Biased eigenvalue problem

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I came across the following "biased eigenvalue problem" when I have tried to solve a quadratic constrained optimization problem.

$$Ax = lambda (x+v), $$

where $A inmathbbS^n$ (the set of $ntimes n$ positive semidefinite matrices), and $v in mathbbR^n$ a given unit vector.

My thoughts

By substituting $y =x+v$ into the problem, we have
$$A(y-v) = lambda y implies Ay = lambda y + u,$$ where $u =Av$. Here, $lambda$ is only multiplied by the vector $y$, but still it is a biased problem.

How to find the eigenpairs $(x,lambda)$ or $(y,lambda)$? Is there a closed form or an algorithm to solve this problem?

I just need some hints or link for papers. Thanks in advance!

eigenvalues-eigenvectors positive-semidefinite

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edited Jul 27 at 7:17

asked Jul 27 at 7:01

Alex Silva

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up vote
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down vote

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I came across the following "biased eigenvalue problem" when I have tried to solve a quadratic constrained optimization problem.

$$Ax = lambda (x+v), $$

where $A inmathbbS^n$ (the set of $ntimes n$ positive semidefinite matrices), and $v in mathbbR^n$ a given unit vector.

My thoughts

By substituting $y =x+v$ into the problem, we have
$$A(y-v) = lambda y implies Ay = lambda y + u,$$ where $u =Av$. Here, $lambda$ is only multiplied by the vector $y$, but still it is a biased problem.

How to find the eigenpairs $(x,lambda)$ or $(y,lambda)$? Is there a closed form or an algorithm to solve this problem?

I just need some hints or link for papers. Thanks in advance!

eigenvalues-eigenvectors positive-semidefinite

share|cite|improve this question

edited Jul 27 at 7:17

asked Jul 27 at 7:01

Alex Silva

2,68231232

add a comment | 

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

I came across the following "biased eigenvalue problem" when I have tried to solve a quadratic constrained optimization problem.

$$Ax = lambda (x+v), $$

where $A inmathbbS^n$ (the set of $ntimes n$ positive semidefinite matrices), and $v in mathbbR^n$ a given unit vector.

My thoughts

By substituting $y =x+v$ into the problem, we have
$$A(y-v) = lambda y implies Ay = lambda y + u,$$ where $u =Av$. Here, $lambda$ is only multiplied by the vector $y$, but still it is a biased problem.

How to find the eigenpairs $(x,lambda)$ or $(y,lambda)$? Is there a closed form or an algorithm to solve this problem?

I just need some hints or link for papers. Thanks in advance!

eigenvalues-eigenvectors positive-semidefinite

share|cite|improve this question

edited Jul 27 at 7:17

asked Jul 27 at 7:01

Alex Silva

2,68231232

I came across the following "biased eigenvalue problem" when I have tried to solve a quadratic constrained optimization problem.

$$Ax = lambda (x+v), $$

where $A inmathbbS^n$ (the set of $ntimes n$ positive semidefinite matrices), and $v in mathbbR^n$ a given unit vector.

My thoughts

By substituting $y =x+v$ into the problem, we have
$$A(y-v) = lambda y implies Ay = lambda y + u,$$ where $u =Av$. Here, $lambda$ is only multiplied by the vector $y$, but still it is a biased problem.

How to find the eigenpairs $(x,lambda)$ or $(y,lambda)$? Is there a closed form or an algorithm to solve this problem?

I just need some hints or link for papers. Thanks in advance!

eigenvalues-eigenvectors positive-semidefinite

share|cite|improve this question

edited Jul 27 at 7:17

asked Jul 27 at 7:01

Alex Silva

2,68231232

share|cite|improve this question

share|cite|improve this question

edited Jul 27 at 7:17

edited Jul 27 at 7:17

edited Jul 27 at 7:17

asked Jul 27 at 7:01

Alex Silva

2,68231232

asked Jul 27 at 7:01

Alex Silva

2,68231232

asked Jul 27 at 7:01

Alex Silva

2,68231232

2,68231232

add a comment | 

add a comment | 

1 Answer
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The problem can be written

$$(A-lambda I)x=lambda u.$$

In general, $A-lambda I$ is invertible (unless $lambda$ is an Eigenvalue of $A$) and the solution

$$x=lambda (A-lambda I)^-1u$$ holds.

The "trajectory" of the solution is a rational expression in $lambda$, nothing simple.

With a random $3times3$ example:

share|cite|improve this answer

edited Jul 27 at 9:02

answered Jul 27 at 8:08

Yves Daoust

110k665203

add a comment | 

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1 Answer
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1 Answer
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active

oldest

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active

oldest

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up vote
1
down vote

The problem can be written

$$(A-lambda I)x=lambda u.$$

In general, $A-lambda I$ is invertible (unless $lambda$ is an Eigenvalue of $A$) and the solution

$$x=lambda (A-lambda I)^-1u$$ holds.

The "trajectory" of the solution is a rational expression in $lambda$, nothing simple.

With a random $3times3$ example:

share|cite|improve this answer

edited Jul 27 at 9:02

answered Jul 27 at 8:08

Yves Daoust

110k665203

add a comment | 

up vote
1
down vote

The problem can be written

$$(A-lambda I)x=lambda u.$$

In general, $A-lambda I$ is invertible (unless $lambda$ is an Eigenvalue of $A$) and the solution

$$x=lambda (A-lambda I)^-1u$$ holds.

The "trajectory" of the solution is a rational expression in $lambda$, nothing simple.

With a random $3times3$ example:

share|cite|improve this answer

edited Jul 27 at 9:02

answered Jul 27 at 8:08

Yves Daoust

110k665203

add a comment | 

up vote
1
down vote

up vote
1
down vote

The problem can be written

$$(A-lambda I)x=lambda u.$$

In general, $A-lambda I$ is invertible (unless $lambda$ is an Eigenvalue of $A$) and the solution

$$x=lambda (A-lambda I)^-1u$$ holds.

The "trajectory" of the solution is a rational expression in $lambda$, nothing simple.

With a random $3times3$ example:

share|cite|improve this answer

edited Jul 27 at 9:02

answered Jul 27 at 8:08

Yves Daoust

110k665203

The problem can be written

$$(A-lambda I)x=lambda u.$$

In general, $A-lambda I$ is invertible (unless $lambda$ is an Eigenvalue of $A$) and the solution

$$x=lambda (A-lambda I)^-1u$$ holds.

The "trajectory" of the solution is a rational expression in $lambda$, nothing simple.

With a random $3times3$ example:

share|cite|improve this answer

edited Jul 27 at 9:02

answered Jul 27 at 8:08

Yves Daoust

110k665203

share|cite|improve this answer

share|cite|improve this answer

edited Jul 27 at 9:02

edited Jul 27 at 9:02

edited Jul 27 at 9:02

answered Jul 27 at 8:08

Yves Daoust

110k665203

answered Jul 27 at 8:08

Yves Daoust

110k665203

answered Jul 27 at 8:08

Yves Daoust

110k665203

110k665203

add a comment | 

add a comment | 

 

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