Proof that spectrum of a matrix is subset of the positive real numbers
$begingroup$
So my given problem is:
$Let,, A in mathbb{C}^{n ,times ,n},, be,, such,, that,,\ forall xin mathbb{C}^n : langle,Ax,xrangle geq 0 \ where ,, langle,cdot,cdotrangle ,, is ,, the ,, standard ,, inner ,, product ,, of ,, mathbb{C}^n.,, Show ,, that,,\ spec(A) subset [0, infty).$
Know I don't have any approach how to tackle the problem.
Thank you in advance
$textbf{EDIT:}$
I came this far:
Let $lambda$ be e-value of A and x the corresponding e-vector.
$Ax = lambda x ,,, multiply ,, x^T\ x^TAx = lambda x^Tx$
The left hand side of the equation is greater or equal to zero since
$x^TAx=langle,Ax,xrangle geq 0$
Then
$ = lambda sum_{i=1}^{n} x^{2}_i $
Since $x^Tx=sum_{i=1}^{n} x^{2}_i$
The sum is positive therefore $lambda$ has to be greater or equal to 0 so that the whole term is greater or equal to zero.
But now I dont know how to show that all eigenvalues are real.
linear-algebra proof-writing diagonalization
asked Jan 9 at 18:38
Fo Young Areal LoFo Young Areal Lo
345
$endgroup$
|
show 4 more comments
$begingroup$
So my given problem is:
$Let,, A in mathbb{C}^{n ,times ,n},, be,, such,, that,,\ forall xin mathbb{C}^n : langle,Ax,xrangle geq 0 \ where ,, langle,cdot,cdotrangle ,, is ,, the ,, standard ,, inner ,, product ,, of ,, mathbb{C}^n.,, Show ,, that,,\ spec(A) subset [0, infty).$
Know I don't have any approach how to tackle the problem.
Thank you in advance
$textbf{EDIT:}$
I came this far:
Let $lambda$ be e-value of A and x the corresponding e-vector.
$Ax = lambda x ,,, multiply ,, x^T\ x^TAx = lambda x^Tx$
The left hand side of the equation is greater or equal to zero since
$x^TAx=langle,Ax,xrangle geq 0$
Then
$ = lambda sum_{i=1}^{n} x^{2}_i $
Since $x^Tx=sum_{i=1}^{n} x^{2}_i$
The sum is positive therefore $lambda$ has to be greater or equal to 0 so that the whole term is greater or equal to zero.
But now I dont know how to show that all eigenvalues are real.
linear-algebra proof-writing diagonalization
asked Jan 9 at 18:38
Fo Young Areal LoFo Young Areal Lo
345
$endgroup$
$begingroup$
Think about eigenvectors.
$endgroup$
– Lord Shark the Unknown
Jan 9 at 18:43
$begingroup$
I mean I know that it has to do with the eigenvectors since I'm supposed to show that the set of all eigenvalues is a subset of the positive real numbers. I was looking for some more specific information.
$endgroup$
– Fo Young Areal Lo
Jan 9 at 18:47
$begingroup$
Start with the eigenvector equation and see if you can use that to figure something out about the eigenvalue.
$endgroup$
– tch
Jan 9 at 18:48
1
$begingroup$
This seems a lot like a homework problem so I'm hesitant to give more hints without an explanation of what you have tried and why it hasn't worked.
$endgroup$
– tch
Jan 9 at 18:53
1
$begingroup$
The crucial information is that $langle Ax,x rangle $ is real for all $x$.
$endgroup$
– copper.hat
Jan 9 at 19:21
|
show 4 more comments
$begingroup$
So my given problem is:
$Let,, A in mathbb{C}^{n ,times ,n},, be,, such,, that,,\ forall xin mathbb{C}^n : langle,Ax,xrangle geq 0 \ where ,, langle,cdot,cdotrangle ,, is ,, the ,, standard ,, inner ,, product ,, of ,, mathbb{C}^n.,, Show ,, that,,\ spec(A) subset [0, infty).$
Know I don't have any approach how to tackle the problem.
Thank you in advance
$textbf{EDIT:}$
I came this far:
Let $lambda$ be e-value of A and x the corresponding e-vector.
