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Alexandre Eremenko
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All real tridiagonal matrices with $b_kc_k>0$, are diagonalizable, and their spectra are real and simple.

See, for example, Gantmakher and Krein, Oscillation matrices and kernels..., AMS 2002.

Sketch of the proof. Expanding the determinant $|A-\lambda I|$ write a recurrent formula for characteristic polynomials of truncated matrices. It is seen from this formula that the eigenvalues depend only on $a_k$ and the products $c_kb_k$. Therefore the symmetric matrix with $c_k^\prime=b_k^\prime=\sqrt{c_kb_k}$ has the same spectrum.

On the other hand, if you allow all $c_k=0$, for example, you can have a Jordan cell which is not diagonalizable.

Alexandre Eremenko
  • 91.8k
  • 9
  • 259
  • 429