F=ℝ: Normal matrix with real eigenvalues but not diagonalizable

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SUMMARY

The discussion centers on the properties of normal matrices with real eigenvalues, specifically whether such matrices are necessarily diagonalizable. It is established that a normal matrix with real eigenvalues is indeed diagonalizable, and the participants explore the implications of this in the context of the real field ℝ. A counterexample is sought but ultimately deemed non-existent, as the properties of normal matrices guarantee diagonalizability in this case.

PREREQUISITES
  • Understanding of normal matrices in linear algebra
  • Knowledge of eigenvalues and eigenvectors
  • Familiarity with diagonalization concepts
  • Basic principles of inner product spaces
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  • Study the proof of diagonalizability for normal matrices over the real field
  • Explore the properties of complex normal matrices and their diagonalizability
  • Investigate the implications of the spectral theorem in linear algebra
  • Learn about counterexamples in linear algebra and their significance
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Students and professionals in mathematics, particularly those studying linear algebra, matrix theory, and eigenvalue problems.

Bipolarity
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I am going through Friedberg and came up with a rather difficult problem I can't seem to resolve.
If ## F = ℝ ## and A is a normal matrix with real eigenvalues, then does it follow that A is diagonalizable? If not, can I find a counterexample?

I'm trying to find a counterexample, by constructing a matrix A that is normal and has real eigenvalues, but is not diagonalizable. It is giving me some problems! Any ideas?

If I did not have the assumption that A has real eigenvalues, the rotation matrix would suffice as a counterexample.

If I had the complex field instead of the real field, then easily A is diagonalizable since normality implies orthogonal diagonalizibility in a complex inner product space.

BiP
 
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Just a hint: there is no counterexample, look for proof.
 

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