$Ax = lambda x ,,, multiply ,, x^T\ x^TAx = lambda x^Tx$
The left hand side of the equation is greater or equal to zero since
$x^TAx=langle,Ax,xrangle geq 0$
Then
$ = lambda sum_{i=1}^{n} x^{2}_i $
Since $x^Tx=sum_{i=1}^{n} x^{2}_i$
The sum is positive therefore $lambda$ has to be greater or equal to 0 so that the whole term is greater or equal to zero.
But now I dont know how to show that all eigenvalues are real.
linear-algebra proof-writing diagonalization
asked Jan 9 at 18:38
Fo Young Areal LoFo Young Areal Lo
345
$endgroup$
So my given problem is:
$Let,, A in mathbb{C}^{n ,times ,n},, be,, such,, that,,\ forall xin mathbb{C}^n : langle,Ax,xrangle geq 0 \ where ,, langle,cdot,cdotrangle ,, is ,, the ,, standard ,, inner ,, product ,, of ,, mathbb{C}^n.,, Show ,, that,,\ spec(A) subset [0, infty).$
Know I don't have any approach how to tackle the problem.
Thank you in advance
$textbf{EDIT:}$
I came this far:
Let $lambda$ be e-value of A and x the corresponding e-vector.
$Ax = lambda x ,,, multiply ,, x^T\ x^TAx = lambda x^Tx$
The left hand side of the equation is greater or equal to zero since
$x^TAx=langle,Ax,xrangle geq 0$
Then
$ = lambda sum_{i=1}^{n} x^{2}_i $
Since $x^Tx=sum_{i=1}^{n} x^{2}_i$
The sum is positive therefore $lambda$ has to be greater or equal to 0 so that the whole term is greater or equal to zero.
But now I dont know how to show that all eigenvalues are real.
linear-algebra proof-writing diagonalization
linear-algebra proof-writing diagonalization
asked Jan 9 at 18:38
Fo Young Areal LoFo Young Areal Lo
345
asked Jan 9 at 18:38
Fo Young Areal LoFo Young Areal Lo
345
edited Jan 10 at 8:03
Fo Young Areal Lo
asked Jan 9 at 18:38
Fo Young Areal LoFo Young Areal Lo
345
asked Jan 9 at 18:38
Fo Young Areal LoFo Young Areal Lo
345
asked Jan 9 at 18:38
Fo Young Areal LoFo Young Areal Lo
345
345
$begingroup$
Think about eigenvectors.
$endgroup$
– Lord Shark the Unknown
Jan 9 at 18:43
$begingroup$
I mean I know that it has to do with the eigenvectors since I'm supposed to show that the set of all eigenvalues is a subset of the positive real numbers. I was looking for some more specific information.
$endgroup$
– Fo Young Areal Lo
Jan 9 at 18:47
$begingroup$
Start with the eigenvector equation and see if you can use that to figure something out about the eigenvalue.
$endgroup$
– tch
Jan 9 at 18:48
1
$begingroup$
This seems a lot like a homework problem so I'm hesitant to give more hints without an explanation of what you have tried and why it hasn't worked.
$endgroup$
– tch
Jan 9 at 18:53
1
$begingroup$
The crucial information is that $langle Ax,x rangle $ is real for all $x$.
$endgroup$
– copper.hat
Jan 9 at 19:21
|
show 4 more comments
$begingroup$
Think about eigenvectors.
$endgroup$
– Lord Shark the Unknown
Jan 9 at 18:43
$begingroup$
I mean I know that it has to do with the eigenvectors since I'm supposed to show that the set of all eigenvalues is a subset of the positive real numbers. I was looking for some more specific information.
$endgroup$
– Fo Young Areal Lo
Jan 9 at 18:47
$begingroup$
Start with the eigenvector equation and see if you can use that to figure something out about the eigenvalue.
$endgroup$
– tch
Jan 9 at 18:48
1
$begingroup$
This seems a lot like a homework problem so I'm hesitant to give more hints without an explanation of what you have tried and why it hasn't worked.
$endgroup$
– tch
Jan 9 at 18:53
1
$begingroup$
The crucial information is that $langle Ax,x rangle $ is real for all $x$.
