0
$\begingroup$

Let $A=(a_{kl})$ be a matrix in $M_n(\mathbb{R})$ when $n$ is even. Let $B=(b_{kl})$ be the symmetric $2n$ by $2n$ matrix whose entries are given by,

  • $b_{k,l}=a_{kl}$ if $1\leq k,l\leq n$.
  • $b_{n+k,l}=b_{kl}$ if $l$ is odd.
  • $b_{n+k,l}=-b_{kl}$ if $l$ is even.

Q. Let $V$ be a basis for the eigenvectors of $B$. Is there any approach to derive a basis for the eigenvectors of $A$ from $V$?

Improve this question
$\endgroup$
8
  • 1
    $\begingroup$ I would assume the opposite. If you know the eigenvectors of $A$ then you could try to build eigenvectors for $B$, changing the sign of certain components. With a bit of luck, you could use one eigenvector for $A$ to generate two different eigenvectors for $B$. Anyways, did you try $n=2$ and test some simple cases by hand?? $\endgroup$
    – Beni Bogosel
    Commented Dec 9, 2022 at 9:10
  • $\begingroup$ Yes, extending eigenvectors of the submatrix $A$ to $B$ would be also considered as a question. But, the converse process is my concern for now. $\endgroup$
    – ABB
    Commented Dec 9, 2022 at 9:35
  • $\begingroup$ What is $b_{n+k,n+l}$? Also it seems to me that if $A$ isn't symmetric then B won't be neither since the top left $n \times n$ block of $B$ is exactly $A$. $\endgroup$
    – Antoine Labelle
    Commented Dec 9, 2022 at 14:33
  • $\begingroup$ $B$ is supposed to be symmetric, so to find $B$ we need only to know left side entries that are known for us. $\endgroup$
    – ABB
    Commented Dec 9, 2022 at 15:25
  • $\begingroup$ I don't think you defined the fourth block as $B=\begin{pmatrix}A&AU\\AU&*\end{pmatrix}$? $\endgroup$
    – Toni Mhax
    Commented Dec 10, 2022 at 13:29

0

Reset to default

You must log in to answer this question.