Does Axler's Spectral Theorem Imply Normal Matrices?

[boo]

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.

 

[mathwonk]

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.

 

[boo]

Does the spectral theorem guarantee that algebraic and geometric multiplicity for all eigenvalues are always equal for normal operators?

 

[boo]

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.