$endgroup$
– copper.hat
Jan 9 at 19:21
$begingroup$
Think about eigenvectors.
$endgroup$
– Lord Shark the Unknown
Jan 9 at 18:43
$begingroup$
Think about eigenvectors.
$endgroup$
– Lord Shark the Unknown
Jan 9 at 18:43
$begingroup$
I mean I know that it has to do with the eigenvectors since I'm supposed to show that the set of all eigenvalues is a subset of the positive real numbers. I was looking for some more specific information.
$endgroup$
– Fo Young Areal Lo
Jan 9 at 18:47
$begingroup$
I mean I know that it has to do with the eigenvectors since I'm supposed to show that the set of all eigenvalues is a subset of the positive real numbers. I was looking for some more specific information.
$endgroup$
– Fo Young Areal Lo
Jan 9 at 18:47
$begingroup$
Start with the eigenvector equation and see if you can use that to figure something out about the eigenvalue.
$endgroup$
– tch
Jan 9 at 18:48
$begingroup$
Start with the eigenvector equation and see if you can use that to figure something out about the eigenvalue.
$endgroup$
– tch
Jan 9 at 18:48
1
1
$begingroup$
This seems a lot like a homework problem so I'm hesitant to give more hints without an explanation of what you have tried and why it hasn't worked.
$endgroup$
– tch
Jan 9 at 18:53
$begingroup$
This seems a lot like a homework problem so I'm hesitant to give more hints without an explanation of what you have tried and why it hasn't worked.
$endgroup$
– tch
Jan 9 at 18:53
1
1
$begingroup$
The crucial information is that $langle Ax,x rangle $ is real for all $x$.
$endgroup$
– copper.hat
Jan 9 at 19:21
$begingroup$
The crucial information is that $langle Ax,x rangle $ is real for all $x$.
$endgroup$
– copper.hat
Jan 9 at 19:21
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
If
$lambda in text{spec}(A), tag 0$
we have a vector $x in Bbb C^n$ such that
$Ax = lambda x, ; x ne 0; tag 1$
then
$bar lambda langle x, x rangle = langle lambda x, x rangle = langle Ax, x rangle ge 0; tag 2$
we recall that the assertion $langle Ax, x rangle ge 0$ (tacitly, by definition) implies $langle Ax, x rangle in [0, infty) subset Bbb R$, and since
$0 ne langle x, x rangle in (0, infty) subset Bbb R, tag 3$
(2) yields
$0 le bar lambda in [0, infty) subset Bbb R; tag 4$
now
$bar lambda in [0, infty) subset Bbb R Longleftrightarrow lambda in [0, infty) subset Bbb R; tag 5$
therefore
$text{spec}(A) subset [0, infty), tag 6$
as was to be shown. $OEDelta$.
answered Jan 9 at 19:10
Robert LewisRobert Lewis
47.2k23067
$endgroup$
$begingroup$
what does that $lambda$ with the bar above denote? The complex conjugate of $lambda$?
$endgroup$
– Fo Young Areal Lo
Jan 10 at 8:01
$begingroup$
@FoYoungArealLo: yes, $bar lambda$ is as you say!
$endgroup$
– Robert Lewis
Jan 10 at 8:10
$begingroup$
Ok, can you explain to me why $bar{lambda} langle,x,xrangle = langle,lambda x,xrangle $ holds? I would expect it to be $bar{lambda} langle,x,xrangle = langle,bar{lambda} x,xrangle$
$endgroup$
– Fo Young Areal Lo
Jan 10 at 8:31
$begingroup$
@FoYoungArealLo: it depends on how one defines the standard inner product on $Bbb C^n$; one argument always gets conjugated; I chose the first. $langle u, v rangle = sum bar u_i v_i$ where $u = (u_1, ldots, u_n)$ etc.
