Eigenvalues of symmetric tridiagonal matrices with complex entries In this paper the authors proved that for a real symmetric tridiagonal matrix $T_n$, where $b_i \neq 0$, as follows
$$T_n = \begin{bmatrix}
a_1&b_1&0&0&0&0&0&0&\cdots&0\\
b_1&a_2&b_2&0&0&0&0&0&\cdots&0\\
0&b_2&a_3&b_3&0&0&0&0&\cdots&0\\
0&0&b_3&a_4&b_4&0&0&0&\cdots&0\\
0&0&0&b_4&a_5&b_5&0&0&\cdots&0\\
0&0&0&\ddots&\ddots&\ddots&\ddots&\ddots&\ddots&0\\
0&0&0&0&0&b_{n-4}&a_{n-3}&b_{n-3}&0&0\\ 
0&0&0&0&0&0&b_{n-3}&a_{n-2}&b_{n-2}&0\\
0&0&0&0&0&0&0&b_{n-2}&a_{n-1}&b_{n-1}\\
0&0&0&0&0&0&0&0&b_{n-1}&a_n\\
\end{bmatrix},
$$
no two successive leading principal minors of $T_n$ have the same eigenvalue. The $i$th leading principal minor of $T_n$ is denoted by $T_i$ and its characteristic polynomial is denoted by $P_i(\lambda) = \det(\lambda I - T_i)$. For example $T_4$ is as follows
$$
T_4 = \begin{bmatrix}
a_1&b_1&0&0\\
b_1&a_2&b_2&0\\
0&b_2&a_3&b_3\\
0&0&b_3&a_4
\end{bmatrix}.
$$
In fact they proved there is no $x \in \Bbb R$ that will be root of any $P_i(\lambda)$ and $P_{i+1}(\lambda)$. The proof is as follows.

Proof. It is well known that the following recursive relation of $P_n(\lambda)$ exists:
\begin{align*}
P_1(\lambda) &= \lambda-a_1\\
P_j(\lambda) &= (\lambda-a_j)P_{j-1}(\lambda) - b_{j-1}^2P_{j-2}(\lambda), 2 \leq j \leq n.
\end{align*}
If $P_1(\lambda) = 0 = P_2 (\lambda)$, then $(\lambda − a_2 ) P_1 (\lambda) − b_1^2 = 0$, which implies $b_1 = 0$, but this contradicts the restriction on $T_n$ that $b_1 \neq 0$. Once again, for $2 \lt j \leq n$, if $P_{j−1} (\lambda) = 0 = P_j (\lambda)$, then the recurrence $(\lambda − a_{j+1} ) P_j (\lambda) − b^2_j P_{j−1} (\lambda) = 0$, which gives $P_{j−1} (\lambda) = 0$. This will in turn imply that $P_{j−2} (\lambda) = 0$. Thus, we will end up with $P_2 (\lambda) = 0$, implying that $b_1 = 0$ which is a contradiction.

The proof is for real symmetric tridiagonal matrices which have real eigenvalues. I would like to know what if the symmetric matrix has some complex entries? In this way some of its eigenvalues will be complex, but I don't see any problem to apply the proof given above to complex  case.
I would like to know if it is "legal" to apply the proof given above to the complex case? Please note that I am interested in the case where the complex tridiagonal matrices are symmetric rather than Hermitian.
 A: Yes.
More generally, the same conclusion and the same proof are valid if we replace $\Bbb R$ with any commutative ring (with zero and identity throughout this answer), such as $\Bbb C$, $\Bbb Z$ or $\Bbb Q_p$ or $\Bbb Z/2022\Bbb Z$.

In the case of $\Bbb C$, the conclusion and the proof are literally the same. As long as you do not use positivity of a number (such as the square of a nonzero number is positive or the Archimedean property etc.), a proof  for $\Bbb R$ is usually automatically valid for $\Bbb C$. So, we could say that the case for $\Bbb C$ is trivially true once we have proved for $\Bbb R$. (Of course, we should double check that every relevant concept and every step is valid in $\Bbb C$. I have.)

In the case of a general commutative ring $\mathfrak R$, we need to define all relevant concepts over $\mathfrak R$ first. This is easy to do, although it might take a while to become familiar and comfortable with them such as the determinant of a matrix over $\mathfrak R$. Then all the steps will just as easy as the case of $\Bbb R$ except two places.

*

*The recursive relation of $P_n(\lambda)$ still holds over $\mathfrak R$. This one can be proven directly by the definition of the determinant as the algebraic sum of $n!$ products.


*The Cayley–Hamilton theorem that says every square matrix satisfies its own characteristic equation also holds over $\mathfrak R$. This  fact is proven here.
This theorem ensures that any eigenvalue is a root of the characteristic equation.
