Let $A$ be complex matrices and $x$ be eigenvalue of $Aoverline{A}$.












1












$begingroup$


Let $A$ be non-singular $ntimes n$ complex matrices and $x$ be negative eigenvalue of $Aoverline{A}$. Show that algebraic multiplicity of $x$ is even number.



I show that the $det(Aoverline{A})>0$ so numbers of negative eigenvalue is even
. But I stuck to prove algebraic multiplicity of $x$ is even number.






linear-algebra matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    What happens if $A$ is $1times 1$?
    $endgroup$
    – kimchi lover
    Jan 18 at 13:57










  • $begingroup$
    @i707107: Perhaps $(A^*)^T$ means $overline A$.?
    $endgroup$
    – hardmath
    Jan 18 at 17:25










  • $begingroup$
    Properties like this usually appearance in the discussion of "consimilarity". For the current problem, see exercise 7(c) in chapter 4.6 on p.253 of the first edition of Horn and Johnson's Matrix Analysis. Their proof makes use of eigenvectors. I believe there is a proof without eigenvectors, but I haven't time to delve into it.
    $endgroup$
    – user1551
    Jan 18 at 18:30








  • 1




    $begingroup$
    @hardmath yes, because I don’t know how to write like that
    $endgroup$
    – lio
    Jan 19 at 1:18










  • $begingroup$
    @lio. It's best to explain in words. Is it the transpose of the conjugate transpose (so just the conjugate?) or is it just the conjugate transpose of $A$?
    $endgroup$
    – hardmath
    Jan 19 at 1:56
















1












$begingroup$


Let $A$ be non-singular $ntimes n$ complex matrices and $x$ be negative eigenvalue of $Aoverline{A}$. Show that algebraic multiplicity of $x$ is even number.



I show that the $det(Aoverline{A})>0$ so numbers of negative eigenvalue is even
. But I stuck to prove algebraic multiplicity of $x$ is even number.






linear-algebra matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    What happens if $A$ is $1times 1$?
    $endgroup$
    – kimchi lover
    Jan 18 at 13:57










  • $begingroup$
    @i707107: Perhaps $(A^*)^T$ means $overline A$.?
    $endgroup$
    – hardmath
    Jan 18 at 17:25










  • $begingroup$
    Properties like this usually appearance in the discussion of "consimilarity". For the current problem, see exercise 7(c) in chapter 4.6 on p.253 of the first edition of Horn and Johnson's Matrix Analysis. Their proof makes use of eigenvectors. I believe there is a proof without eigenvectors, but I haven't time to delve into it.
    $endgroup$
    – user1551
    Jan 18 at 18:30








  • 1




    $begingroup$
    @hardmath yes, because I don’t know how to write like that
    $endgroup$
    – lio
    Jan 19 at 1:18










  • $begingroup$
    @lio. It's best to explain in words. Is it the transpose of the conjugate transpose (so just the conjugate?) or is it just the conjugate transpose of $A$?
    $endgroup$
    – hardmath
    Jan 19 at 1:56














1












1








1





$begingroup$


Let $A$ be non-singular $ntimes n$ complex matrices and $x$ be negative eigenvalue of $Aoverline{A}$. Show that algebraic multiplicity of $x$ is even number.



I show that the $det(Aoverline{A})>0$ so numbers of negative eigenvalue is even
. But I stuck to prove algebraic multiplicity of $x$ is even number.






linear-algebra matrices







share|cite|improve this question











$endgroup$




Let $A$ be non-singular $ntimes n$ complex matrices and $x$ be negative eigenvalue of $Aoverline{A}$. Show that algebraic multiplicity of $x$ is even number.



I show that the $det(Aoverline{A})>0$ so numbers of negative eigenvalue is even
. But I stuck to prove algebraic multiplicity of $x$ is even number.






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 20 at 15:43









i707107

12.6k21647




12.6k21647










asked Jan 18 at 13:34









liolio

20516




20516












  • $begingroup$
    What happens if $A$ is $1times 1$?
    $endgroup$
    – kimchi lover
    Jan 18 at 13:57










  • $begingroup$
    @i707107: Perhaps $(A^*)^T$ means $overline A$.?
    $endgroup$
    – hardmath
    Jan 18 at 17:25










  • $begingroup$
    Properties like this usually appearance in the discussion of "consimilarity". For the current problem, see exercise 7(c) in chapter 4.6 on p.253 of the first edition of Horn and Johnson's Matrix Analysis. Their proof makes use of eigenvectors. I believe there is a proof without eigenvectors, but I haven't time to delve into it.
    $endgroup$
    – user1551
    Jan 18 at 18:30








