Proving T is Normal Operator on Finite Dim Inner Prod Space

In summary, the conversation discusses proving that if T is a normal operator on a finite dimensional inner product space, then it has the same image as its adjoint. The person has succeeded in the case where F=C but is struggling with the general case. They mention using the basic facts about normal operators and exercises related to orthogonal complements and direct sums to prove this. They also mention using the book Linear Algebra Done Right by Axler.
  • #1
Treadstone 71
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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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  • #2
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
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
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).
 

FAQ: Proving T is Normal Operator on Finite Dim Inner Prod Space

What is a normal operator on a finite dimensional inner product space?

A normal operator on a finite dimensional inner product space is a linear operator that commutes with its adjoint. In other words, the operator and its adjoint have the same eigenvectors and the eigenvalues of the operator and its adjoint are complex conjugates of each other. Normal operators have many useful properties and are widely used in mathematics and physics.

How do you prove that T is a normal operator on a finite dimensional inner product space?

To prove that T is a normal operator on a finite dimensional inner product space, we need to show that T and its adjoint T* commute. This can be done by showing that T*T* = T*T, where T* is the adjoint of T. This implies that T and T* have the same eigenvectors and their eigenvalues are complex conjugates of each other, thus satisfying the definition of a normal operator.

What are some examples of normal operators on finite dimensional inner product spaces?

Some examples of normal operators on finite dimensional inner product spaces include self-adjoint operators (where T = T*) and unitary operators (where T*T = TT* = I, the identity operator). Other examples include diagonalizable operators with real eigenvalues and rotations in a complex vector space.

Why are normal operators important?

Normal operators are important because they have many useful properties that make them easier to study and work with. For example, normal operators are always diagonalizable, which makes it easier to find their eigenvalues and eigenvectors. They also have a spectral theorem, which states that a normal operator can be represented as a diagonal matrix with respect to an orthonormal basis of eigenvectors.

Can a non-normal operator on a finite dimensional inner product space have real eigenvalues?

Yes, a non-normal operator on a finite dimensional inner product space can still have real eigenvalues. This is because the property of having real eigenvalues is not dependent on the operator being normal, but rather on the operator being diagonalizable. A non-normal operator can still be diagonalizable, but it will not have the additional properties and advantages that come with being a normal operator.

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