt = 'Please input a symmetric matrix A.'
A = input(prompt);
if (A == A'),
eig(A)
else
disp('A is not a symmetric matrix. Please input a symmetric matrix.')
end
Note that the script hopefully verifies that A is symmetric, and I have the eigenvalues for A, but I am not sure where to go from here to $1$. find the eigenvectors, $2$. get the two largest eigenvectors, and $3$. write a useful function to fit into the script. I would be very grateful for any help given. Thanks!
matlab
edited Nov 24 at 5:03
Rócherz
2,7262721
asked Dec 2 '13 at 22:13
Heath Huffman
164517
|
show 1 more comment
up vote
0
down vote
favorite
For a final project in my linear algebra intro, I have been tasked with writing a script that finds the largest and the second largest eigenvectors of a symmetric matrix in Matlab. For the best possible grade, it must include a function as well. So far, I have been able to get my script to verify that a matrix is symmetric, and am feeling a little bit stuck. I need some guidance for finishing this assignment, as my Matlab experience is extremely limited.
Here is what I have so far:
prompt = 'Please input a symmetric matrix A.'
A = input(prompt);
if (A == A'),
eig(A)
else
disp('A is not a symmetric matrix. Please input a symmetric matrix.')
end
Note that the script hopefully verifies that A is symmetric, and I have the eigenvalues for A, but I am not sure where to go from here to $1$. find the eigenvectors, $2$. get the two largest eigenvectors, and $3$. write a useful function to fit into the script. I would be very grateful for any help given. Thanks!
matlab
edited Nov 24 at 5:03
Rócherz
2,7262721
asked Dec 2 '13 at 22:13
Heath Huffman
164517
3
This may be better suited for stackoverflow.com/questions/tagged/matlab
– Henry Swanson
Dec 2 '13 at 22:14
I'll x-post it there. I appreciate it.
– Heath Huffman
Dec 2 '13 at 22:17
Yeah, they'll probably know the specific syntax. If there's not a syntax for getting eigenvectors, you can look at the nullspace of $A - lambda I$, where $lambda$ is an eigenvalue. That'll give you the eigenvectors. The other two steps are just MATLAB syntax somehow.
– Henry Swanson
Dec 2 '13 at 22:21
1
To reinforce(?) @HenrySwanson 's Comment, if the Question is about an approach (Matlab: full eigensolution followed by picking out the two largest eigenvalues) already decided upon, it would certainly be ripe for SO. A mathematical (but computational) question would be how to best approximate the two top eigenvalues without getting all of them (such questions are also handled at SciComp.SE).
– hardmath
Dec 2 '13 at 22:27
1
It's the $largett Power Method$ algorithm or $largett Von Mises$ one which yields the eigenvalue of largest magnitude. Once you get this, you can "remove" that eigenvalue and repeat the algorithm for the next one. See ---> en.wikipedia.org/wiki/Power_iteration
– Felix Marin
Dec 2 '13 at 23:04
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
For a final project in my linear algebra intro, I have been tasked with writing a script that finds the largest and the second largest eigenvectors of a symmetric matrix in Matlab. For the best possible grade, it must include a function as well. So far, I have been able to get my script to verify that a matrix is symmetric, and am feeling a little bit stuck. I need some guidance for finishing this assignment, as my Matlab experience is extremely limited.
Here is what I have so far:
prompt = 'Please input a symmetric matrix A.'
A = input(prompt);
if (A == A'),
eig(A)
else
disp('A is not a symmetric matrix. Please input a symmetric matrix.')
end
Note that the script hopefully verifies that A is symmetric, and I have the eigenvalues for A, but I am not sure where to go from here to $1$. find the eigenvectors, $2$. get the two largest eigenvectors, and $3$. write a useful function to fit into the script. I would be very grateful for any help given. Thanks!
matlab
edited Nov 24 at 5:03
Rócherz
2,7262721
asked Dec 2 '13 at 22:13
Heath Huffman
164517
For a final project in my linear algebra intro, I have been tasked with writing a script that finds the largest and the second largest eigenvectors of a symmetric matrix in Matlab. For the best possible grade, it must include a function as well. So far, I have been able to get my script to verify that a matrix is symmetric, and am feeling a little bit stuck. I need some guidance for finishing this assignment, as my Matlab experience is extremely limited.
Here is what I have so far:
prompt = 'Please input a symmetric matrix A.'