$endgroup$
– Robert Lewis
Jan 10 at 8:37
$begingroup$
In my original post I put in an edit in which I show my approach to proof the statement. Maybe you could have a look on it and tell me what I have to do now or what I did wrong
$endgroup$
– Fo Young Areal Lo
Jan 10 at 10:10
add a comment |
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1 Answer
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1 Answer
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oldest
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oldest
votes
$begingroup$
If
$lambda in text{spec}(A), tag 0$
we have a vector $x in Bbb C^n$ such that
$Ax = lambda x, ; x ne 0; tag 1$
then
$bar lambda langle x, x rangle = langle lambda x, x rangle = langle Ax, x rangle ge 0; tag 2$
we recall that the assertion $langle Ax, x rangle ge 0$ (tacitly, by definition) implies $langle Ax, x rangle in [0, infty) subset Bbb R$, and since
$0 ne langle x, x rangle in (0, infty) subset Bbb R, tag 3$
(2) yields
$0 le bar lambda in [0, infty) subset Bbb R; tag 4$
now
$bar lambda in [0, infty) subset Bbb R Longleftrightarrow lambda in [0, infty) subset Bbb R; tag 5$
therefore
$text{spec}(A) subset [0, infty), tag 6$
as was to be shown. $OEDelta$.
answered Jan 9 at 19:10
Robert LewisRobert Lewis
47.2k23067
$endgroup$
$begingroup$
what does that $lambda$ with the bar above denote? The complex conjugate of $lambda$?
$endgroup$
– Fo Young Areal Lo
Jan 10 at 8:01
$begingroup$
@FoYoungArealLo: yes, $bar lambda$ is as you say!
$endgroup$
– Robert Lewis
Jan 10 at 8:10
$begingroup$
Ok, can you explain to me why $bar{lambda} langle,x,xrangle = langle,lambda x,xrangle $ holds? I would expect it to be $bar{lambda} langle,x,xrangle = langle,bar{lambda} x,xrangle$
$endgroup$
– Fo Young Areal Lo
Jan 10 at 8:31
$begingroup$
@FoYoungArealLo: it depends on how one defines the standard inner product on $Bbb C^n$; one argument always gets conjugated; I chose the first. $langle u, v rangle = sum bar u_i v_i$ where $u = (u_1, ldots, u_n)$ etc.
$endgroup$
– Robert Lewis
Jan 10 at 8:37
$begingroup$
In my original post I put in an edit in which I show my approach to proof the statement. Maybe you could have a look on it and tell me what I have to do now or what I did wrong
$endgroup$
– Fo Young Areal Lo
Jan 10 at 10:10
add a comment |
$begingroup$
If
$lambda in text{spec}(A), tag 0$
we have a vector $x in Bbb C^n$ such that
$Ax = lambda x, ; x ne 0; tag 1$
then
$bar lambda langle x, x rangle = langle lambda x, x rangle = langle Ax, x rangle ge 0; tag 2$
we recall that the assertion $langle Ax, x rangle ge 0$ (tacitly, by definition) implies $langle Ax, x rangle in [0, infty) subset Bbb R$, and since
$0 ne langle x, x rangle in (0, infty) subset Bbb R, tag 3$
(2) yields
$0 le bar lambda in [0, infty) subset Bbb R; tag 4$
now
$bar lambda in [0, infty) subset Bbb R Longleftrightarrow lambda in [0, infty) subset Bbb R; tag 5$
therefore
$text{spec}(A) subset [0, infty), tag 6$
as was to be shown. $OEDelta$.
answered Jan 9 at 19:10
Robert LewisRobert Lewis
47.2k23067
$endgroup$
$begingroup$
what does that $lambda$ with the bar above denote? The complex conjugate of $lambda$?
$endgroup$
– Fo Young Areal Lo
Jan 10 at 8:01
$begingroup$
@FoYoungArealLo: yes, $bar lambda$ is as you say!
$endgroup$
– Robert Lewis
Jan 10 at 8:10
$begingroup$
Ok, can you explain to me why $bar{lambda} langle,x,xrangle = langle,lambda x,xrangle $ holds? I would expect it to be $bar{lambda} langle,x,xrangle = langle,bar{lambda} x,xrangle$
$endgroup$
– Fo Young Areal Lo
Jan 10 at 8:31
$begingroup$
@FoYoungArealLo: it depends on how one defines the standard inner product on $Bbb C^n$; one argument always gets conjugated; I chose the first. $langle u, v rangle = sum bar u_i v_i$ where $u = (u_1, ldots, u_n)$ etc.
