The following matrix $A_{n \times n}$ is given

\begin{bmatrix} a_1&b_1&0&0&0&\cdots&0&0\\ b_1&a_2&b_2&0&0&\cdots&0&0\\ 0&b_2&a_3&b_3&0&\cdots&0&0\\ \vdots&\vdots&\ddots&\ddots&\ddots&\vdots&\vdots&\vdots\\ 0&0&0&0&0&\cdots&a_{n-1}&b_{n-1}\\ 0&0&0&0&0&\cdots&b_{n-1}&a_{n}\\ \end{bmatrix}

$A_i$ denotes the $i$th leading principle submatrix of $A$ and $P_i(\lambda) = det(\lambda I_i - A_i)$ is the characteristic polynomial of $A_i$. It has been claimed that

  • $P_1(\lambda) = \lambda - a_1$ (which is obvious)
  • $P_j(\lambda) = (\lambda - a_j)P_{j-1}(\lambda) - b_{j-1}^2P_{j-2}(\lambda), \text{ }2\leq j \leq n$

I tried to prove the second one but I failed, how can I prove that?


  • $\begingroup$ Try to prove by induction. $\endgroup$
    – user371838
    Dec 8 '17 at 7:04
  • $\begingroup$ @Rohan I tried that by the definition of determinant, let this holds for $A_m, \text{ then tried to prove for }$ $det(\lambda I_{m+1}-A_{m+1})$. Is it right? but I didn't get the desired result $\endgroup$
    – M a m a D
    Dec 8 '17 at 7:07

For $n>1$, the matrix $A_{n}$ looks like this:

$$A_{n} = \begin{bmatrix}A_{n-1}&&\vec{0}\\&&b_{n-1}\\\vec{0}&b_{n-1}&a_{n}\end{bmatrix}$$

And so

$$\lambda I_n - A_{n} = \begin{bmatrix}\lambda I_{n-1} - A_{n-1}&&\vec{0}\\&&b_{n-1}\\\vec{0}&b_{n-1}&\lambda - a_{n}\end{bmatrix}$$

We can compute the determinant by expanding on the bottom row: the determinant of a matrix is equal to the alternating-signed sum over each entry in the bottom row times the determinant of the corresponding minor.


$$\det(\lambda I_n - A_n) = (\lambda-a_n)\det(\lambda I_{n-1}-A_{n-1}) - b_{n-1} \det\begin{bmatrix}B_{n-1} & \vec{0}\\\ldots&b_{n-1}\end{bmatrix}$$

Here I use the shorthand notation $B_{n-1}$ to refer to the matrix $I_{n-1} - A_{n-1}$ with the last column deleted. (We delete the last column because we're forming a minor of $A_n$ by deleting the row and column containing the entry $b_{n-1}$.)

The first determinant above is just $\det(\lambda I_{n-1}-A_{n-1})\equiv P_{n-1}$ We can apply the same expansion trick (but this time along the last column) to compute the second determinant above:

$$\det\begin{bmatrix} \vdots & \vec{0}\\B_{n-1} &b_{n-1}\end{bmatrix} = b_{n-1} \det(\lambda I_{n-2} - A_{n-2})$$

We get this result because by expanding upon the last column, the only potentially nonzero term is the last entry in the last column, $b_{n-1}$. The minor formed by deleting $b_{n-1}$ is the matrix $B_{n-1}$ with its last row removed— but since $B_{n-1}$ is itself $A_{n-1}$ with its last column removed, we get $A_{n-1}$ with its last row and column removed, which is just $A_{n-2}$.

Hence overall:

$$\det(\lambda I_n - A_n) = (\lambda - a_n)P_{n-1} - b_{n-1}^2 P_{n-2}$$


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