determinant recursive formula of a specific matrix
For a field $K, n \in \mathbb{N}_{>0}$ and $\lambda \in K$ let $A_{n, \lambda} \in \textrm{Mat} (n,K) $ be the following matrix with entries $\lambda$ on the diagonal, $-1$ on both minor diagonals and $0$ elsewhere.
$A_{n, \lambda}=\begin {pmatrix} \lambda & -1\\ -1 & \lambda & -1\\ & -1 & \lambda & -1\\ & & \ddots& \ddots &\ddots \\ & & & & -1 \\ & & & -1 & \lambda & \end{pmatrix}$
I trying to deduce a recursion formula for $d_{n,\lambda}:=\det (A_{n,\lambda})$ i.e. i want to constitute for $n > 1 $ the number $d_{n,\lambda}$ through numbers $d_{k,\lambda}$ with $0<k<n$.
Furthermore i want to show, that for all $ k \in \mathbb{N}$
$d_{2+3k,1}=0$
My idea was to calculate a few values and then conclude using mathematical induction, but my problem is to find the correct pattern... The values a calculated are $d_{2,\lambda} = \lambda ^2-1, d_{3,\lambda}= \lambda(\lambda^2-1), d_{4,\lambda}= \lambda^4-3\lambda^2+1$
$\endgroup$ 31 Answer
$\begingroup$Expand with respect to the first column we get $$ d_n = \lambda d_{n-1} - d_{n-2} $$ which must be complemented with $$d_0=1, \quad d_1 =\lambda $$
Now $$ d_{n+3} = \lambda d_{n+2} - d_{n+1} = \lambda (\lambda d_{n+1} - d_{n}) - d_{n+1} = (\lambda^2-1) d_{n+1} - \lambda d_{n} $$ and so, considering $d_n$ as a polynomial in $\lambda$, we have by induction that $d_{2+3k}$ is a multiple of $\lambda^2-1$ because $d_2 = \lambda^2-1$.
In particular, $d_{2+3k}(\pm 1)=0$.
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