$endgroup$
– Robert Lewis
Jan 10 at 8:37
$begingroup$
In my original post I put in an edit in which I show my approach to proof the statement. Maybe you could have a look on it and tell me what I have to do now or what I did wrong
$endgroup$
– Fo Young Areal Lo
Jan 10 at 10:10
add a comment |
$begingroup$
If
$lambda in text{spec}(A), tag 0$
we have a vector $x in Bbb C^n$ such that
$Ax = lambda x, ; x ne 0; tag 1$
then
$bar lambda langle x, x rangle = langle lambda x, x rangle = langle Ax, x rangle ge 0; tag 2$
we recall that the assertion $langle Ax, x rangle ge 0$ (tacitly, by definition) implies $langle Ax, x rangle in [0, infty) subset Bbb R$, and since
$0 ne langle x, x rangle in (0, infty) subset Bbb R, tag 3$
(2) yields
$0 le bar lambda in [0, infty) subset Bbb R; tag 4$
now
$bar lambda in [0, infty) subset Bbb R Longleftrightarrow lambda in [0, infty) subset Bbb R; tag 5$
therefore
$text{spec}(A) subset [0, infty), tag 6$
as was to be shown. $OEDelta$.
answered Jan 9 at 19:10
Robert LewisRobert Lewis
47.2k23067
$endgroup$
If
$lambda in text{spec}(A), tag 0$
we have a vector $x in Bbb C^n$ such that
$Ax = lambda x, ; x ne 0; tag 1$
then
$bar lambda langle x, x rangle = langle lambda x, x rangle = langle Ax, x rangle ge 0; tag 2$
we recall that the assertion $langle Ax, x rangle ge 0$ (tacitly, by definition) implies $langle Ax, x rangle in [0, infty) subset Bbb R$, and since
$0 ne langle x, x rangle in (0, infty) subset Bbb R, tag 3$
(2) yields
$0 le bar lambda in [0, infty) subset Bbb R; tag 4$
now
$bar lambda in [0, infty) subset Bbb R Longleftrightarrow lambda in [0, infty) subset Bbb R; tag 5$
therefore
$text{spec}(A) subset [0, infty), tag 6$
as was to be shown. $OEDelta$.
answered Jan 9 at 19:10
Robert LewisRobert Lewis
47.2k23067
edited Jan 9 at 19:15
answered Jan 9 at 19:10
Robert LewisRobert Lewis
47.2k23067
answered Jan 9 at 19:10
Robert LewisRobert Lewis
47.2k23067
answered Jan 9 at 19:10
Robert LewisRobert Lewis
47.2k23067
47.2k23067
$begingroup$
what does that $lambda$ with the bar above denote? The complex conjugate of $lambda$?
$endgroup$
– Fo Young Areal Lo
Jan 10 at 8:01
$begingroup$
@FoYoungArealLo: yes, $bar lambda$ is as you say!
$endgroup$
– Robert Lewis
Jan 10 at 8:10
$begingroup$
Ok, can you explain to me why $bar{lambda} langle,x,xrangle = langle,lambda x,xrangle $ holds? I would expect it to be $bar{lambda} langle,x,xrangle = langle,bar{lambda} x,xrangle$
$endgroup$
– Fo Young Areal Lo
Jan 10 at 8:31
$begingroup$
@FoYoungArealLo: it depends on how one defines the standard inner product on $Bbb C^n$; one argument always gets conjugated; I chose the first. $langle u, v rangle = sum bar u_i v_i$ where $u = (u_1, ldots, u_n)$ etc.
$endgroup$
– Robert Lewis
Jan 10 at 8:37
$begingroup$
In my original post I put in an edit in which I show my approach to proof the statement. Maybe you could have a look on it and tell me what I have to do now or what I did wrong
$endgroup$
– Fo Young Areal Lo
Jan 10 at 10:10
add a comment |
$begingroup$
what does that $lambda$ with the bar above denote? The complex conjugate of $lambda$?
$endgroup$
– Fo Young Areal Lo
Jan 10 at 8:01
$begingroup$
@FoYoungArealLo: yes, $bar lambda$ is as you say!