A = input(prompt);
if (A == A'),
eig(A)
else
disp('A is not a symmetric matrix. Please input a symmetric matrix.')
end
Note that the script hopefully verifies that A is symmetric, and I have the eigenvalues for A, but I am not sure where to go from here to $1$. find the eigenvectors, $2$. get the two largest eigenvectors, and $3$. write a useful function to fit into the script. I would be very grateful for any help given. Thanks!
matlab
matlab
edited Nov 24 at 5:03
Rócherz
2,7262721
asked Dec 2 '13 at 22:13
Heath Huffman
164517
edited Nov 24 at 5:03
Rócherz
2,7262721
asked Dec 2 '13 at 22:13
Heath Huffman
164517
edited Nov 24 at 5:03
Rócherz
2,7262721
edited Nov 24 at 5:03
Rócherz
2,7262721
edited Nov 24 at 5:03
Rócherz
2,7262721
2,7262721
asked Dec 2 '13 at 22:13
Heath Huffman
164517
asked Dec 2 '13 at 22:13
Heath Huffman
164517
asked Dec 2 '13 at 22:13
Heath Huffman
164517
164517
3
This may be better suited for stackoverflow.com/questions/tagged/matlab
– Henry Swanson
Dec 2 '13 at 22:14
I'll x-post it there. I appreciate it.
– Heath Huffman
Dec 2 '13 at 22:17
Yeah, they'll probably know the specific syntax. If there's not a syntax for getting eigenvectors, you can look at the nullspace of $A - lambda I$, where $lambda$ is an eigenvalue. That'll give you the eigenvectors. The other two steps are just MATLAB syntax somehow.
– Henry Swanson
Dec 2 '13 at 22:21
1
To reinforce(?) @HenrySwanson 's Comment, if the Question is about an approach (Matlab: full eigensolution followed by picking out the two largest eigenvalues) already decided upon, it would certainly be ripe for SO. A mathematical (but computational) question would be how to best approximate the two top eigenvalues without getting all of them (such questions are also handled at SciComp.SE).
– hardmath
Dec 2 '13 at 22:27
1
It's the $largett Power Method$ algorithm or $largett Von Mises$ one which yields the eigenvalue of largest magnitude. Once you get this, you can "remove" that eigenvalue and repeat the algorithm for the next one. See ---> en.wikipedia.org/wiki/Power_iteration
– Felix Marin
Dec 2 '13 at 23:04
|
show 1 more comment
3
This may be better suited for stackoverflow.com/questions/tagged/matlab
– Henry Swanson
Dec 2 '13 at 22:14
I'll x-post it there. I appreciate it.
– Heath Huffman
Dec 2 '13 at 22:17
Yeah, they'll probably know the specific syntax. If there's not a syntax for getting eigenvectors, you can look at the nullspace of $A - lambda I$, where $lambda$ is an eigenvalue. That'll give you the eigenvectors. The other two steps are just MATLAB syntax somehow.
– Henry Swanson
Dec 2 '13 at 22:21
1
To reinforce(?) @HenrySwanson 's Comment, if the Question is about an approach (Matlab: full eigensolution followed by picking out the two largest eigenvalues) already decided upon, it would certainly be ripe for SO. A mathematical (but computational) question would be how to best approximate the two top eigenvalues without getting all of them (such questions are also handled at SciComp.SE).
– hardmath
Dec 2 '13 at 22:27
1
It's the $largett Power Method$ algorithm or $largett Von Mises$ one which yields the eigenvalue of largest magnitude. Once you get this, you can "remove" that eigenvalue and repeat the algorithm for the next one. See ---> en.wikipedia.org/wiki/Power_iteration
– Felix Marin
Dec 2 '13 at 23:04
3
3
This may be better suited for stackoverflow.com/questions/tagged/matlab
– Henry Swanson
Dec 2 '13 at 22:14
This may be better suited for stackoverflow.com/questions/tagged/matlab
– Henry Swanson
Dec 2 '13 at 22:14
I'll x-post it there. I appreciate it.
– Heath Huffman
Dec 2 '13 at 22:17
I'll x-post it there. I appreciate it.
– Heath Huffman
Dec 2 '13 at 22:17
Yeah, they'll probably know the specific syntax. If there's not a syntax for getting eigenvectors, you can look at the nullspace of $A - lambda I$, where $lambda$ is an eigenvalue. That'll give you the eigenvectors. The other two steps are just MATLAB syntax somehow.
