Proof regarding the image and kernel of a normal operator

In summary, the conversation discusses the proof that if T is normal, then T and T* have the same kernel and the same image. The attempt at a solution involves decomposing T into direct sum subspaces and using eigenvalues, but the solution is found by recognizing that T is unitarily diagonalizable. This leads to the conclusion that the right and left nullspaces of T and T* are identical.
  • #1
Adgorn
130
18

Homework Statement


Show that if T is normal, then T and T* have the same kernel and the same image.

Homework Equations


N/A

The Attempt at a Solution


At first I tried proving that Ker T ⊆ Ker T* and Ker T* ⊆ Ker, thus proving Ker T = Ker T* and doing the same thing with I am T, but could not find a way of doing so. I am relatively certain this has to do with discomposing T into direct sum subspace or doing something with the orthonormal basis of V comprised of eigenvalues of T but I cannot seem to figure it out.
I would love some assistance on the matter.
 
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  • #2
You understand that ##T## is unitarily diagonalizable here, right? You really can't ask for much more as the mutually orthonormal eigenvectors form a partition (and indeed are a proper, length preserving, coordinate system).

##T = U D U^*##

##T^* = \big(U D U^*\big)^* = U D^* U^*##

with respect to the the right nullspace, how would you identify it in ##T##, above? What would it look like in ##T^*##? If you are so interested, what would the left nullspace look like for both of them?
 

Related to Proof regarding the image and kernel of a normal operator

1. What is a normal operator?

A normal operator is a linear operator on a vector space that commutes with its adjoint. In other words, the operator and its adjoint have the same eigenvectors and their corresponding eigenvalues are complex conjugates of each other.

2. How do you prove that an operator is normal?

To prove that an operator is normal, you must show that it commutes with its adjoint. This can be done by showing that the operator and its adjoint have the same eigenvectors and their corresponding eigenvalues are complex conjugates of each other.

3. What is the image of a normal operator?

The image of a normal operator is the set of all vectors that can be obtained by applying the operator to a vector in the vector space. It is a subspace of the vector space.

4. How is the kernel of a normal operator related to its eigenvalues?

The kernel of a normal operator is the set of all vectors that are mapped to the zero vector under the operator. The eigenvalues of a normal operator are the values that, when applied to a vector in the kernel, result in the zero vector. This means that the eigenvalues are the values that make the kernel non-trivial.

5. Can a normal operator have a non-trivial kernel and image?

Yes, a normal operator can have a non-trivial kernel and image. In fact, a normal operator can have a non-trivial kernel and image simultaneously if and only if the operator is not invertible.

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