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I have a tri-diagonal matrix with a specific structure defined by parameters $T(a,b,\alpha)$. Do you how could I calculate eigenvalues and eigenvectors? I am familiar with theoretical results when $\alpha = 0$, but unfortunately this is not the case.

$$T(a,b,\alpha) = \begin{pmatrix} a+\alpha & b & 0 & 0 & 0 & \dots & 0 \\ b & a & b & 0 & 0 & \dots & 0 \\ 0 & b & a & b & 0 & \dots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \\ 0 & 0 & 0 & 0 & \dots & b & a + \alpha \\ \end{pmatrix} $$

I.e. the $a + \alpha$ is only in the positions $T_{11}$ and $T_{nn}$.

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    i dont see any simple formula. you can relate the eigenvalues/vectors in terms of tridiagonal matrix of half the size. – 2017-01-04
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    can you hint me how could I do it? – 2017-01-04

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i will take $n = 4, a = 0 \mbox{ and } b = 1 $ and the case for general even $n, a \mbox{ and } b$ should be similar. let us look at the $4 \times 4$ matrix $$A = \pmatrix{\alpha&1&0&0\\1&0&1&0\\0&1&0&1\\0&0&0&\alpha},\ B_+ = \pmatrix{\alpha&1\\1&1}\, B_- = \pmatrix{\alpha&1\\1&-1}.$$

suppose $\lambda, \pmatrix{x_1\\x_2}$ is the eigenvalue/vector pair of $B_+,$ then you can verify that $\lambda, \pmatrix{x_1\\x_2\\x_2\\x_1}$ is an eigenvalue/vector of $A.$ in a similar way if $\lambda, \pmatrix{x_1\\x_2}$ is the eigenvalue/vector pair of $B_-,$ u can verify that $\lambda, \pmatrix{x_1\\x_2\\-x_2\\x_1}$ is an eigenvalue/vector of $A.$

let $\sigma(A),$ spectrum of $A$ is the set of all eigenvalues of $A.$ you can show that $\sigma(B_-) = -\sigma(B_+), \sigma(A) = \sigma(B_+) \, \cup \sigma(B_-)$ and that $\sigma(B+) = \frac{\alpha+1 \pm \sqrt{(\alpha - 1)^2 + 4}}2.$