 Solution

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.