Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is there any theorem to find the eigenvalues of any anti-circulant matrix using the equivalent (with the same first row) circulant matrix. I found out that, for any anti-circulant matrix, the eigenvalues (taken as $\mu$) of the anti-circulant matrix can be written as, \begin{equation} \mu = \pm \mid{\lambda_j}\mid \label{mu_alpha} \end{equation} where $\lambda_j$ is an eigenvalue of 1-circulant matrix with the same first row. This seems valid since any anti-circulant matrix should be symmetric resulting in real eigenvalues.

Can anyone send me a link to any reference which has this proof..? or can you please comment if you think that this should not be correct ?

share|improve this question
5  
Please don't cross-post questions within minutes of each other. Pick one (usually this one first), and try in the other if you haven't gotten answers after a good amount of time has passed. –  J. M. Sep 7 '11 at 3:00
    
Thanks for info.. J.M. I thought that these are two different communities, and posted the same at the same time since I need some urgent help with this proof. –  Udara Sep 7 '11 at 6:09
    
A further question: If we have an irreducible and apperiodic circulant matrix $A$ with $N$ diffent rows. We can then define $N$ different anti-circulant matrices $\{B_n\}$ such that $B_n$ has $n-$th row of matrix $A$ as its first row. Q: Is there will exists at least one $B_n$ positive definite? –  user34757 Jun 29 '12 at 16:36
    
@yiwei Welcome to math.SE! Please ask new questions through the Ask Question page, and not via answers to other questions. –  Asaf Karagila Jun 30 '12 at 22:38

1 Answer 1

Let $A$ be a circulant matrix and $B$ be the anticirculant matrix such that $A$ and $B$ have identical first rows. Then $B=PA$ for some permutation matrix $P$. Hence $A$ and $B$ have identical singular values. However, circulant matrices and anticirculant matrices are normal matrices, so they can be unitarily diagonalized and their singular values are the moduli of their eigenvalues. Therefore, the eigenvalues $A$ and $B$ have identical moduli. Hence your assertion.

share|improve this answer
    
Thanks for your explanation. Is there any reference (book, research paper) which I can look into this proof? –  Udara Sep 7 '11 at 6:08
    
I don't know about such references, sorry. –  user1551 Sep 7 '11 at 7:31

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.