Finding eigen values of an antidiagonal matrix.

Asked Apr 20 '24 at 14:57

Active Apr 20 '24 at 15:18

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Let, $A=\begin{bmatrix}0&0&0&\dots&0&100&\\ 0&0&0&\dots&99&0&\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&2&0&\dots&0&0\\ 1&0&0&\dots&0&0 \end{bmatrix}_{100\times 100}$
I want to find eigenvalues of this matirx.

Finding $A-\lambda I$ and then its determinant does not seem to be a practical approach.

This is how i tried to solve:
Since, $A^2=\begin{bmatrix}100\times 1&0&0&\dots&0&0&\\ 0&98\times 2&0&\dots&0&0&\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\dots&2\times 98&0\\ 0&0&0&\dots&0&1\times 100 \end{bmatrix}_{100\times 100}$
Now as eigenvalues of a diagonal matrix are precisely diagonal entries,so if I am allowed to take positive or negative square roots of eigenvalues of $A^2$ in order to find eigenvalues of A, then I can find eigenvalues of $A$ easily. Or is there any better method to find eigenvalues of an antidiagonal matrix of order $n$?

edited Apr 20 '24 at 15:18

asked Apr 20 '24 at 14:57

2

if you relabel the row/columns, you can turn your matrix into a block diagonal matrix with $2\times 2$ blocks of the form $[\begin{smallmatrix}0 & k\ 101-k & 0\end{smallmatrix}]$. So its eigenvalues do have the form $\pm \sqrt{k(101-k)}$. The generalization to general anti-diagonal matrix is obvious. – achille hui Apr 20 '24 at 15:25
@achillehui Sir,can you please elaborate relabeling. – Apr 20 '24 at 17:03
@ECC refer to the answers and comments to this question – Viera Čerňanová Apr 23 '24 at 13:29

Finding eigen values of an antidiagonal matrix.

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