Proving T is Normal Operator on Finite Dim Inner Prod Space

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Homework Help Overview

The discussion revolves around proving properties of normal operators in finite dimensional inner product spaces, specifically that a normal operator T has the same image as its adjoint T*. The original poster mentions success in a specific case but struggles with the general proof.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are discussing the implications of specific properties of normal operators and referencing theorems and exercises from a textbook. Questions arise regarding the definitions and implications of the terms used, such as the underlying field and the nature of the operator.

Discussion Status

The conversation is ongoing, with participants seeking clarification on previous attempts and discussing relevant theorems and exercises that may aid in the proof. There is a focus on understanding the relationships between the null space and range of the operator and its adjoint.

Contextual Notes

There is mention of specific theorems and exercises from a textbook that are relevant to the problem, indicating a structured approach to exploring the properties of normal operators. The original poster's reference to a successful case suggests a partial understanding, but the general case remains unresolved.

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"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.
 
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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?
 
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 exercise 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'.
 
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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