Does Axler's Spectral Theorem Imply Normal Matrices?

Click For Summary

Discussion Overview

The discussion revolves around the implications of Axler's spectral theorem concerning normal matrices and diagonalizability in both complex and real inner product spaces. Participants explore whether a linear operator can be diagonalizable with non-orthogonal eigenvectors if it is not normal, and the relationship between normality and the properties of eigenvalues.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether a linear operator T can be diagonalizable with non-orthogonal eigenvectors if T is not normal, particularly in complex inner product spaces.
  • Another participant suggests constructing a 2x2 example of a diagonalizable matrix with non-orthogonal eigenvectors to explore the conjecture.
  • A participant inquires if the spectral theorem guarantees that algebraic and geometric multiplicities for all eigenvalues are equal for normal operators.
  • A later reply asserts that in a complex inner product space, every normal operator has a full orthonormal set of basis vectors, while in a real inner product space, every self-adjoint operator does as well.

Areas of Agreement / Disagreement

Participants express differing views on the implications of the spectral theorem, particularly regarding diagonalizability and the properties of eigenvectors in relation to normality. The discussion remains unresolved on some points, particularly the conjecture about diagonalizability without normality.

Contextual Notes

The discussion includes assumptions about the definitions of normal and self-adjoint operators, as well as the conditions under which eigenvalues exist in real inner product spaces. There are unresolved questions about the implications of the spectral theorem in different contexts.

boo
Messages
26
Reaction score
8
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.
 
Physics news on Phys.org
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.
 
Does the spectral theorem guarantee that algebraic and geometric multiplicity for all eigenvalues are always equal for normal operators?
 
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.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K