I want to find largest eigenvalue using power iteration. Is it correct?
$begingroup$
$$A = begin{pmatrix}4 &–1 & 0 \-1& 4& -1\ 0 &-1& 4 end{pmatrix} $$ $X_0 = (1, 1, 1)$. Perform 5 steps of power iterations to find the largest eigenvalue of A.
And my solution is this.
enter image description here
But, some people says that eigenvalue is 5. This is because when last step, it is based on the middle value. Why?
linear-algebra eigenvalues-eigenvectors
edited Dec 16 '18 at 17:33
Wesley Strik
2,172424
asked Dec 16 '18 at 17:19
Genius ParkGenius Park
111
$endgroup$
add a comment |
$begingroup$
$$A = begin{pmatrix}4 &–1 & 0 \-1& 4& -1\ 0 &-1& 4 end{pmatrix} $$ $X_0 = (1, 1, 1)$. Perform 5 steps of power iterations to find the largest eigenvalue of A.
And my solution is this.
enter image description here
But, some people says that eigenvalue is 5. This is because when last step, it is based on the middle value. Why?
linear-algebra eigenvalues-eigenvectors
edited Dec 16 '18 at 17:33
Wesley Strik
2,172424
asked Dec 16 '18 at 17:19
Genius ParkGenius Park
111
$endgroup$
$begingroup$
Remember that is is just an approximation, if you want you can take the average of the ratio of the entries of the two vectors. As you apply the matrix multiple times your values will (possibly, if it has eigenvalues) get better and better.
$endgroup$
– Wesley Strik
Dec 16 '18 at 17:37
$begingroup$
It is not hard for this matrix to determine the actual eigenvalue - although maybe a bit messy. You get a polynomial of the form $(x-4)(x-(4+ sqrt{2}))(x-(4- sqrt{2}))=0$
$endgroup$
– Wesley Strik
Dec 16 '18 at 17:40
$begingroup$
Your method is fine. You assume that $Av = lambda v$ where $v$ an eigenvector, at some point. This assumption need not be always met at all or even after a certain amount of iterations, but, you know, these are approximate methods.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:26
add a comment |
$begingroup$
$$A = begin{pmatrix}4 &–1 & 0 \-1& 4& -1\ 0 &-1& 4 end{pmatrix} $$ $X_0 = (1, 1, 1)$. Perform 5 steps of power iterations to find the largest eigenvalue of A.
And my solution is this.
enter image description here
But, some people says that eigenvalue is 5. This is because when last step, it is based on the middle value. Why?
linear-algebra eigenvalues-eigenvectors
edited Dec 16 '18 at 17:33
Wesley Strik
2,172424
asked Dec 16 '18 at 17:19
Genius ParkGenius Park
111
$endgroup$
$$A = begin{pmatrix}4 &–1 & 0 \-1& 4& -1\ 0 &-1& 4 end{pmatrix} $$ $X_0 = (1, 1, 1)$. Perform 5 steps of power iterations to find the largest eigenvalue of A.
And my solution is this.
enter image description here
But, some people says that eigenvalue is 5. This is because when last step, it is based on the middle value. Why?
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
edited Dec 16 '18 at 17:33
Wesley Strik
2,172424
asked Dec 16 '18 at 17:19
Genius ParkGenius Park
111
edited Dec 16 '18 at 17:33
Wesley Strik
2,172424
asked Dec 16 '18 at 17:19
Genius ParkGenius Park
111
edited Dec 16 '18 at 17:33
Wesley Strik
2,172424
edited Dec 16 '18 at 17:33
Wesley Strik
2,172424
edited Dec 16 '18 at 17:33
Wesley Strik
2,172424
2,172424
asked Dec 16 '18 at 17:19
Genius ParkGenius Park
111
asked Dec 16 '18 at 17:19
Genius ParkGenius Park
111
asked Dec 16 '18 at 17:19
Genius ParkGenius Park
111
111
$begingroup$
Remember that is is just an approximation, if you want you can take the average of the ratio of the entries of the two vectors. As you apply the matrix multiple times your values will (possibly, if it has eigenvalues) get better and better.
