Question

6.22 (Extension of the previous exercise) Let be a complex vector space with a positive definite scalar product and T E L(U). If T is self-adjoint, then the coefficients of the characteristic polynomial of T are all real.

Previous Exercise:

Answer 1

Given that U is complex vector space with positive definite scalar product T: U rightarrow U and T = T (T^-)^t = T (since T is self adjoin) let f(x) elementof e[x] and T elementof Mat_n times n (c) is Hermitian let f(n) = a_0 + a_1 x + a_2 x^2 +-----+ a_n x^n elementof c[x] Then f(T) = a_0 + a_1 T + a_2 T^2 + ----- + a_n T^n Now, f(T)^- = a_0 + a_1 T + a_2 T^2 + ----- + a_n T^n^- since we know that aA + bB^- = a^- A^- + b^- B^- we have f(T)^- = a_0^- + a_1 T^- + a_2 T^2 + -----+ a_n^T^n^- since a_1 T^- = a_1 T^- f(T)^- = a_0^- + a_1 T^- + a_2 T^2 + ----+ a_n^- T^-n \r\n (f(T))^- t = a_0^- + a_1 T^- t + a_2^- (T^- 2)^t + -----+ a_n^- (T^- n)^t since (T^-)^t = T (f(T)^-)^t = a_0^- + a_1 T^- + a_2 T^2 + -----+ a_n^T^n (T^- n)^t = T. T. T. ------T^--t = T^-t T^-t -----T^-t = T^n since that complex vector poetics scalar product i.e. a^- = a from (f(T)^-)^t = a_0 + a_1 T + a_2 T^2 +---- + a_n T^n