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Let $A$ be a matrix with $\lambda_1,...,\lambda_n$ eigenvalues, and $v_1,...,v_n$ the corresponding eigvenvectors. Let $B= \begin{pmatrix} 0& A \\ A& 0 \end{pmatrix}$. It's known that the eigenvalues of $B$ are $\{\pm \lambda_1,..., \pm \lambda_n\}$. Find the eigenvectors of $B$.

I've managed to find $n$ of the $2n$ eigenvectors of $B$, those who correspond to the eigenvalues: $\{ \lambda_1,..., \lambda_n\}$: $u_1= \begin{pmatrix} v_1 \\ v_1 \end{pmatrix},..., u_n = \begin{pmatrix} v_n \\ v_n \end{pmatrix}$. I assumed that the new eigenvectors should be built of the original eigenvectors, thus I started playing with combinations of the original eigenvectors, such as: $\begin{pmatrix} v_i \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ v_i \end{pmatrix}$ however those vectors didn't help at all. Eventually, looking at how $B$ is defined, I understood that since $B$ simply works the same as $A$ works on vectors for the first half and then again for the second half, I found that $u_1= \begin{pmatrix} v_1 \\ v_1 \end{pmatrix},..., u_n = \begin{pmatrix} v_n \\ v_n \end{pmatrix}$ can work.

However, I couldn't manage to find the corresponding eigenvectors for the $\{-\lambda_1,...,-\lambda_n\}$ eigenvalues. I can't see how applying the original matrix $A$ in any kind of form will get us to the negative of its' eigenvalues.

Any hint/clue?

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  • $\begingroup$ Why not, to the last question? What have you tried? "I couldn't manage to find .." What did you find, What specifically do you not understand, or what, specifically do you need to proceed. (And how did you "manage to find the first $n$ eigenvectors of B? What worked for you?) $\endgroup$
    – amWhy
    Commented Feb 5, 2021 at 15:50
  • $\begingroup$ @amWhy Edited.. $\endgroup$
    – ChikChak
    Commented Feb 5, 2021 at 15:58
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    $\begingroup$ Try to fill in the gaps $$B\begin{pmatrix} v_i \\ -v_i \end{pmatrix} = \dots = -\lambda_i \begin{pmatrix} v_i \\ -v_i \end{pmatrix}.$$ $\endgroup$ Commented Feb 5, 2021 at 16:03
  • $\begingroup$ @GNUSupporter8964民主女神地下教會 $B \begin{pmatrix} v_i \\ - v_i \end{pmatrix} = \begin{pmatrix} 0& A \\ A& 0 \end{pmatrix} \begin{pmatrix} v_i \\ - v_i \end{pmatrix} =\begin{pmatrix} A(-v_i) \\ Av_i \end{pmatrix} = \begin{pmatrix} -\lambda_i v_i \\ \lambda_i v_i \end{pmatrix} = -\lambda_i \begin{pmatrix} v_i \\ - v_i \end{pmatrix} $ Thanks :) $\endgroup$
    – ChikChak
    Commented Feb 5, 2021 at 16:09
  • $\begingroup$ You did it. Now you may answer your own question, so that the system knows that your problem has been resolved. $\endgroup$ Commented Feb 5, 2021 at 16:11

1 Answer 1

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Let's look at $u_i =\begin{pmatrix} v_i \\ -v_i \end{pmatrix}$:

$B \begin{pmatrix} v_i \\ -v_i \end{pmatrix} = \begin{pmatrix} 0& A \\ A& 0 \end{pmatrix} \begin{pmatrix} v_i \\ -v_i \end{pmatrix}= \begin{pmatrix} A(-v_i) \\ Av_i \end{pmatrix} = \begin{pmatrix} -\lambda_iv_i \\ \lambda_iv_i \end{pmatrix} = -\lambda_i\begin{pmatrix} v_i \\ -v_i \end{pmatrix}$

Thus we get the other $n$ eigenvectors, ending with total of $2n$ eigenvectors.

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