Eigenvalues of complex matrices












3












$begingroup$


I have a question regarding eigenvalues of complex matrices. Let's assume that $Z in mathbb{C}^{n times n}$ with eigenpair $(lambda_i,gamma_i)$. We now look at



$Bgamma_i^*=lambda_i^* gamma_i^* tag{1}$



where $(bullet)^*$ denotes complex conjugate. If $Z$ is diagonalizable, then $Z$ has a unique set of eigenpairs, so in this case we know that $B=Z^*$ (please correct me if I'm wrong). My question is: if $Z$ is defective, can I find a matrix $B = Z$ such (1) is valid? In other words, can I form a defective matrix that shares eigenpair set with its own conjugate?



Thanks in advance.






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    I'm not sure what you mean by "$Z$ has a unique set of eigenpairs"
    $endgroup$
    – Omnomnomnom
    Jan 2 at 22:03






  • 1




    $begingroup$
    Also, why not simply take $B = Z^*$ when $Z$ is defective?
    $endgroup$
    – Omnomnomnom
    Jan 2 at 22:05






  • 1




    $begingroup$
    By unique I, in general, mean that there does not exist a matrix different from $Z$ that has the same eigenvalues and eigenvectors as $Z$. Maybe I formulated my question poorly (or don't understand your point); my question is if I can form some matrix (with complex values) that has the same eigenvectors and eigenvalues as this matrix' conjugate.
    $endgroup$
    – Martin
    Jan 2 at 22:13








  • 1




    $begingroup$
    so a matrix not equal to $Z^*$ that has the same eigenvectors and eigenvalues as $Z^*$?
    $endgroup$
    – Omnomnomnom
    Jan 2 at 23:05






  • 1




    $begingroup$
    Note that for any $A$ and $v$: $(Av)_i = sum_k A_{i,k}v_k$ so $(Av)^* = A^*v^*$. Therefore you can just take the complex conjugate of the entire eigen equation $Zgamma_ = lambda gamma_i$ to get $B = Z^*$ as @Omnomnomnom suggested.
    $endgroup$
    – tch
    Jan 3 at 0:45


















3












$begingroup$


I have a question regarding eigenvalues of complex matrices. Let's assume that $Z in mathbb{C}^{n times n}$ with eigenpair $(lambda_i,gamma_i)$. We now look at



$Bgamma_i^*=lambda_i^* gamma_i^* tag{1}$



where $(bullet)^*$ denotes complex conjugate. If $Z$ is diagonalizable, then $Z$ has a unique set of eigenpairs, so in this case we know that $B=Z^*$ (please correct me if I'm wrong). My question is: if $Z$ is defective, can I find a matrix $B = Z$ such (1) is valid? In other words, can I form a defective matrix that shares eigenpair set with its own conjugate?



Thanks in advance.






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    I'm not sure what you mean by "$Z$ has a unique set of eigenpairs"
    $endgroup$
    – Omnomnomnom
    Jan 2 at 22:03






  • 1




    $begingroup$
    Also, why not simply take $B = Z^*$ when $Z$ is defective?
    $endgroup$
    – Omnomnomnom
    Jan 2 at 22:05






  • 1




    $begingroup$
    By unique I, in general, mean that there does not exist a matrix different from $Z$ that has the same eigenvalues and eigenvectors as $Z$. Maybe I formulated my question poorly (or don't understand your point); my question is if I can form some matrix (with complex values) that has the same eigenvectors and eigenvalues as this matrix' conjugate.
    $endgroup$
    – Martin
    Jan 2 at 22:13








  • 1




    $begingroup$
    so a matrix not equal to $Z^*$ that has the same eigenvectors and eigenvalues as $Z^*$?
    $endgroup$
    – Omnomnomnom
    Jan 2 at 23:05






  • 1




    $begingroup$
    Note that for any $A$ and $v$: $(Av)_i = sum_k A_{i,k}v_k$ so $(Av)^* = A^*v^*$. Therefore you can just take the complex conjugate of the entire eigen equation $Zgamma_ = lambda gamma_i$ to get $B = Z^*$ as @Omnomnomnom suggested.
    $endgroup$
    – tch
    Jan 3 at 0:45
















3












3








3


1



$begingroup$


I have a question regarding eigenvalues of complex matrices. Let's assume that $Z in mathbb{C}^{n times n}$ with eigenpair $(lambda_i,gamma_i)$. We now look at



$Bgamma_i^*=lambda_i^* gamma_i^* tag{1}$



where $(bullet)^*$ denotes complex conjugate. If $Z$ is diagonalizable, then $Z$ has a unique set of eigenpairs, so in this case we know that $B=Z^*$ (please correct me if I'm wrong). My question is: if $Z$ is defective, can I find a matrix $B = Z$ such (1) is valid? In other words, can I form a defective matrix that shares eigenpair set with its own conjugate?



Thanks in advance.






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$




I have a question regarding eigenvalues of complex matrices. Let's assume that $Z in mathbb{C}^{n times n}$ with eigenpair $(lambda_i,gamma_i)$. We now look at



$Bgamma_i^*=lambda_i^* gamma_i^* tag{1}$



where $(bullet)^*$ denotes complex conjugate. If $Z$ is diagonalizable, then $Z$ has a unique set of eigenpairs, so in this case we know that $B=Z^*$ (please correct me if I'm wrong). My question is: if $Z$ is defective, can I find a matrix $B = Z$ such (1) is valid? In other words, can I form a defective matrix that shares eigenpair set with its own conjugate?



