0
$\begingroup$

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 ?

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ Hint: the square of your anti-circulant matrix has the same eigenvectors as the corresponding circulant matrix. $\endgroup$
    – Gjergji Zaimi
    Commented Sep 7, 2011 at 5:38
  • 1
    $\begingroup$ Simul-posted, in a major breach of etiquette, to math.stackexchange, math.stackexchange.com/questions/62466/… $\endgroup$
    – Gerry Myerson
    Commented Sep 7, 2011 at 6:02
  • 2
    $\begingroup$ @Gerry: please don't bite the newbies. While it is better not to simul-post, neither the mathoverflow faq nor the math.stackexchange faq says that explicitly. In any case, a friendly hint would suffice. $\endgroup$
    – Neil Strickland
    Commented Sep 7, 2011 at 9:03
  • $\begingroup$ Extremely sorry for simul-post in math.stackexchange, as I didn't know that these two are similar. Can anyone help me to understand what math.stackexchange and math.overflow.. and the differences of them, so that I could post my questions correctly according to the rules.. $\endgroup$
    – Udara
    Commented Sep 9, 2011 at 1:12

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .