exercise a5adjoints let v rx with inner productpq jypxqx dxlet r be a polynomial in rr show that

In this question we have been given a differential equation we need to find out the solution of

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Suppose that V is a finite dimensional inner product space and T is a self -adjoint linear opera

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Exercise A5 (Adjoints) Let V = R[x] with inner product (p|q) = âˆ«[a,b] p(x)q(x) dx. Let r be a polynomial in R[x]. Show that the linear transformation T(p) := x + r(x)p(x) is self-adjoint. b) Let V = P2(R) with the inner product (p|q) = âˆ«[c,d] p(x)q(x) dx (note that the integral bounds are different from above). (i) Show that the linear operator T : P2(R) â†’ P2(R) defined via T(a+bx+cx^2) = bx is not self-adjoint. (ii) Show that the matrix T with respect to the basis {1, x, x^2} of P2(R) is: 0 0 0 0 (iii) Notice that the matrix of T in part (ii) is equal to its conjugate transpose, but we showed in part (i) that T is not self-adjoint. Why is this not a contradiction?

          Exercise A5 (Adjoints)

Let V = R[x] with inner product (p|q) = âˆ«[a,b] p(x)q(x) dx. Let r be a polynomial in R[x]. Show that the linear transformation T(p) := x + r(x)p(x) is self-adjoint.

b) Let V = P2(R) with the inner product (p|q) = âˆ«[c,d] p(x)q(x) dx (note that the integral bounds are different from above).

(i) Show that the linear operator T : P2(R) â†’ P2(R) defined via T(a+bx+cx^2) = bx is not self-adjoint.

(ii) Show that the matrix T with respect to the basis {1, x, x^2} of P2(R) is:

0 0
0 0

(iii) Notice that the matrix of T in part (ii) is equal to its conjugate transpose, but we showed in part (i) that T is not self-adjoint. Why is this not a contradiction?

exercise a5adjoints let v rx with inner productpq jypxqx dxlet r be a polynomial in rr show that the linear transformation tp x rxp is self adjoint b let v p2r with the inner product pq fpq 57418

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