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Given is a symmetric tridiagonal $n\times n$ matrix $A$ over $\mathbb{R}$ of the following form: $A=\begin{Bmatrix} 0 & a_{1} & & & \\ a_{1} & 0 &a_{2} & & \\ & a_{2} & 0 &\ddots & \\ & &\ddots &\ddots & a_{n-1} \\ & & & a_{n-1} & 0 \end{Bmatrix}$

I am struggling with showing that if $\lambda$ is an eigenvalue, that $-\lambda$ is an eigenvalue too.

Well, first i tried to see how the characteristic polynomial looks like for the cases $n=2,3$. Then i saw in Wikipedia that there exists a recursive formula for the determinant, namely for computing the determinant $\begin{vmatrix} -\lambda &a_{1} & & & \\ a_{1} &-\lambda &a_{2} & & \\ &a_{2} &-\lambda & \ddots & \\ & &\ddots &\ddots & a_{n-1} \\ & & &a_{n-1} &-\lambda \end{vmatrix}$ it will be $\Delta _{n}=(-\lambda)\Delta _{n-1} - (a_{n-2})^{2}\Delta_{n-2}$.

Unfortunately, i am stuck in the last step of the induction... Can anybody help me, please? Is this the correct approach? Thank you in advance!

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Why don't you try to use induction to show that for a $(n\times n)$ matrix $A_n$, the eigenvalue must satisfy: $$ \lambda^2 - \sum_{i=2}^n a_i^2 = 0$$ This will automatically show that $-\lambda$ is also an eigenvalue.

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Hint. You don't need mathematical induction. Let $D=D^{-1}=\operatorname{diag}(1,-1,1,-1,\ldots)$. What is $D^{-1}AD$?

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