Relation between determinant of a Jacobi matrix and its minor Let
$$
A = \begin{bmatrix}
a_1 & b_1 \\
b_1 & a_2 & b_2 \\
& b_2 & \ddots & \ddots \\
& & \ddots & \ddots & b_{n-1} \\
& & & b_{n-1} & a_n
\end{bmatrix},\ B = \begin{bmatrix}
a_2 & b_2 \\
b_2& a_3 & \ddots & \\
&\ddots  & \ddots &  b_{n-1} \\
& & b_{n-1} & a_n
\end{bmatrix},
$$
where $b_i \neq 0$.
Can we prove that for any real value of $\lambda$, $\det(\lambda I_n-A) \neq \det(\lambda I_{n-1}-B)$ where $I_k$ is identity matrix of order $k$? It has been a few days I am trying to prove it but I failed. I would be appreciated for any help.
I could only prove that no two immediate consecutive minors of a Jacobi matrix share eigenvalues. I mean there is no $\lambda \in \mathbb{R}$ such that $\det(\lambda I_n-A) = 0$ and $\det(\lambda I_{n-1}-B) = 0$. I proved it by contradiction which will leads to some $b_i = 0$, while $b_i \neq 0$. 
If $P_{n}(\lambda) = \det(\lambda I_n-A)$ then $P_{n}(\lambda) = (\lambda-a_n)P_{n-1}(\lambda)-b_{n-1}^2P_{n-2}(\lambda)$. By this equation I proved the contradiction.
Thanks in advance.

This proposition is not necessarily true in general.

Lemma: If $B$ and $C$ are square matrices and $a \in \mathbb{R}$, then there exists $b \in \mathbb{R}^*$ such that the equation$$
(λ - a) |λI - B| = b^2 |λI - C|
$$
  has a real solution.

Proof: It suffices to prove that there exist $λ \in \mathbb{R}$ and $b \in \mathbb{R}^*$ satisfying the equation. Note that $|λI - B|$ is a polynomial of $λ$ with a positive leading coefficient. If it has real roots, then the largest one $λ_0 \leqslant ρ(B)$ and $|λI - B|$ is positive for $λ > λ_0$. Otherwise $|λI - B| > 0$ for all $λ \in \mathbb{R}$. Thus $|λI - B|$ is always positive for $λ > ρ(B)$.
Now take an arbitrary $λ > \max(a, ρ(B), ρ(C))$, then $λ - a > 0$, $|λI - B| > 0$, $|λI - C| > 0$, and take $b = \sqrt{(λ - a)\dfrac{|λI - B|}{|λI - C|}} > 0$. Thus $(λ - a) |λI - B| = b^2 |λI - C|$.
Now back to the question. Denote$$
C = \begin{pmatrix}
a_3 & b_3 &&\\
b_3 & \ddots & \ddots &\\
& \ddots & \ddots & b_{n - 1}\\
&& b_{n - 1} & a_n
\end{pmatrix},
$$
then$$
|λI - A| = (λ - a_1) |λI - B| - b_1^2 |λI - C|.
$$
Thus$$
|λI - A| = |λI - B| \Longleftrightarrow (λ - a_1 - 1) |λI - B| = b_1^2 |λI - C|.
$$
The lemma implies for any $a_1, \cdots, a_n$ and $b_2, \cdots, b_{n - 1}$, there exists $b_1 \in \mathbb{R}^*$ such that $|λI - A| = |λI - B|$ has a real solution. Therefore, $|λI - A| ≠ |λI - B|\ (\forall λ\in \mathbb{R})$ is not necessarily true.
