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How to compute eigenvector for rank 1 matrix without high complexity SVD?
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I am sorry for asking probably well known or trivial problem.
How to compute eigenvector for rank 1 matrix, say $A = ab^* in M_n(mathbb{C})$ without performing high complexity SVD ($A = U Sigma V^*$)?
I understand that the eigenvalue $lambda $ of such rank 1 matrix is either $b^*a$ or $0$. But how to find the appropriate eigenvector $Ax = lambda x$, where $x$ is an eigenvector?
linear-algebra
asked Jan 9 at 9:51
learninglearning
526
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add a comment |
$begingroup$
I am sorry for asking probably well known or trivial problem.
How to compute eigenvector for rank 1 matrix, say $A = ab^* in M_n(mathbb{C})$ without performing high complexity SVD ($A = U Sigma V^*$)?
I understand that the eigenvalue $lambda $ of such rank 1 matrix is either $b^*a$ or $0$. But how to find the appropriate eigenvector $Ax = lambda x$, where $x$ is an eigenvector?
linear-algebra
asked Jan 9 at 9:51
learninglearning
526
$endgroup$
add a comment |
$begingroup$
I am sorry for asking probably well known or trivial problem.
How to compute eigenvector for rank 1 matrix, say $A = ab^* in M_n(mathbb{C})$ without performing high complexity SVD ($A = U Sigma V^*$)?
I understand that the eigenvalue $lambda $ of such rank 1 matrix is either $b^*a$ or $0$. But how to find the appropriate eigenvector $Ax = lambda x$, where $x$ is an eigenvector?
linear-algebra
asked Jan 9 at 9:51
learninglearning
526
$endgroup$
I am sorry for asking probably well known or trivial problem.
How to compute eigenvector for rank 1 matrix, say $A = ab^* in M_n(mathbb{C})$ without performing high complexity SVD ($A = U Sigma V^*$)?
I understand that the eigenvalue $lambda $ of such rank 1 matrix is either $b^*a$ or $0$. But how to find the appropriate eigenvector $Ax = lambda x$, where $x$ is an eigenvector?
linear-algebra
linear-algebra
asked Jan 9 at 9:51
learninglearning
526
asked Jan 9 at 9:51
learninglearning
526
asked Jan 9 at 9:51
learninglearning
526
asked Jan 9 at 9:51
learninglearning
526
asked Jan 9 at 9:51
learninglearning
526
526
add a comment |
add a comment |
1 Answer
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$begingroup$
The eigenvector corresponding to the eigenvalue $b^*a$ is just $a$; $Aa=(ab^*)a=a(b^*a)$. That's just the associative law for matrix products - note that the product of a $ntimes 1$ times a $1times 1$ is the same as vector space scalar multiplication.
In practice, if we don't have it written that way already? Pick a column of the rank-1 matrix (preferably one that's not identically zero). It's a multiple of the eigenvector.
answered Jan 9 at 10:01
jmerryjmerry
6,457718
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1 Answer
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1 Answer
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$begingroup$
The eigenvector corresponding to the eigenvalue $b^*a$ is just $a$; $Aa=(ab^*)a=a(b^*a)$. That's just the associative law for matrix products - note that the product of a $ntimes 1$ times a $1times 1$ is the same as vector space scalar multiplication.
In practice, if we don't have it written that way already? Pick a column of the rank-1 matrix (preferably one that's not identically zero). It's a multiple of the eigenvector.
answered Jan 9 at 10:01
jmerryjmerry
6,457718
$endgroup$
add a comment |
$begingroup$
The eigenvector corresponding to the eigenvalue $b^*a$ is just $a$; $Aa=(ab^*)a=a(b^*a)$. That's just the associative law for matrix products - note that the product of a $ntimes 1$ times a $1times 1$ is the same as vector space scalar multiplication.
In practice, if we don't have it written that way already? Pick a column of the rank-1 matrix (preferably one that's not identically zero). It's a multiple of the eigenvector.
answered Jan 9 at 10:01
jmerryjmerry
6,457718
$endgroup$
add a comment |
$begingroup$
The eigenvector corresponding to the eigenvalue $b^*a$ is just $a$; $Aa=(ab^*)a=a(b^*a)$. That's just the associative law for matrix products - note that the product of a $ntimes 1$ times a $1times 1$ is the same as vector space scalar multiplication.
In practice, if we don't have it written that way already? Pick a column of the rank-1 matrix (preferably one that's not identically zero). It's a multiple of the eigenvector.
answered Jan 9 at 10:01
jmerryjmerry
6,457718
$endgroup$
The eigenvector corresponding to the eigenvalue $b^*a$ is just $a$; $Aa=(ab^*)a=a(b^*a)$. That's just the associative law for matrix products - note that the product of a $ntimes 1$ times a $1times 1$ is the same as vector space scalar multiplication.
In practice, if we don't have it written that way already? Pick a column of the rank-1 matrix (preferably one that's not identically zero). It's a multiple of the eigenvector.
answered Jan 9 at 10:01
jmerryjmerry
6,457718
answered Jan 9 at 10:01
jmerryjmerry
6,457718
answered Jan 9 at 10:01
jmerryjmerry
6,457718
answered Jan 9 at 10:01
jmerryjmerry
6,457718
6,457718
add a comment |
add a comment |
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