Mixing
Can I use the power method to find all eigenvalues?
$begingroup$
With Power Method, I can find the largest eigenvalue and the corresponding eigenvector. But what if I have and $n,n$ matrix e.g $4,4$
$$A = begin{bmatrix}
2 & 0 & 0 & 0 \
1 & 2 & 0 & 0 \
0 & 1 & 3 & 0 \
0 & 0 & 1 & 3
end{bmatrix}$$
And then I want to compute the eigenvalues.
$$|A - lambda I| = begin{vmatrix}
2 - lambda & 0 & 0 & 0 \
1 & 2- lambda & 0 & 0 \
0 & 1 & 3- lambda & 0 \
0 & 0 & 1 & 3- lambda
end{vmatrix} =
(2 - lambda)^2(3 - lambda)^2$$
Here it's easy to find that the eigenvalues are 2 and 3. But with the power method, I would only get number 3. Is a way to re-compute the power method with that eigenvalue 3 and find eigenvalue 2?
eigenvalues-eigenvectors
asked Jan 31 at 1:12
Daniel MårtenssonDaniel Mårtensson
993419
$endgroup$
add a comment |
$begingroup$
With Power Method, I can find the largest eigenvalue and the corresponding eigenvector. But what if I have and $n,n$ matrix e.g $4,4$
$$A = begin{bmatrix}
2 & 0 & 0 & 0 \
1 & 2 & 0 & 0 \
0 & 1 & 3 & 0 \
0 & 0 & 1 & 3
end{bmatrix}$$
And then I want to compute the eigenvalues.
$$|A - lambda I| = begin{vmatrix}
2 - lambda & 0 & 0 & 0 \
1 & 2- lambda & 0 & 0 \
0 & 1 & 3- lambda & 0 \
0 & 0 & 1 & 3- lambda
end{vmatrix} =
(2 - lambda)^2(3 - lambda)^2$$
Here it's easy to find that the eigenvalues are 2 and 3. But with the power method, I would only get number 3. Is a way to re-compute the power method with that eigenvalue 3 and find eigenvalue 2?
eigenvalues-eigenvectors
asked Jan 31 at 1:12
Daniel MårtenssonDaniel Mårtensson
993419
$endgroup$
1
$begingroup$
If you've found the largest eigenvalue $3$, then the matrix $3I - A$ has all the same eigenvectors, and eigenvalues equal to $3 - lambda$ where $lambda$ is an eigenvalue for $A$. So, in this case, the power method could be used to identify $1$ as an eigenvalue of $3I - A$, which implies $2$ is an eigenvalue of $A$.
$endgroup$
– Theo Bendit
Jan 31 at 1:17
$begingroup$
Great question -- look up the Arnoldi method.
$endgroup$
– Neal
Jan 31 at 1:43
$begingroup$
As an alternative to @TheoBendit's suggestion, there are algorithms that 'deflate' the matrix to a smaller matrix without the largest eigenvalue. Then the power method can be repeated.
$endgroup$
– Klaas van Aarsen
Jan 31 at 8:59
add a comment |
$begingroup$
With Power Method, I can find the largest eigenvalue and the corresponding eigenvector. But what if I have and $n,n$ matrix e.g $4,4$
$$A = begin{bmatrix}
2 & 0 & 0 & 0 \
1 & 2 & 0 & 0 \
0 & 1 & 3 & 0 \
0 & 0 & 1 & 3
end{bmatrix}$$
And then I want to compute the eigenvalues.
$$|A - lambda I| = begin{vmatrix}
2 - lambda & 0 & 0 & 0 \
1 & 2- lambda & 0 & 0 \
0 & 1 & 3- lambda & 0 \
0 & 0 & 1 & 3- lambda
end{vmatrix} =
(2 - lambda)^2(3 - lambda)^2$$
Here it's easy to find that the eigenvalues are 2 and 3. But with the power method, I would only get number 3. Is a way to re-compute the power method with that eigenvalue 3 and find eigenvalue 2?
eigenvalues-eigenvectors
asked Jan 31 at 1:12
Daniel MårtenssonDaniel Mårtensson
993419
$endgroup$
With Power Method, I can find the largest eigenvalue and the corresponding eigenvector. But what if I have and $n,n$ matrix e.g $4,4$
$$A = begin{bmatrix}
2 & 0 & 0 & 0 \
1 & 2 & 0 & 0 \
0 & 1 & 3 & 0 \
0 & 0 & 1 & 3
end{bmatrix}$$
And then I want to compute the eigenvalues.
$$|A - lambda I| = begin{vmatrix}
2 - lambda & 0 & 0 & 0 \
1 & 2- lambda & 0 & 0 \
0 & 1 & 3- lambda & 0 \
0 & 0 & 1 & 3- lambda
end{vmatrix} =
(2 - lambda)^2(3 - lambda)^2$$
Here it's easy to find that the eigenvalues are 2 and 3. But with the power method, I would only get number 3. Is a way to re-compute the power method with that eigenvalue 3 and find eigenvalue 2?