  • 1




    $begingroup$
    @hardmath yes, because I don’t know how to write like that
    $endgroup$
    – lio
    Jan 19 at 1:18










  • $begingroup$
    @lio. It's best to explain in words. Is it the transpose of the conjugate transpose (so just the conjugate?) or is it just the conjugate transpose of $A$?
    $endgroup$
    – hardmath
    Jan 19 at 1:56


















  • $begingroup$
    What happens if $A$ is $1times 1$?
    $endgroup$
    – kimchi lover
    Jan 18 at 13:57










  • $begingroup$
    @i707107: Perhaps $(A^*)^T$ means $overline A$.?
    $endgroup$
    – hardmath
    Jan 18 at 17:25










  • $begingroup$
    Properties like this usually appearance in the discussion of "consimilarity". For the current problem, see exercise 7(c) in chapter 4.6 on p.253 of the first edition of Horn and Johnson's Matrix Analysis. Their proof makes use of eigenvectors. I believe there is a proof without eigenvectors, but I haven't time to delve into it.
    $endgroup$
    – user1551
    Jan 18 at 18:30








  • 1




    $begingroup$
    @hardmath yes, because I don’t know how to write like that
    $endgroup$
    – lio
    Jan 19 at 1:18










  • $begingroup$
    @lio. It's best to explain in words. Is it the transpose of the conjugate transpose (so just the conjugate?) or is it just the conjugate transpose of $A$?
    $endgroup$
    – hardmath
    Jan 19 at 1:56
















$begingroup$
What happens if $A$ is $1times 1$?
$endgroup$
– kimchi lover
Jan 18 at 13:57




$begingroup$
What happens if $A$ is $1times 1$?
$endgroup$
– kimchi lover
Jan 18 at 13:57












$begingroup$
@i707107: Perhaps $(A^*)^T$ means $overline A$.?
$endgroup$
– hardmath
Jan 18 at 17:25




$begingroup$
@i707107: Perhaps $(A^*)^T$ means $overline A$.?
$endgroup$
– hardmath
Jan 18 at 17:25












$begingroup$
Properties like this usually appearance in the discussion of "consimilarity". For the current problem, see exercise 7(c) in chapter 4.6 on p.253 of the first edition of Horn and Johnson's Matrix Analysis. Their proof makes use of eigenvectors. I believe there is a proof without eigenvectors, but I haven't time to delve into it.
$endgroup$
– user1551
Jan 18 at 18:30






$begingroup$
Properties like this usually appearance in the discussion of "consimilarity". For the current problem, see exercise 7(c) in chapter 4.6 on p.253 of the first edition of Horn and Johnson's Matrix Analysis. Their proof makes use of eigenvectors. I believe there is a proof without eigenvectors, but I haven't time to delve into it.
$endgroup$
– user1551
Jan 18 at 18:30






1




1




$begingroup$
@hardmath yes, because I don’t know how to write like that
$endgroup$
– lio
Jan 19 at 1:18




$begingroup$
@hardmath yes, because I don’t know how to write like that
$endgroup$
– lio
Jan 19 at 1:18












$begingroup$
@lio. It's best to explain in words. Is it the transpose of the conjugate transpose (so just the conjugate?) or is it just the conjugate transpose of $A$?
$endgroup$
– hardmath
Jan 19 at 1:56




$begingroup$
@lio. It's best to explain in words. Is it the transpose of the conjugate transpose (so just the conjugate?) or is it just the conjugate transpose of $A$?
$endgroup$
– hardmath
Jan 19 at 1:56










1 Answer
1






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$begingroup$

$textbf{Proposition.}$ If $Ain M_n(mathbb{C})$ and $lambdain spectrum(Aoverline{A})cap (-infty,0)$, then $lambda$ has an even algebraic multiplicity.



$textbf{Proof}.$ The key is: "there are $Sin GL_n(mathbb{C}),Rin M_n(mathbb{R})$ s.t. $A=S{Roverline{S}};^{-1}$" (Corollary 4.6.15 in Matrix Analysis by Horn,Johnson, cited by user1551 -who has good readings-).



Then $Aoverline{A}=SR^2S^{-1}$ and $spectrum(Aoverline{A})={mu^2;muin spectrum(R)}$ (equality of lists).



Thus $pm isqrt{-lambda}$ are eigenvalues of the real matrix $R$; since they have same multiplicity, $lambda$ has even algebraic multiplicity. $square$



$textbf{Remark}.$ That implies the famous result $det(I+Aoverline{A})geq 0$.



