# solutions to exercises 1.26

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.