Exercise: Find the determinants of Jacobi matrices with \(a=1,bc=-1\) and \(a=-1,bc=-1\).

Solution: First assume \(a=1,bc=-1\), and use the regression obtained in the lecture notes, i.e $$j_{k+1}=aj_k-bcj_{k-1} = j_k+j_{k-1}$$

Those are Fibonacci numbers with initial conditions obtained by the first two determinants: $$ j_1 = \text{det}(a) = a = 1,~~~ ~~~ j_2 = \left| \begin{matrix} a & b \\ c & a \end{matrix} \right| = a^2-bc = 1+1 = 2, $$ and the rest follows the standard Fibonacci sequence (\(1,2,3,5,8,13,\ldots\)). For the second case, \(a=-1,bc=-1\), we similarly get $$ j_{k+1}=aj_k-bcj_{k-1} = -j_k+j_{k-1}$$ with initial conditions \(j_1 = -1\) and \(j_2 = 1+1 = 2 \), which is the alternating Fibonacci sequence (\(-1,2,-3,5,-8,13,\ldots\)).

Exercise: Find the determinant of

$$

A(x_1,\ldots,x_n)=\left( \begin{matrix} 1 & 1 & 1 & \ldots & 1 \\ x_0 & x_2 & x_3 & & x_n \\ x_0^2 & x_2^2 & x_3^2 & & x_n^2 \\ \vdots & & & \ddots & \\

x_0^{n-2} & x_2^{n-2} & x_3^{n-2} & \ldots & x_n^{n-2}\\

x_0^n & x_2^n & x_3^n & \ldots & x_n^n \end{matrix} \right)

$$

Solution: Experiments with small matrices should convince you that this deperminant is equal to

\[

(x_1+\ldots+x_n)V(x_1,\ldots,x_n),

\]

where \(V(x_1,\ldots,x_n)\) is the standard Vandermonde determinant.

This can be proved by multiplying the Vandermonde matrix from the left by

\[

\left(

\begin{array}{cccc}

1&0&\ldots&0&0&0\\

0&1&\ldots&0&0&0\\

&&\ddots&&\vdots&\\

0&0&\ldots&0&1&0\\

(-1)^ns_n &(-1)^{n-1}s_{n-1} & \ldots &s_3 &-s_2 & s_1\\

%& & -s_2 & s_1

\end{array}

\right)

\]

where

\[

s_1=\sum_{i=1}^n x_i, s_2=\sum_{i<j}x_i x_j, s_3=\sum_{i<j<k}x_ix_jx_k,\ldots

\]

are *elementary symmetric functions*. One can easily see that this product is \(A(x_1,\ldots,x_n)\), whence the result.

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