Existence of a real eigenvalue

Question

I have a matrix $M \in \mathbb{R}^{(n+1) \times (n+1)}$ that is tridiagonal. In numerical computations I found out that I always find a real eigenvalue. My question is: Is there a theorem that guarantees me that there is a real eigenvalue? Or is there a way to prove the existence for all $n \in \mathbb{N}$? Of course in the case that (n+1) is odd, the existence is trivial, but what can we say in the even case?

$$\begin{Bmatrix} 1 & 1 & 2...0& 0 \\ 1 & 2 & ...n & 0 \\ 2 & 0 & ...n+1 & n+1 \\ 0 & 0 & ...n+1 & n+2 \end{Bmatrix}$$

Answer 1

It will be better if you write the definition of your matrix in a more readable way. From what you wrote, it seems that your matrix satisfies $M(k,k+1)M(k+1,k)\geq 0$. With this condition, all eigenvalues are real.

In general the eigenvalues of a Jacobi (3-diagonal) matrix will not change if you replace both off diagonal elements $M(k,k+1)$ and $M(k+1,k)$ by square root of their product. So if the product of off diagonal elements is positive you obtain a symmetric matrix.

Ref. R. Gantmakher and M. Krein, Oscillation matrices..., MR1908601.

EDIT. One way to prove this is indicated in the comment below by Anthony Quas. Another way (used in Gantmakher and Krein) is to write explicitly the characteristic polynomial, and to notice that off-diagonal elements enter only in the form of products $M(k,k+1)M(k+1,k)$.

One more remark. This argument shows that there always exists an explicitly written quadratic form with respect to which our operator is Hermitean. If the products of off diagonal elements are all positive, this quadratic form is positive definite, and we have all real eigenvalues. But even if the products are not all positive, and the quadratic form is indefinite, one can sometimes obtain existence of SOME real eigenvalues by using a theorem of Pontryagin on Hermitean operators with respect to indefinite form.