– Henry Swanson
Dec 2 '13 at 22:21
Yeah, they'll probably know the specific syntax. If there's not a syntax for getting eigenvectors, you can look at the nullspace of $A - lambda I$, where $lambda$ is an eigenvalue. That'll give you the eigenvectors. The other two steps are just MATLAB syntax somehow.
– Henry Swanson
Dec 2 '13 at 22:21
1
1
To reinforce(?) @HenrySwanson 's Comment, if the Question is about an approach (Matlab: full eigensolution followed by picking out the two largest eigenvalues) already decided upon, it would certainly be ripe for SO. A mathematical (but computational) question would be how to best approximate the two top eigenvalues without getting all of them (such questions are also handled at SciComp.SE).
– hardmath
Dec 2 '13 at 22:27
To reinforce(?) @HenrySwanson 's Comment, if the Question is about an approach (Matlab: full eigensolution followed by picking out the two largest eigenvalues) already decided upon, it would certainly be ripe for SO. A mathematical (but computational) question would be how to best approximate the two top eigenvalues without getting all of them (such questions are also handled at SciComp.SE).
– hardmath
Dec 2 '13 at 22:27
1
1
It's the $largett Power Method$ algorithm or $largett Von Mises$ one which yields the eigenvalue of largest magnitude. Once you get this, you can "remove" that eigenvalue and repeat the algorithm for the next one. See ---> en.wikipedia.org/wiki/Power_iteration
– Felix Marin
Dec 2 '13 at 23:04
It's the $largett Power Method$ algorithm or $largett Von Mises$ one which yields the eigenvalue of largest magnitude. Once you get this, you can "remove" that eigenvalue and repeat the algorithm for the next one. See ---> en.wikipedia.org/wiki/Power_iteration
– Felix Marin
Dec 2 '13 at 23:04
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
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Given the $n$-iterated vector $Psi_{n}$, we get the net one $Psi_{n + 1}$ with
$$
Psi_{n + 1} = {varphi_{n} over lambda_{n}},,quad
mbox{where}quad varphi_{n} equiv APsi_{n}
$$
$lambda_{n}$ is the component of $varphi_{n}$ with the largest magnitude. After $N$ ( "many" ) iterations, we get the eigenvalue as $lambda_{N}$. This is the eigenvalue with the largest magnitude. Next, we normalize the eigenvector:
$ds{varphi_{N} to {varphi_{N} over varphi_{N}^{sf T},varphi_{N}}}$. "Reduce" the matrix as
$$
tilde{A} equiv A - lambda_{N} varphi_{N}varphi_{N}^{sf T}
$$
Repeat the procedure with $tilde{A}$ and so on.
See this link.
answered Dec 2 '13 at 23:06
Felix Marin
66.9k7107139
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
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newcommand{ul}[1]{underline{#1}}%
newcommand{verts}[1]{leftvert, #1 ,rightvert}$
Given the $n$-iterated vector $Psi_{n}$, we get the net one $Psi_{n + 1}$ with
$$
Psi_{n + 1} = {varphi_{n} over lambda_{n}},,quad
mbox{where}quad varphi_{n} equiv APsi_{n}
$$
$lambda_{n}$ is the component of $varphi_{n}$ with the largest magnitude. After $N$ ( "many" ) iterations, we get the eigenvalue as $lambda_{N}$. This is the eigenvalue with the largest magnitude. Next, we normalize the eigenvector:
$ds{varphi_{N} to {varphi_{N} over varphi_{N}^{sf T},varphi_{N}}}$. "Reduce" the matrix as
$$
tilde{A} equiv A - lambda_{N} varphi_{N}varphi_{N}^{sf T}
$$
Repeat the procedure with $tilde{A}$ and so on.
See this link.
answered Dec 2 '13 at 23:06
Felix Marin
66.9k7107139
add a comment |
up vote
2
down vote
accepted
$newcommand{+}{^{dagger}}%
newcommand{angles}[1]{leftlangle #1 rightrangle}%
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newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}%
newcommand{verts}[1]{leftvert, #1 ,rightvert}$
Given the $n$-iterated vector $Psi_{n}$, we get the net one $Psi_{n + 1}$ with
$$
Psi_{n + 1} = {varphi_{n} over lambda_{n}},,quad
mbox{where}quad varphi_{n} equiv APsi_{n}
$$
$lambda_{n}$ is the component of $varphi_{n}$ with the largest magnitude. After $N$ ( "many" ) iterations, we get the eigenvalue as $lambda_{N}$. This is the eigenvalue with the largest magnitude. Next, we normalize the eigenvector:
$ds{varphi_{N} to {varphi_{N} over varphi_{N}^{sf T},varphi_{N}}}$. "Reduce" the matrix as
$$
tilde{A} equiv A - lambda_{N} varphi_{N}varphi_{N}^{sf T}
$$
Repeat the procedure with $tilde{A}$ and so on.