$endgroup$
– Robert Lewis
Jan 10 at 8:10
$begingroup$
Ok, can you explain to me why $bar{lambda} langle,x,xrangle = langle,lambda x,xrangle $ holds? I would expect it to be $bar{lambda} langle,x,xrangle = langle,bar{lambda} x,xrangle$
$endgroup$
– Fo Young Areal Lo
Jan 10 at 8:31
$begingroup$
@FoYoungArealLo: it depends on how one defines the standard inner product on $Bbb C^n$; one argument always gets conjugated; I chose the first. $langle u, v rangle = sum bar u_i v_i$ where $u = (u_1, ldots, u_n)$ etc.
$endgroup$
– Robert Lewis
Jan 10 at 8:37
$begingroup$
In my original post I put in an edit in which I show my approach to proof the statement. Maybe you could have a look on it and tell me what I have to do now or what I did wrong
$endgroup$
– Fo Young Areal Lo
Jan 10 at 10:10
$begingroup$
what does that $lambda$ with the bar above denote? The complex conjugate of $lambda$?
$endgroup$
– Fo Young Areal Lo
Jan 10 at 8:01
$begingroup$
what does that $lambda$ with the bar above denote? The complex conjugate of $lambda$?
$endgroup$
– Fo Young Areal Lo
Jan 10 at 8:01
$begingroup$
@FoYoungArealLo: yes, $bar lambda$ is as you say!
$endgroup$
– Robert Lewis
Jan 10 at 8:10
$begingroup$
@FoYoungArealLo: yes, $bar lambda$ is as you say!
$endgroup$
– Robert Lewis
Jan 10 at 8:10
$begingroup$
Ok, can you explain to me why $bar{lambda} langle,x,xrangle = langle,lambda x,xrangle $ holds? I would expect it to be $bar{lambda} langle,x,xrangle = langle,bar{lambda} x,xrangle$
$endgroup$
– Fo Young Areal Lo
Jan 10 at 8:31
$begingroup$
Ok, can you explain to me why $bar{lambda} langle,x,xrangle = langle,lambda x,xrangle $ holds? I would expect it to be $bar{lambda} langle,x,xrangle = langle,bar{lambda} x,xrangle$
$endgroup$
– Fo Young Areal Lo
Jan 10 at 8:31
$begingroup$
@FoYoungArealLo: it depends on how one defines the standard inner product on $Bbb C^n$; one argument always gets conjugated; I chose the first. $langle u, v rangle = sum bar u_i v_i$ where $u = (u_1, ldots, u_n)$ etc.
$endgroup$
– Robert Lewis
Jan 10 at 8:37
$begingroup$
@FoYoungArealLo: it depends on how one defines the standard inner product on $Bbb C^n$; one argument always gets conjugated; I chose the first. $langle u, v rangle = sum bar u_i v_i$ where $u = (u_1, ldots, u_n)$ etc.
$endgroup$
– Robert Lewis
Jan 10 at 8:37
$begingroup$
In my original post I put in an edit in which I show my approach to proof the statement. Maybe you could have a look on it and tell me what I have to do now or what I did wrong
$endgroup$
– Fo Young Areal Lo
Jan 10 at 10:10
$begingroup$
In my original post I put in an edit in which I show my approach to proof the statement. Maybe you could have a look on it and tell me what I have to do now or what I did wrong
$endgroup$
– Fo Young Areal Lo
Jan 10 at 10:10
add a comment |
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$begingroup$
Think about eigenvectors.
$endgroup$
– Lord Shark the Unknown
Jan 9 at 18:43
$begingroup$
I mean I know that it has to do with the eigenvectors since I'm supposed to show that the set of all eigenvalues is a subset of the positive real numbers. I was looking for some more specific information.
$endgroup$
– Fo Young Areal Lo
Jan 9 at 18:47
$begingroup$
Start with the eigenvector equation and see if you can use that to figure something out about the eigenvalue.
$endgroup$
– tch
Jan 9 at 18:48
1
$begingroup$
This seems a lot like a homework problem so I'm hesitant to give more hints without an explanation of what you have tried and why it hasn't worked.
$endgroup$
– tch
Jan 9 at 18:53
1
$begingroup$
The crucial information is that $langle Ax,x rangle $ is real for all $x$.
$endgroup$
– copper.hat
Jan 9 at 19:21