Characteristic polynomial of a symmetric tridiagonal matrix after removing first row and column

Question

Given matrix $A$

$$ A = \begin{bmatrix} a_1&b_1&0&0&0&0& \cdots &0\\ b_1&a_2&b_2&0&0&0&\cdots&0\\ 0&b_2&a_3&b_3&0&0&\cdots&0\\ 0&0&b_3&a_4&b_4&0&\cdots&0\\ 0&0&0&b_4&a_5&b_5&\cdots&0\\ 0&0&0&\ddots&\ddots&\ddots&\ddots&\vdots\\ 0&0&0&0&0&b_{n-2}&a_{n-1}&b_{n-1}\\ 0&0&0&0&0&0&b_{n-1}&a_n\\ \end{bmatrix} $$

The recursive relation of characteristic polynomial of $A$ is

$$ P_1(\lambda) = (a_1 - \lambda) , P_0(\lambda) = 1, b_0=0 $$ $$ P_j(\lambda) = (a_j - \lambda)P_{j-1}(\lambda) - b_{j-1}^2P_{j-2}(\lambda), 2 \leq j \leq n $$ where $P_j(\lambda) = det(A_j - \lambda I_j)$ is characteristic polynomial of the $j$-th leading principal minor of $A$.

If we let the $B$ to be

$$ B = \begin{bmatrix} a_2&b_2&0&0&0&\cdots&0\\ b_2&a_3&b_3&0&0&\cdots&0\\ 0&b_3&a_4&b_4&0&\cdots&0\\ 0&0&b_4&a_5&b_5&\cdots&0\\ 0&0&\ddots&\ddots&\ddots&\ddots&\vdots\\ 0&0&0&0&b_{n-2}&a_{n-1}&b_{n-1}\\ 0&0&0&0&0&b_{n-1}&a_n\\ \end{bmatrix} $$

Is there any recursive relationship between characteristic polynomial of $A$ and $B$?

Thanks in advance.

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What I tried

The recursive relation for both of them are the same, the only difference is in the starting value of the recurrence. Let $Q(\lambda)$ be the characteristic polynomial of $B$ then:

$$Q_3(\lambda) = (a_3 - \lambda)(a_2-\lambda)-b_2^2$$ $$ P_3(\lambda) = (a_3-\lambda)\bigg[(a_2-\lambda)(a_1-\lambda)-b_1^2 \bigg]- b_2^2(a_1-\lambda) $$

I am trying to write $Q_3(\lambda)$ based on $P_3(\lambda)$

Answer 1

You have a formula for the determinant of a tridiagonal matrix, whether symmetric of not: if you expand the determinant of $$A_n = \begin{bmatrix} a_1-\lambda&b_1&0&0&0&0& \cdots &0\\ b_1&a_2-\lambda&b_2&0&0&0&\cdots&0\\ 0&b_2&a_3-\lambda&b_3&0&0&\cdots&0\\ 0&0&b_3&a_4-\lambda&b_4&0&\cdots&0\\ 0&0&0&b_4&a_5-\lambda&b_5&\cdots&0\\ 0&0&0&\ddots&\ddots&\ddots&\ddots&\vdots\\ 0&0&0&0&0&b_{n-2}&a_{n-1}-\lambda&b_{n-1}\\ 0&0&0&0&0&0&b_{n-1}&a_n-\lambda\\ \end{bmatrix}$$ by the last column (or the last row), you obtain readily the order $2$ recurrence relation $$P_n(\lambda)=(a_n-\lambda)P_{n-1}(\lambda)-b_{n-1}^2P_{n-2}(\lambda).$$