# 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.

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.

• Is your statement trivial or it needs some proof? Nov 22, 2022 at 19:44
• Depending on your background, it can be considered trivial or nontrivial for the case of a general commutative ring. It can be considered trivial for $\Bbb C$ if $\Bbb R$ is done Nov 22, 2022 at 21:00
• Thank you. I come from computer science and only need it for $\Bbb C$ Nov 23, 2022 at 9:44