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.


1 Answer 1



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.

  • $\begingroup$ Is your statement trivial or it needs some proof? $\endgroup$
    – M a m a D
    Nov 22, 2022 at 19:44
  • $\begingroup$ 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 $\endgroup$
    – Apass.Jack
    Nov 22, 2022 at 21:00
  • $\begingroup$ Thank you. I come from computer science and only need it for $\Bbb C$ $\endgroup$
    – M a m a D
    Nov 23, 2022 at 9:44

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