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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.

RobPratt
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VicaYang
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  • 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... – VicaYang Nov 29 '20 at 13:52
  • Try reading this. –  Nov 29 '20 at 14:12
  • 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? – VicaYang Nov 29 '20 at 14:23
  • 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. –  Nov 29 '20 at 14:29
  • Yes, I agree with you, but it is also costly with lots of calculations. In fact, an easier method is to construct the eigenvector given the eigenvalue (which should be faster because each component of the eigenvector can be solved one by one immediately). But I think both are not elegant compared with the simple form of the eigenvalues. – VicaYang Nov 29 '20 at 14:43
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    That's a Sylvester-Kac matrix pretty known in the litterature. – Toni Mhax Nov 29 '20 at 17:48
  • Thanks, I think I got it! Here is a good survey of this matrix. – VicaYang Nov 29 '20 at 18:13

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