See this link.
answered Dec 2 '13 at 23:06
Felix Marin
66.9k7107139
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$newcommand{+}{^{dagger}}%
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newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}%
newcommand{verts}[1]{leftvert, #1 ,rightvert}$
Given the $n$-iterated vector $Psi_{n}$, we get the net one $Psi_{n + 1}$ with
$$
Psi_{n + 1} = {varphi_{n} over lambda_{n}},,quad
mbox{where}quad varphi_{n} equiv APsi_{n}
$$
$lambda_{n}$ is the component of $varphi_{n}$ with the largest magnitude. After $N$ ( "many" ) iterations, we get the eigenvalue as $lambda_{N}$. This is the eigenvalue with the largest magnitude. Next, we normalize the eigenvector:
$ds{varphi_{N} to {varphi_{N} over varphi_{N}^{sf T},varphi_{N}}}$. "Reduce" the matrix as
$$
tilde{A} equiv A - lambda_{N} varphi_{N}varphi_{N}^{sf T}
$$
Repeat the procedure with $tilde{A}$ and so on.
See this link.
answered Dec 2 '13 at 23:06
Felix Marin
66.9k7107139
$newcommand{+}{^{dagger}}%
newcommand{angles}[1]{leftlangle #1 rightrangle}%
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newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}%
newcommand{verts}[1]{leftvert, #1 ,rightvert}$
Given the $n$-iterated vector $Psi_{n}$, we get the net one $Psi_{n + 1}$ with
$$
Psi_{n + 1} = {varphi_{n} over lambda_{n}},,quad
mbox{where}quad varphi_{n} equiv APsi_{n}
$$
$lambda_{n}$ is the component of $varphi_{n}$ with the largest magnitude. After $N$ ( "many" ) iterations, we get the eigenvalue as $lambda_{N}$. This is the eigenvalue with the largest magnitude. Next, we normalize the eigenvector:
$ds{varphi_{N} to {varphi_{N} over varphi_{N}^{sf T},varphi_{N}}}$. "Reduce" the matrix as
$$
tilde{A} equiv A - lambda_{N} varphi_{N}varphi_{N}^{sf T}
$$
Repeat the procedure with $tilde{A}$ and so on.
See this link.
answered Dec 2 '13 at 23:06
Felix Marin
66.9k7107139
edited Dec 2 '13 at 23:17
answered Dec 2 '13 at 23:06
Felix Marin
66.9k7107139
answered Dec 2 '13 at 23:06
Felix Marin
66.9k7107139
answered Dec 2 '13 at 23:06
Felix Marin
66.9k7107139
66.9k7107139
add a comment |
add a comment |
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3
This may be better suited for stackoverflow.com/questions/tagged/matlab
– Henry Swanson
Dec 2 '13 at 22:14
I'll x-post it there. I appreciate it.
– Heath Huffman
Dec 2 '13 at 22:17
Yeah, they'll probably know the specific syntax. If there's not a syntax for getting eigenvectors, you can look at the nullspace of $A - lambda I$, where $lambda$ is an eigenvalue. That'll give you the eigenvectors. The other two steps are just MATLAB syntax somehow.
– Henry Swanson
Dec 2 '13 at 22:21
1
To reinforce(?) @HenrySwanson 's Comment, if the Question is about an approach (Matlab: full eigensolution followed by picking out the two largest eigenvalues) already decided upon, it would certainly be ripe for SO. A mathematical (but computational) question would be how to best approximate the two top eigenvalues without getting all of them (such questions are also handled at SciComp.SE).
– hardmath
Dec 2 '13 at 22:27
1
It's the $largett Power Method$ algorithm or $largett Von Mises$ one which yields the eigenvalue of largest magnitude. Once you get this, you can "remove" that eigenvalue and repeat the algorithm for the next one. See ---> en.wikipedia.org/wiki/Power_iteration
– Felix Marin
Dec 2 '13 at 23:04