How to compute the Jacobi matrix (tridiagonal matrix) of a polynomial with a recurrence relationship?

Question

I am looking at Trefethen and Bau Exercise 37.1: I have two normalizations of the Legendre polynomials with corresponding recurrence relations: $$P_n(1)=1$$ which follows $$P_n(x) = \frac{2n-1}{n} x P_{n-1}(x) -\frac{n-1}{n}P_{n-2}(x)$$ and $$\|q_n\|=1$$ which follows $$xq_n(x)=\beta_{n-1}q_{n-1}(x)+\alpha_n q_n(x) + \beta_n q_{n+1}(x)$$ where $$\beta_n = \tfrac12 (1-(2n)^{-2})^{-1/2}$$

Apparently, $q_{n+1}$ is proportional to $P_n$. I know that the entries of the Jacobi matrix $t_{ij} = (q_i,xq_j)$ in the $L^2([-1,1])$ inner-product space, and I managed to get the right answer for the $q_j$ with $0$ along the main diagonal, $\beta_{j-1}$ on the upper diagonal and $\beta_j$ along the lower diagonal. However, it doesn't seem that I am getting the right answer for the 1st normalization.

Also, I am asked to find the relationship between the two Jacobi matrices--my intuition is that they are equal but I can't quite justify it--I think it has to do with the fact that the characteristic polynomials are the same monic polynomial.

Furthermore, I am asked to find a formula for $q_{n+1}(1)=\|P_n\|$ using the Jacobi matrices. I've tried to solve this by equating the entries for the Jacobi matrices (where one includes a term of $||P_n||^2$) and solving. But I can't get the answer $\|P_n\|=\sqrt{\frac{2}{2n+1}}$.

Answer 1

I'm just going to sketch it out. I think the key mistake is assuming that the two matrices are going to be equal---the exercise itself notes that $P_n$ and $q_{n+1}$ are proportional but normalized differently. Also, I'm not sure how exactly you defined the second matrix — you mention "two Jacobi matrices" but the exercise just says "the two tridiagonal matrices corresponding to the formulas".

The matrix $T_{i,j} = (q_i, xq_j)$ has elements $T_{n,n+1} = T_{n+1,n} = \beta_n$.

The tridiagonal matrix $P_{i,j}$ defined by the recurrence for $P_n$ is obtained from rewriting the recurrence as $$ x P_n(x) = P_{n,n-1} P_{n-1}(x) + P_{n,n+1} P_{n+1}(x). $$

This is not the same as "$P_{ij} = (P_i, xP_j)$" — the tridiagonal matrices $P$ and $T$ are the matrix representations of "multiplication by $x$" in their respective polynomial basis ($q$ or $P$). So the relationship between them comes from the fact that it's the same linear operator applied to different bases. To check, the eigenvalues of both $P$ and $T$ should be the same and be identical to the roots of the appropriate polynomial—the two matrices are similar.

So it is sufficient to use the fact that $P_n(x) = \|P_n\| q_{n+1}(x)$: $$ \beta_n = T_{n,n+1} = (q_n, xq_{n+1}) = \left( \frac{P_{n-1}}{\|P_{n-1}\|}, \frac{x P_n}{\|P_n\|} \right) = \frac{\|P_{n-1}\|}{\|P_n\|}\, P_{n,n-1}. $$

From here getting the value of $\|P_n\|$ is just straightforward algebra and for me it came out to $\sqrt{\frac{2}{2n+1}}$.