eigenvalues-eigenvectors
eigenvalues-eigenvectors
asked Jan 31 at 1:12
Daniel MårtenssonDaniel Mårtensson
993419
asked Jan 31 at 1:12
Daniel MårtenssonDaniel Mårtensson
993419
asked Jan 31 at 1:12
Daniel MårtenssonDaniel Mårtensson
993419
asked Jan 31 at 1:12
Daniel MårtenssonDaniel Mårtensson
993419
asked Jan 31 at 1:12
Daniel MårtenssonDaniel Mårtensson
993419
993419
1
$begingroup$
If you've found the largest eigenvalue $3$, then the matrix $3I - A$ has all the same eigenvectors, and eigenvalues equal to $3 - lambda$ where $lambda$ is an eigenvalue for $A$. So, in this case, the power method could be used to identify $1$ as an eigenvalue of $3I - A$, which implies $2$ is an eigenvalue of $A$.
$endgroup$
– Theo Bendit
Jan 31 at 1:17
$begingroup$
Great question -- look up the Arnoldi method.
$endgroup$
– Neal
Jan 31 at 1:43
$begingroup$
As an alternative to @TheoBendit's suggestion, there are algorithms that 'deflate' the matrix to a smaller matrix without the largest eigenvalue. Then the power method can be repeated.
$endgroup$
– Klaas van Aarsen
Jan 31 at 8:59
add a comment |
1
$begingroup$
If you've found the largest eigenvalue $3$, then the matrix $3I - A$ has all the same eigenvectors, and eigenvalues equal to $3 - lambda$ where $lambda$ is an eigenvalue for $A$. So, in this case, the power method could be used to identify $1$ as an eigenvalue of $3I - A$, which implies $2$ is an eigenvalue of $A$.
$endgroup$
– Theo Bendit
Jan 31 at 1:17
$begingroup$
Great question -- look up the Arnoldi method.
$endgroup$
– Neal
Jan 31 at 1:43
$begingroup$
As an alternative to @TheoBendit's suggestion, there are algorithms that 'deflate' the matrix to a smaller matrix without the largest eigenvalue. Then the power method can be repeated.
$endgroup$
– Klaas van Aarsen
Jan 31 at 8:59
1
1
$begingroup$
If you've found the largest eigenvalue $3$, then the matrix $3I - A$ has all the same eigenvectors, and eigenvalues equal to $3 - lambda$ where $lambda$ is an eigenvalue for $A$. So, in this case, the power method could be used to identify $1$ as an eigenvalue of $3I - A$, which implies $2$ is an eigenvalue of $A$.
$endgroup$
– Theo Bendit
Jan 31 at 1:17
$begingroup$
If you've found the largest eigenvalue $3$, then the matrix $3I - A$ has all the same eigenvectors, and eigenvalues equal to $3 - lambda$ where $lambda$ is an eigenvalue for $A$. So, in this case, the power method could be used to identify $1$ as an eigenvalue of $3I - A$, which implies $2$ is an eigenvalue of $A$.
$endgroup$
– Theo Bendit
Jan 31 at 1:17
$begingroup$
Great question -- look up the Arnoldi method.
$endgroup$
– Neal
Jan 31 at 1:43
$begingroup$
Great question -- look up the Arnoldi method.
$endgroup$
– Neal
Jan 31 at 1:43
$begingroup$
As an alternative to @TheoBendit's suggestion, there are algorithms that 'deflate' the matrix to a smaller matrix without the largest eigenvalue. Then the power method can be repeated.
$endgroup$
– Klaas van Aarsen
Jan 31 at 8:59
$begingroup$
As an alternative to @TheoBendit's suggestion, there are algorithms that 'deflate' the matrix to a smaller matrix without the largest eigenvalue. Then the power method can be repeated.
$endgroup$
– Klaas van Aarsen
Jan 31 at 8:59
add a comment |
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1
$begingroup$
If you've found the largest eigenvalue $3$, then the matrix $3I - A$ has all the same eigenvectors, and eigenvalues equal to $3 - lambda$ where $lambda$ is an eigenvalue for $A$. So, in this case, the power method could be used to identify $1$ as an eigenvalue of $3I - A$, which implies $2$ is an eigenvalue of $A$.
$endgroup$
– Theo Bendit
Jan 31 at 1:17
$begingroup$
Great question -- look up the Arnoldi method.
$endgroup$
– Neal
Jan 31 at 1:43
$begingroup$
As an alternative to @TheoBendit's suggestion, there are algorithms that 'deflate' the matrix to a smaller matrix without the largest eigenvalue. Then the power method can be repeated.
$endgroup$
– Klaas van Aarsen
Jan 31 at 8:59