Does Axler's Spectral Theorem Imply Normal Matrices?

In summary, the complex spectral theorem for operators on vector space V states that T is normal if and only if V has an orthonormal basis consisting of eigenvectors of T and the matrix representation of T is diagonal with respect to some orthonormal basis of V. It is not possible for T to be diagonalizable with non-orthogonal eigenvectors if it is not normal, both in a complex and real inner product space. Additionally, in a complex inner product space, every normal operator is guaranteed to have a full orthonormal set of basis vectors, while in a real inner product space, every self-adjoint operator does as well.
  • #1
boo
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Going through Axler's awful book on linear algebra. The complex spectral theorem (for operator T on vector space V) states that the following are equivalent: 1) T is normal 2) V has an orthonormal basis consisting of eigenvectors of T and 3) the matrix representation of T is diagonal with respect to some orthonormal basis of V. The question I have is: does that imply that T is only diagonalizable if T is normal (on a COMPLEX inner product space)? I.e., is it possible for T to still be diagonalizable with NON-ORTHOGONAL eigenvectors of T if T is not normal? Same question for a real inner product space (R.I.P.S.) and T being not self-adjoint. Now if T is not self-adjoint (on a R.I.P.S.) there is no guarantee that real eigenvalues even exist so I assume that answer must be no.
 
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  • #2
you might try to cook up a 2x2 example of a diagonalizable matrix with non orthogonal eigenvectors (and different eigenvalues). then check that a vector orthogonal to an eigenvector is not an eigenvector, to prove your matrix is not normal, hence proving your conjecture.
 
  • #3
Does the spectral theorem guarantee that algebraic and geometric multiplicity for all eigenvalues are always equal for normal operators?
 
  • #4
OK I found the answer. In a complex inner product space every normal operator (commutes with it's adjoint) is guaranteed to have a full orthonormal set of basis vectors for that space. In a real inner product space every self adjoint operator does likewise.
 

FAQ: Does Axler's Spectral Theorem Imply Normal Matrices?

1. What is Axler's Spectral Theorem?

Axler's Spectral Theorem is a theorem in linear algebra that states that any normal matrix can be diagonalized by a unitary matrix. This means that a normal matrix can be expressed as a diagonal matrix with its eigenvalues on the diagonal.

2. How does Axler's Spectral Theorem relate to normal matrices?

Axler's Spectral Theorem is specifically applicable to normal matrices. It states that any normal matrix can be diagonalized, which means that the matrix can be expressed in terms of its eigenvalues and eigenvectors.

3. What are normal matrices?

Normal matrices are square matrices that satisfy the condition that the matrix multiplied by its conjugate transpose is equal to the conjugate transpose multiplied by the matrix. In simpler terms, a normal matrix is a matrix that commutes with its conjugate transpose.

4. Does Axler's Spectral Theorem hold true for all normal matrices?

Yes, Axler's Spectral Theorem is a general theorem that applies to all normal matrices. It is a fundamental result in linear algebra and has been proven to hold true for all normal matrices.

5. Why is Axler's Spectral Theorem important?

Axler's Spectral Theorem is important because it provides a way to easily diagonalize normal matrices, which simplifies calculations and makes it easier to understand the properties of these matrices. It is also a crucial result in quantum mechanics, where normal matrices are used to represent observables.

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