$endgroup$
– Wesley Strik
Dec 16 '18 at 17:37
$begingroup$
It is not hard for this matrix to determine the actual eigenvalue - although maybe a bit messy. You get a polynomial of the form $(x-4)(x-(4+ sqrt{2}))(x-(4- sqrt{2}))=0$
$endgroup$
– Wesley Strik
Dec 16 '18 at 17:40
$begingroup$
Your method is fine. You assume that $Av = lambda v$ where $v$ an eigenvector, at some point. This assumption need not be always met at all or even after a certain amount of iterations, but, you know, these are approximate methods.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:26
add a comment |
$begingroup$
Remember that is is just an approximation, if you want you can take the average of the ratio of the entries of the two vectors. As you apply the matrix multiple times your values will (possibly, if it has eigenvalues) get better and better.
$endgroup$
– Wesley Strik
Dec 16 '18 at 17:37
$begingroup$
It is not hard for this matrix to determine the actual eigenvalue - although maybe a bit messy. You get a polynomial of the form $(x-4)(x-(4+ sqrt{2}))(x-(4- sqrt{2}))=0$
$endgroup$
– Wesley Strik
Dec 16 '18 at 17:40
$begingroup$
Your method is fine. You assume that $Av = lambda v$ where $v$ an eigenvector, at some point. This assumption need not be always met at all or even after a certain amount of iterations, but, you know, these are approximate methods.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:26
$begingroup$
Remember that is is just an approximation, if you want you can take the average of the ratio of the entries of the two vectors. As you apply the matrix multiple times your values will (possibly, if it has eigenvalues) get better and better.
$endgroup$
– Wesley Strik
Dec 16 '18 at 17:37
$begingroup$
Remember that is is just an approximation, if you want you can take the average of the ratio of the entries of the two vectors. As you apply the matrix multiple times your values will (possibly, if it has eigenvalues) get better and better.
$endgroup$
– Wesley Strik
Dec 16 '18 at 17:37
$begingroup$
It is not hard for this matrix to determine the actual eigenvalue - although maybe a bit messy. You get a polynomial of the form $(x-4)(x-(4+ sqrt{2}))(x-(4- sqrt{2}))=0$
$endgroup$
– Wesley Strik
Dec 16 '18 at 17:40
$begingroup$
It is not hard for this matrix to determine the actual eigenvalue - although maybe a bit messy. You get a polynomial of the form $(x-4)(x-(4+ sqrt{2}))(x-(4- sqrt{2}))=0$
$endgroup$
– Wesley Strik
Dec 16 '18 at 17:40
$begingroup$
Your method is fine. You assume that $Av = lambda v$ where $v$ an eigenvector, at some point. This assumption need not be always met at all or even after a certain amount of iterations, but, you know, these are approximate methods.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:26
$begingroup$
Your method is fine. You assume that $Av = lambda v$ where $v$ an eigenvector, at some point. This assumption need not be always met at all or even after a certain amount of iterations, but, you know, these are approximate methods.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can check with any entry you want, it is indeed approximately $5$ (if you work it out analytically you notice that largest eigenvalue is $4+ sqrt{2}$), but remember that you are using an approximation method, so there is no "correct"way to do it. If you would use more iterations we expect the error to get smaller and to approach the "true" eigenvalue, if it exists of course.
Another, better way to do it is by looking at:
http://college.cengage.com/mathematics/larson/elementary_linear/5e/students/ch08-10/chap_10_3.pdf
Where they show that a good way to do it is by computing the so-called Rayleigh quotient, which if $A^5 x= bar{x} $ is defined to be:
$$ frac{Abar{x} cdotbar{x}}{x cdot x} $$
Give it a try and let us know which $lambda$ you find ?
answered Dec 16 '18 at 17:35
Wesley StrikWesley Strik
2,172424
$endgroup$
$begingroup$
when I solve a problem like this, can I get the criteria any row of matrix to find eigenvalue?
$endgroup$
– Genius Park
Dec 17 '18 at 4:51
$begingroup$
I matrix might have imaginary eigenvalues and there is no such thing as a greatest imaginary eigenvalue, because $mathbb{C}$ is not completely ordered. You can find an eigenvalue of largest modulus though. I'm afraid I do not completely grasp your question.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:21
$begingroup$
If the question is, does it matter which row I pick, no, not really.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:22
$begingroup$
The assumption is that at some point the vector stabilises and is an eigenvector. Therefore we have that it will be some multiple of itself after applying the matrix. Any entry will be a scalar multiple of the previous entry of the previous vector.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:24
add a comment |
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$begingroup$
You can check with any entry you want, it is indeed approximately $5$ (if you work it out analytically you notice that largest eigenvalue is $4+ sqrt{2}$), but remember that you are using an approximation method, so there is no "correct"way to do it. If you would use more iterations we expect the error to get smaller and to approach the "true" eigenvalue, if it exists of course.