Thanks in advance.






linear-algebra eigenvalues-eigenvectors




linear-algebra eigenvalues-eigenvectors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 22:02









Gregory Nisbet

714612




714612










asked Jan 2 at 21:35









MartinMartin

253




253








  • 2




    $begingroup$
    I'm not sure what you mean by "$Z$ has a unique set of eigenpairs"
    $endgroup$
    – Omnomnomnom
    Jan 2 at 22:03






  • 1




    $begingroup$
    Also, why not simply take $B = Z^*$ when $Z$ is defective?
    $endgroup$
    – Omnomnomnom
    Jan 2 at 22:05






  • 1




    $begingroup$
    By unique I, in general, mean that there does not exist a matrix different from $Z$ that has the same eigenvalues and eigenvectors as $Z$. Maybe I formulated my question poorly (or don't understand your point); my question is if I can form some matrix (with complex values) that has the same eigenvectors and eigenvalues as this matrix' conjugate.
    $endgroup$
    – Martin
    Jan 2 at 22:13








  • 1




    $begingroup$
    so a matrix not equal to $Z^*$ that has the same eigenvectors and eigenvalues as $Z^*$?
    $endgroup$
    – Omnomnomnom
    Jan 2 at 23:05






  • 1




    $begingroup$
    Note that for any $A$ and $v$: $(Av)_i = sum_k A_{i,k}v_k$ so $(Av)^* = A^*v^*$. Therefore you can just take the complex conjugate of the entire eigen equation $Zgamma_ = lambda gamma_i$ to get $B = Z^*$ as @Omnomnomnom suggested.
    $endgroup$
    – tch
    Jan 3 at 0:45
















  • 2




    $begingroup$
    I'm not sure what you mean by "$Z$ has a unique set of eigenpairs"
    $endgroup$
    – Omnomnomnom
    Jan 2 at 22:03






  • 1




    $begingroup$
    Also, why not simply take $B = Z^*$ when $Z$ is defective?
    $endgroup$
    – Omnomnomnom
    Jan 2 at 22:05






  • 1




    $begingroup$
    By unique I, in general, mean that there does not exist a matrix different from $Z$ that has the same eigenvalues and eigenvectors as $Z$. Maybe I formulated my question poorly (or don't understand your point); my question is if I can form some matrix (with complex values) that has the same eigenvectors and eigenvalues as this matrix' conjugate.
    $endgroup$
    – Martin
    Jan 2 at 22:13








  • 1




    $begingroup$
    so a matrix not equal to $Z^*$ that has the same eigenvectors and eigenvalues as $Z^*$?
    $endgroup$
    – Omnomnomnom
    Jan 2 at 23:05






  • 1




    $begingroup$
    Note that for any $A$ and $v$: $(Av)_i = sum_k A_{i,k}v_k$ so $(Av)^* = A^*v^*$. Therefore you can just take the complex conjugate of the entire eigen equation $Zgamma_ = lambda gamma_i$ to get $B = Z^*$ as @Omnomnomnom suggested.
    $endgroup$
    – tch
    Jan 3 at 0:45










2




2




$begingroup$
I'm not sure what you mean by "$Z$ has a unique set of eigenpairs"
$endgroup$
– Omnomnomnom
Jan 2 at 22:03




$begingroup$
I'm not sure what you mean by "$Z$ has a unique set of eigenpairs"
$endgroup$
– Omnomnomnom
Jan 2 at 22:03




1




1




$begingroup$
Also, why not simply take $B = Z^*$ when $Z$ is defective?
$endgroup$
– Omnomnomnom
Jan 2 at 22:05




$begingroup$
Also, why not simply take $B = Z^*$ when $Z$ is defective?
$endgroup$
– Omnomnomnom
Jan 2 at 22:05




1




1




$begingroup$
By unique I, in general, mean that there does not exist a matrix different from $Z$ that has the same eigenvalues and eigenvectors as $Z$. Maybe I formulated my question poorly (or don't understand your point); my question is if I can form some matrix (with complex values) that has the same eigenvectors and eigenvalues as this matrix' conjugate.
$endgroup$
– Martin
Jan 2 at 22:13






$begingroup$
By unique I, in general, mean that there does not exist a matrix different from $Z$ that has the same eigenvalues and eigenvectors as $Z$. Maybe I formulated my question poorly (or don't understand your point); my question is if I can form some matrix (with complex values) that has the same eigenvectors and eigenvalues as this matrix' conjugate.
$endgroup$
– Martin
Jan 2 at 22:13






1




1




$begingroup$
so a matrix not equal to $Z^*$ that has the same eigenvectors and eigenvalues as $Z^*$?
$endgroup$
– Omnomnomnom
Jan 2 at 23:05




$begingroup$
so a matrix not equal to $Z^*$ that has the same eigenvectors and eigenvalues as $Z^*$?
$endgroup$
– Omnomnomnom
Jan 2 at 23:05




1




1




$begingroup$
Note that for any $A$ and $v$: $(Av)_i = sum_k A_{i,k}v_k$ so $(Av)^* = A^*v^*$. Therefore you can just take the complex conjugate of the entire eigen equation $Zgamma_ = lambda gamma_i$ to get $B = Z^*$ as @Omnomnomnom suggested.
$endgroup$
– tch
Jan 3 at 0:45






$begingroup$
Note that for any $A$ and $v$: $(Av)_i = sum_k A_{i,k}v_k$ so $(Av)^* = A^*v^*$. Therefore you can just take the complex conjugate of the entire eigen equation $Zgamma_ = lambda gamma_i$ to get $B = Z^*$ as @Omnomnomnom suggested.
$endgroup$
– tch
Jan 3 at 0:45












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