Normal Operator

  • #1
"Prove that if T is a normal operator on a finite dimensional inner product space, then it has the same image as its adjoint."

I succeeded with F=C, but I can't get it for the general case.
 

Answers and Replies

  • #2
matt grime
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What did you do for the F=C case, not that you've said what F is (underlying field, perhaps? or the space itself?)

I guess you've tried considering

<Nx,Nx>=<x,N*Nx>=<x,NN*x>=<N*x,N*x.

What did it get you?
 
  • #3
AKG
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My book last year gave this as a problem. It says to prove that N(T) = N(T*), and R(T) = R(T*). It gives a hint: use theorem 6.15 and excercise 12. Theorem 6.15 is just the basic facts about normal operators:

|Tx| = |T*x|
T - cI is normal for all c in F
If Tx = cx, then T*x = c*x, where c* denotes the conjugate of c.
If c and d are distinct eigenvalues of T with corresponding eigenvectors x and y, then x and y are orthogonal

Exercise 12 says to prove, for any linear operator T on any inner product space V

a) the orthogonal complement of R(T*) is N(T)
b) If V is finite dimensional, then R(T*) is the orthogonal complement of N(T)

It gives a hint to use another exercise which says:

V is an inner product space, S and S' are subset of V, W is a finite dimensioanl subspace of V. Prove:

a) S' contained in S implies the orthogonal complement of S is contained in the orthogonal complement of S'
b) S is contained in the orth. compl. of the orth. compl. of S, so span(S) is containd in the orth. compl. of the orth. compl. of S
c) W = the orth. compl. of the orth. compl. of W
d) V = W direct sum with the orth. compl. of W

For c, use the fact that if W is a f.d. subspace of an i.p.s V, and y is in V, then there exist unique u in W and z in orth. compl. of W such that y = u + z. For d, prove the fact that if W and W' are subspaces of a vector space V, then V is their direct sum iff for all v in V, there exist unique w and w' in W and W' respectively such that v = w + w'.
 
  • #4
Tom Mattson
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matt grime said:
What did you do for the F=C case, not that you've said what F is (underlying field, perhaps? or the space itself?)
My guess is that he's using Linear Algebra Done Right by Axler, who uses F to stand for either R or C (the real or complex field, respectively).
 

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