Another, better way to do it is by looking at:
http://college.cengage.com/mathematics/larson/elementary_linear/5e/students/ch08-10/chap_10_3.pdf
Where they show that a good way to do it is by computing the so-called Rayleigh quotient, which if $A^5 x= bar{x} $ is defined to be:
$$ frac{Abar{x} cdotbar{x}}{x cdot x} $$
Give it a try and let us know which $lambda$ you find ?
answered Dec 16 '18 at 17:35
Wesley StrikWesley Strik
2,172424
$endgroup$
$begingroup$
when I solve a problem like this, can I get the criteria any row of matrix to find eigenvalue?
$endgroup$
– Genius Park
Dec 17 '18 at 4:51
$begingroup$
I matrix might have imaginary eigenvalues and there is no such thing as a greatest imaginary eigenvalue, because $mathbb{C}$ is not completely ordered. You can find an eigenvalue of largest modulus though. I'm afraid I do not completely grasp your question.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:21
$begingroup$
If the question is, does it matter which row I pick, no, not really.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:22
$begingroup$
The assumption is that at some point the vector stabilises and is an eigenvector. Therefore we have that it will be some multiple of itself after applying the matrix. Any entry will be a scalar multiple of the previous entry of the previous vector.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:24
add a comment |
$begingroup$
You can check with any entry you want, it is indeed approximately $5$ (if you work it out analytically you notice that largest eigenvalue is $4+ sqrt{2}$), but remember that you are using an approximation method, so there is no "correct"way to do it. If you would use more iterations we expect the error to get smaller and to approach the "true" eigenvalue, if it exists of course.
Another, better way to do it is by looking at:
http://college.cengage.com/mathematics/larson/elementary_linear/5e/students/ch08-10/chap_10_3.pdf
Where they show that a good way to do it is by computing the so-called Rayleigh quotient, which if $A^5 x= bar{x} $ is defined to be:
$$ frac{Abar{x} cdotbar{x}}{x cdot x} $$
Give it a try and let us know which $lambda$ you find ?
answered Dec 16 '18 at 17:35
Wesley StrikWesley Strik
2,172424
$endgroup$
$begingroup$
when I solve a problem like this, can I get the criteria any row of matrix to find eigenvalue?
$endgroup$
– Genius Park
Dec 17 '18 at 4:51
$begingroup$
I matrix might have imaginary eigenvalues and there is no such thing as a greatest imaginary eigenvalue, because $mathbb{C}$ is not completely ordered. You can find an eigenvalue of largest modulus though. I'm afraid I do not completely grasp your question.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:21
$begingroup$
If the question is, does it matter which row I pick, no, not really.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:22
$begingroup$
The assumption is that at some point the vector stabilises and is an eigenvector. Therefore we have that it will be some multiple of itself after applying the matrix. Any entry will be a scalar multiple of the previous entry of the previous vector.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:24
add a comment |
$begingroup$
You can check with any entry you want, it is indeed approximately $5$ (if you work it out analytically you notice that largest eigenvalue is $4+ sqrt{2}$), but remember that you are using an approximation method, so there is no "correct"way to do it. If you would use more iterations we expect the error to get smaller and to approach the "true" eigenvalue, if it exists of course.
Another, better way to do it is by looking at:
http://college.cengage.com/mathematics/larson/elementary_linear/5e/students/ch08-10/chap_10_3.pdf
Where they show that a good way to do it is by computing the so-called Rayleigh quotient, which if $A^5 x= bar{x} $ is defined to be:
$$ frac{Abar{x} cdotbar{x}}{x cdot x} $$
Give it a try and let us know which $lambda$ you find ?
answered Dec 16 '18 at 17:35
Wesley StrikWesley Strik
2,172424
$endgroup$
You can check with any entry you want, it is indeed approximately $5$ (if you work it out analytically you notice that largest eigenvalue is $4+ sqrt{2}$), but remember that you are using an approximation method, so there is no "correct"way to do it. If you would use more iterations we expect the error to get smaller and to approach the "true" eigenvalue, if it exists of course.
