Past exam of the mathematics course of the University of Cambridge / 2001 / ib / Paper 3 / 16E (Numerical Analysis)

The monic orthogonal polynomial $p_n$ is orthogonal to every polynomial of degree less than $n$, since $p_0,\dots ,p_{n-1}$ form a basis of that polynomial space. Let $r_1,\dots ,r_k$ be its distinct interior roots at which its sign changes, and put $q(x)={\prod }_{j=1}^k(x-r_j)$, with $q=1$ if $k=0$. The real polynomial $p_{n}q$ has constant sign throughout $(a,b)$: multiplication by each linear factor removes its corresponding sign change, while even-multiplicity roots do not change sign. It is not identically zero. Since $w>0$, its weighted integral is nonzero.

If $k<n$, orthogonality instead gives ${\int }_a^b{p}_{n}qwdx=0$, a contradiction. Thus $k\ge n$. Degree $n$ also gives $k\le n$, so $k=n$ and all roots have multiplicity one, with no roots left outside the interval. Therefore there are exactly $n$ distinct simple zeros in $(a,b)$. The weighted inner products are understood to exist, as required by the stipulated orthogonality.

Let $d_n(x)=\det (xI-A_n)$, and set $d_0=1$, $d_1=x-a_1$. Expansion of the tridiagonal determinant along its last row givesFor the second term, the last-row entry is $-b_n$, its cofactor has negative sign, and the remaining minor has last-column entry $-b_n$ multiplying $d_{n-2}$; their product is $-b_n^2{d}_{n-2}$. Thus $d_n$ and $p_n$ have identical initial values and recurrence, and induction provesThe eigenvalues of the Jacobi matrix $A_n$ are precisely the roots of this characteristic polynomial. Part(a) therefore gives all $n$ eigenvalues simple and lying strictly in $(a,b)$. This conclusion follows from orthogonality as well as symmetry; symmetry alone would establish real eigenvalues but not their interval location.