Positive Definite Matrices and eigenvalues
Multi tool use
$begingroup$
Problem:
Let $A$ ∈ ${C}^{n×n}$ be, such that for every x ∈ $C^n$ <$A$x$, x$> ≥ 0
Show that all eigenvalues of $A$ are positive or zero
I suppose that from the standart inner product in the problem we can say that $A$ is a positive definite matrix and therefore follows that the eigenvalues of $A$ are positive.
However I am not sure if that is right,could someone give a hint?
linear-algebra eigenvalues-eigenvectors positive-definite
asked Jan 10 at 18:45
KaiKai
636
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add a comment |
$begingroup$
Problem:
Let $A$ ∈ ${C}^{n×n}$ be, such that for every x ∈ $C^n$ <$A$x$, x$> ≥ 0
Show that all eigenvalues of $A$ are positive or zero
I suppose that from the standart inner product in the problem we can say that $A$ is a positive definite matrix and therefore follows that the eigenvalues of $A$ are positive.
However I am not sure if that is right,could someone give a hint?
linear-algebra eigenvalues-eigenvectors positive-definite
asked Jan 10 at 18:45
KaiKai
636
$endgroup$
add a comment |
$begingroup$
Problem:
Let $A$ ∈ ${C}^{n×n}$ be, such that for every x ∈ $C^n$ <$A$x$, x$> ≥ 0
Show that all eigenvalues of $A$ are positive or zero
I suppose that from the standart inner product in the problem we can say that $A$ is a positive definite matrix and therefore follows that the eigenvalues of $A$ are positive.
However I am not sure if that is right,could someone give a hint?
linear-algebra eigenvalues-eigenvectors positive-definite
asked Jan 10 at 18:45
KaiKai
636
$endgroup$
Problem:
Let $A$ ∈ ${C}^{n×n}$ be, such that for every x ∈ $C^n$ <$A$x$, x$> ≥ 0
Show that all eigenvalues of $A$ are positive or zero
I suppose that from the standart inner product in the problem we can say that $A$ is a positive definite matrix and therefore follows that the eigenvalues of $A$ are positive.
However I am not sure if that is right,could someone give a hint?
linear-algebra eigenvalues-eigenvectors positive-definite
linear-algebra eigenvalues-eigenvectors positive-definite
asked Jan 10 at 18:45
KaiKai
636
asked Jan 10 at 18:45
KaiKai
636
asked Jan 10 at 18:45
KaiKai
636
asked Jan 10 at 18:45
KaiKai
636
asked Jan 10 at 18:45
KaiKai
636
636
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It's a straightforward computation. If $lambda$ is an eigenvalue of $A$, choose an eigenvector $v$ with $langle v,vrangle=1$. Then
$$
lambda=langle lambda v,vrangle=langle Av,vranglegeq0.
$$
answered Jan 10 at 18:50
Martin ArgeramiMartin Argerami
128k1182183
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add a comment |
$begingroup$
Using the definition that you wrote, pick your $x = v$, where $v$ is any eigenvector, then
$$<Av,v> = lambda geq 0$$
This tells you that no matter which eigenvector you pick to be your $x$, you will get that its corresponding eigenvalue is non-negative.
answered Jan 10 at 18:53
Ahmad BazziAhmad Bazzi
8,3622824
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
It's a straightforward computation. If $lambda$ is an eigenvalue of $A$, choose an eigenvector $v$ with $langle v,vrangle=1$. Then
$$
lambda=langle lambda v,vrangle=langle Av,vranglegeq0.
$$
answered Jan 10 at 18:50
Martin ArgeramiMartin Argerami
128k1182183
$endgroup$
add a comment |
$begingroup$
It's a straightforward computation. If $lambda$ is an eigenvalue of $A$, choose an eigenvector $v$ with $langle v,vrangle=1$. Then
$$
lambda=langle lambda v,vrangle=langle Av,vranglegeq0.
$$
answered Jan 10 at 18:50
Martin ArgeramiMartin Argerami
128k1182183
$endgroup$
add a comment |
$begingroup$
It's a straightforward computation. If $lambda$ is an eigenvalue of $A$, choose an eigenvector $v$ with $langle v,vrangle=1$. Then
$$
lambda=langle lambda v,vrangle=langle Av,vranglegeq0.
$$
answered Jan 10 at 18:50
Martin ArgeramiMartin Argerami
128k1182183
$endgroup$
It's a straightforward computation. If $lambda$ is an eigenvalue of $A$, choose an eigenvector $v$ with $langle v,vrangle=1$. Then
$$
lambda=langle lambda v,vrangle=langle Av,vranglegeq0.
$$
answered Jan 10 at 18:50
Martin ArgeramiMartin Argerami
128k1182183
answered Jan 10 at 18:50
Martin ArgeramiMartin Argerami
128k1182183
answered Jan 10 at 18:50
Martin ArgeramiMartin Argerami
128k1182183
answered Jan 10 at 18:50
Martin ArgeramiMartin Argerami
128k1182183
128k1182183
add a comment |
add a comment |
$begingroup$
Using the definition that you wrote, pick your $x = v$, where $v$ is any eigenvector, then
$$<Av,v> = lambda geq 0$$
This tells you that no matter which eigenvector you pick to be your $x$, you will get that its corresponding eigenvalue is non-negative.
answered Jan 10 at 18:53
Ahmad BazziAhmad Bazzi
8,3622824
$endgroup$
add a comment |
$begingroup$
Using the definition that you wrote, pick your $x = v$, where $v$ is any eigenvector, then
$$<Av,v> = lambda geq 0$$
This tells you that no matter which eigenvector you pick to be your $x$, you will get that its corresponding eigenvalue is non-negative.
answered Jan 10 at 18:53
Ahmad BazziAhmad Bazzi
8,3622824
$endgroup$
add a comment |
$begingroup$
Using the definition that you wrote, pick your $x = v$, where $v$ is any eigenvector, then
$$<Av,v> = lambda geq 0$$
This tells you that no matter which eigenvector you pick to be your $x$, you will get that its corresponding eigenvalue is non-negative.
answered Jan 10 at 18:53
Ahmad BazziAhmad Bazzi
8,3622824
$endgroup$
Using the definition that you wrote, pick your $x = v$, where $v$ is any eigenvector, then
$$<Av,v> = lambda geq 0$$
This tells you that no matter which eigenvector you pick to be your $x$, you will get that its corresponding eigenvalue is non-negative.
answered Jan 10 at 18:53
Ahmad BazziAhmad Bazzi
8,3622824
answered Jan 10 at 18:53
Ahmad BazziAhmad Bazzi
8,3622824
answered Jan 10 at 18:53
Ahmad BazziAhmad Bazzi
8,3622824
answered Jan 10 at 18:53
Ahmad BazziAhmad Bazzi
8,3622824
8,3622824
add a comment |
add a comment |
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