Question

Let A be an invertible linear operator on a finite-dimensional complex vector space V. Recall that we have shown in class that in this case, there exists a unique unitary operator U such that A=UA. The point of this exercise is to prove the following result: an invertible operator A is normal if and only if U|A= |AU. a) Show that if UA = |A|U, then AA* = A*A. Now, we want to show the other direction, i.e. if AA* = A* A, then UA = AU, which is going to be more difficult. b) Show that if A is normal, then U|A|2 = |A|2U. c) We now want to finish the proof by concluding that U|A|2 = |A|2U implies U|A| = |A|U. For notational convenience, define B:= |A|2 and use the spectral theorem to show that there exists a polynomial g(t) such that g(B) = VB. Use this to conclude UA = |AU.

Answer 1

a)

Suppose U is unitary. then U is invertible and

U(a/b)= (Ua\ UU-1b)=( a/ U-1b) for all a and b

Hence U-1  is the adjoint of U

conversely, suppose U* exists and  UU*=1=U*U. . Then U is invertible with U-1=U*. so we need only to show that U preserves inner products

b)

Consider Cnx1 with the inner product (x/y)= y*x. Let A be an nxn matrix over C, and let U be the linear operator defined by U(X)=AX. then

(UX/UY)=(AX/AY)=Y*A*AX

its for all X,Y

Hence U is unitary if and only if A*A=1

hence the proof