Indeed, $det(I+Aoverline{A})=det(I+R^2)=det((I+iR)(I-iR))$.






share|cite|improve this answer











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    $begingroup$

    $textbf{Proposition.}$ If $Ain M_n(mathbb{C})$ and $lambdain spectrum(Aoverline{A})cap (-infty,0)$, then $lambda$ has an even algebraic multiplicity.



    $textbf{Proof}.$ The key is: "there are $Sin GL_n(mathbb{C}),Rin M_n(mathbb{R})$ s.t. $A=S{Roverline{S}};^{-1}$" (Corollary 4.6.15 in Matrix Analysis by Horn,Johnson, cited by user1551 -who has good readings-).



    Then $Aoverline{A}=SR^2S^{-1}$ and $spectrum(Aoverline{A})={mu^2;muin spectrum(R)}$ (equality of lists).



    Thus $pm isqrt{-lambda}$ are eigenvalues of the real matrix $R$; since they have same multiplicity, $lambda$ has even algebraic multiplicity. $square$



    $textbf{Remark}.$ That implies the famous result $det(I+Aoverline{A})geq 0$.



    Indeed, $det(I+Aoverline{A})=det(I+R^2)=det((I+iR)(I-iR))$.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      $textbf{Proposition.}$ If $Ain M_n(mathbb{C})$ and $lambdain spectrum(Aoverline{A})cap (-infty,0)$, then $lambda$ has an even algebraic multiplicity.



      $textbf{Proof}.$ The key is: "there are $Sin GL_n(mathbb{C}),Rin M_n(mathbb{R})$ s.t. $A=S{Roverline{S}};^{-1}$" (Corollary 4.6.15 in Matrix Analysis by Horn,Johnson, cited by user1551 -who has good readings-).



      Then $Aoverline{A}=SR^2S^{-1}$ and $spectrum(Aoverline{A})={mu^2;muin spectrum(R)}$ (equality of lists).



      Thus $pm isqrt{-lambda}$ are eigenvalues of the real matrix $R$; since they have same multiplicity, $lambda$ has even algebraic multiplicity. $square$



      $textbf{Remark}.$ That implies the famous result $det(I+Aoverline{A})geq 0$.



      Indeed, $det(I+Aoverline{A})=det(I+R^2)=det((I+iR)(I-iR))$.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        $textbf{Proposition.}$ If $Ain M_n(mathbb{C})$ and $lambdain spectrum(Aoverline{A})cap (-infty,0)$, then $lambda$ has an even algebraic multiplicity.



        $textbf{Proof}.$ The key is: "there are $Sin GL_n(mathbb{C}),Rin M_n(mathbb{R})$ s.t. $A=S{Roverline{S}};^{-1}$" (Corollary 4.6.15 in Matrix Analysis by Horn,Johnson, cited by user1551 -who has good readings-).



        Then $Aoverline{A}=SR^2S^{-1}$ and $spectrum(Aoverline{A})={mu^2;muin spectrum(R)}$ (equality of lists).



        Thus $pm isqrt{-lambda}$ are eigenvalues of the real matrix $R$; since they have same multiplicity, $lambda$ has even algebraic multiplicity. $square$



        $textbf{Remark}.$ That implies the famous result $det(I+Aoverline{A})geq 0$.



        Indeed, $det(I+Aoverline{A})=det(I+R^2)=det((I+iR)(I-iR))$.






        share|cite|improve this answer











        $endgroup$



        $textbf{Proposition.}$ If $Ain M_n(mathbb{C})$ and $lambdain spectrum(Aoverline{A})cap (-infty,0)$, then $lambda$ has an even algebraic multiplicity.



        $textbf{Proof}.$ The key is: "there are $Sin GL_n(mathbb{C}),Rin M_n(mathbb{R})$ s.t. $A=S{Roverline{S}};^{-1}$" (Corollary 4.6.15 in Matrix Analysis by Horn,Johnson, cited by user1551 -who has good readings-).



        Then $Aoverline{A}=SR^2S^{-1}$ and $spectrum(Aoverline{A})={mu^2;muin spectrum(R)}$ (equality of lists).



        Thus $pm isqrt{-lambda}$ are eigenvalues of the real matrix $R$; since they have same multiplicity, $lambda$ has even algebraic multiplicity. $square$



        $textbf{Remark}.$ That implies the famous result $det(I+Aoverline{A})geq 0$.



        Indeed, $det(I+Aoverline{A})=det(I+R^2)=det((I+iR)(I-iR))$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 19 at 11:52

























        answered Jan 19 at 11:40









        loup blancloup blanc

        24k21852




        24k21852






























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