Another, better way to do it is by looking at:
http://college.cengage.com/mathematics/larson/elementary_linear/5e/students/ch08-10/chap_10_3.pdf
Where they show that a good way to do it is by computing the so-called Rayleigh quotient, which if $A^5 x= bar{x} $ is defined to be:
$$ frac{Abar{x} cdotbar{x}}{x cdot x} $$
Give it a try and let us know which $lambda$ you find ?
answered Dec 16 '18 at 17:35
Wesley StrikWesley Strik
2,172424
edited Dec 16 '18 at 17:43
answered Dec 16 '18 at 17:35
Wesley StrikWesley Strik
2,172424
answered Dec 16 '18 at 17:35
Wesley StrikWesley Strik
2,172424
answered Dec 16 '18 at 17:35
Wesley StrikWesley Strik
2,172424
2,172424
$begingroup$
when I solve a problem like this, can I get the criteria any row of matrix to find eigenvalue?
$endgroup$
– Genius Park
Dec 17 '18 at 4:51
$begingroup$
I matrix might have imaginary eigenvalues and there is no such thing as a greatest imaginary eigenvalue, because $mathbb{C}$ is not completely ordered. You can find an eigenvalue of largest modulus though. I'm afraid I do not completely grasp your question.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:21
$begingroup$
If the question is, does it matter which row I pick, no, not really.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:22
$begingroup$
The assumption is that at some point the vector stabilises and is an eigenvector. Therefore we have that it will be some multiple of itself after applying the matrix. Any entry will be a scalar multiple of the previous entry of the previous vector.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:24
add a comment |
$begingroup$
when I solve a problem like this, can I get the criteria any row of matrix to find eigenvalue?
$endgroup$
– Genius Park
Dec 17 '18 at 4:51
$begingroup$
I matrix might have imaginary eigenvalues and there is no such thing as a greatest imaginary eigenvalue, because $mathbb{C}$ is not completely ordered. You can find an eigenvalue of largest modulus though. I'm afraid I do not completely grasp your question.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:21
$begingroup$
If the question is, does it matter which row I pick, no, not really.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:22
$begingroup$
The assumption is that at some point the vector stabilises and is an eigenvector. Therefore we have that it will be some multiple of itself after applying the matrix. Any entry will be a scalar multiple of the previous entry of the previous vector.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:24
$begingroup$
when I solve a problem like this, can I get the criteria any row of matrix to find eigenvalue?
$endgroup$
– Genius Park
Dec 17 '18 at 4:51
$begingroup$
when I solve a problem like this, can I get the criteria any row of matrix to find eigenvalue?
$endgroup$
– Genius Park
Dec 17 '18 at 4:51
$begingroup$
I matrix might have imaginary eigenvalues and there is no such thing as a greatest imaginary eigenvalue, because $mathbb{C}$ is not completely ordered. You can find an eigenvalue of largest modulus though. I'm afraid I do not completely grasp your question.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:21
$begingroup$
I matrix might have imaginary eigenvalues and there is no such thing as a greatest imaginary eigenvalue, because $mathbb{C}$ is not completely ordered. You can find an eigenvalue of largest modulus though. I'm afraid I do not completely grasp your question.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:21
$begingroup$
If the question is, does it matter which row I pick, no, not really.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:22
$begingroup$
If the question is, does it matter which row I pick, no, not really.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:22
$begingroup$
The assumption is that at some point the vector stabilises and is an eigenvector. Therefore we have that it will be some multiple of itself after applying the matrix. Any entry will be a scalar multiple of the previous entry of the previous vector.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:24
$begingroup$
The assumption is that at some point the vector stabilises and is an eigenvector. Therefore we have that it will be some multiple of itself after applying the matrix. Any entry will be a scalar multiple of the previous entry of the previous vector.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:24
add a comment |
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$begingroup$
Remember that is is just an approximation, if you want you can take the average of the ratio of the entries of the two vectors. As you apply the matrix multiple times your values will (possibly, if it has eigenvalues) get better and better.
$endgroup$
– Wesley Strik
Dec 16 '18 at 17:37
$begingroup$
It is not hard for this matrix to determine the actual eigenvalue - although maybe a bit messy. You get a polynomial of the form $(x-4)(x-(4+ sqrt{2}))(x-(4- sqrt{2}))=0$
$endgroup$
– Wesley Strik
Dec 16 '18 at 17:40
$begingroup$
Your method is fine. You assume that $Av = lambda v$ where $v$ an eigenvector, at some point. This assumption need not be always met at all or even after a certain amount of iterations, but, you know, these are approximate methods.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:26