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The $n$-order tridiagonal matrix is defined by $A(i,i)=0, A(i,i+1)=n-i, A(i+1,i)=i$, i.e.,

$$A=\begin{pmatrix}0&n-1\\1&0&n-2\\&2&0&n-3\\&&\ddots&\ddots&\ddots\\&&&n-2&0&1\\&&&&n-1&0\end{pmatrix}$$

All eigenvalues of $A$ are $-(n-1),-(n-3),\ldots,(n-3),(n-1)$, however, I could not figure out a simple way to show that.

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  • $\begingroup$ I have some observations, e.g., $n-1$ is the eigenvalue; and if $\lambda$ is the eigenvalue, then $-\lambda$ is also an eigenvalue. However, I think these two observations do not help so much... $\endgroup$
    – VicaYang
    Nov 29, 2020 at 13:52
  • $\begingroup$ Try reading this. $\endgroup$ Nov 29, 2020 at 14:12
  • $\begingroup$ Thanks Peter. But I think this page discussed the tridiagonal with the constant number? There is a page discussing non-constant case, but there are no useful conclusion. Also, I think calculating the determinant is not enough to show all the eigen values? $\endgroup$
    – VicaYang
    Nov 29, 2020 at 14:23
  • $\begingroup$ Using this generalization you can compute the determinant for each $n \in \mathbb{N}$ and since eigenvalues are roots of the characteristic polynomial - which is defined as $ \det (A - \lambda \mathbb{I}_{n} )$ - you can then compute them form it. $\endgroup$ Nov 29, 2020 at 14:29
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    $\begingroup$ That's a Sylvester-Kac matrix pretty known in the litterature. $\endgroup$
    – Toni Mhax
    Nov 29, 2020 at 17:48

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