Proof that spectrum of a matrix is subset of the positive real numbers












0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 8:03







Fo Young Areal Lo

















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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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. $OEDelta$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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. $OEDelta$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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


















1












$begingroup$

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. $OEDelta$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















1












1








1





$begingroup$

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. $OEDelta$.






share|cite|improve this answer











$endgroup$



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. $OEDelta$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 9 at 19:15

























answered Jan 9 at 19:10









Robert LewisRobert Lewis

48.6k23167




48.6k23167












  • $begingroup$
    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










  • $begingroup$
    @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




















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