To see that the eigenvalues are the roots, consider vectors of the form U=(x0(x),r1(x),…,丌n-1(x)) The recurrence(3)implies that(Tu)i=TTi-la)=avi for i n-1, regardless of the value of Now, if a is a root of Tn(), then(3)specializes to x丌n-1(x)=bn-1丌n-2(x)+an丌n-1x TUn=bn-1Un-1+anUn =(Tu)n Therefore, every root of Tn(a)is an eigenvalue of Tn 2 Hermite polynomials Let w(a)=e- on R. According to mathworld. wolfram. com, the system of polynomials Hn defined by Hm Hnw(a)dr=5mn2nIVT satisfies the recurrence H0(x)=1 Hn+1(r)=2c Hn(a)-2nHn-l(a (5) Let the orthonormal Hermite polynomials be defined by Hn(a)=5/2-1zzHn(a), so that and substitute 2n/(n!)1/21/AHn(a)for Hn(r)in(5)to find zHn(r)=VaVnHn-1(z)+Vn+lHn+1(r) These equations are exactly of the form (1)-3), so the eigenvalues of 10 are the roots of Hn Exericse: Write a MATLAB function that generates this tridiagonal matrix. Use the eig command to compute its eigenvalues for any choice of n. Histogram the roots of Hn for n= 50, 100, 250, 300, 500 generalized ensemble is real or complex? the semi-circular law. Does it make a difference whether the the sing the hist functio on. CompareTo see that the eigenvalues are the roots, consider vectors of the form v = (π0(x), π1(x), . . . , πn−1(x))T . The recurrence (3) implies that (T v)i = xπi−1(x) = xvi for i = 1, . . . , n − 1, regardless of the value of x. Now, if x is a root of πn(x), then (3) specializes to xπn−1(x) = bn−1πn−2(x) + anπn−1x i.e. xvn = bn−1vn−1 + anvn = (T v)n. Therefore, every root of πn(x) is an eigenvalue of Tn. 2 Hermite polynomials Let w(x) = e −x 2 on R. According to mathworld.wolfram.com, the system of polynomials H˜ n defined by Z ∞ −∞ H˜mH˜ nw(x) dx = δmn2 nn! √ π satisfies the recurrence H˜ 0(x) = 1 (4) H˜ n+1(x) = 2xH˜ n(x) − 2nH˜ n−1(x). (5) Let the orthonormal Hermite polynomials be defined by Hn(x) = 1 2n/2 √ n!π1/4 H˜ n(x), so that Z ∞ −∞ HmHnw dx = δmn, and substitute 2n/2 (n!)1/2π 1/4Hn(x) for H˜ n(x) in (5) to find H0(x) = π 1/4 xHn(x) = √ 1 2 √ nHn−1(x) + √ 1 2 √ n + 1Hn+1(x). These equations are exactly of the form (1)-(3), so the eigenvalues of 1 √ 2          0 √ 1 √ 1 0 √ 2 0 √ 2 0 √ 3 . . . . . . . . . √ n − 2 0 √ n − 1 √ n − 1 0          are the roots of Hn. Exericse: Write a MATLAB function that generates this tridiagonal matrix. Use the eig command to compute its eigenvalues for any choice of n . Histogram the roots of Hn for n = 50, 100, 250, 300, 500 using the histn function. Compare it to the semi-circular law. Does it make a difference whether the the generalized ensemble is real or complex?