# Does Axler's Spectral Theorem Imply Normal Matrices?

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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.

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.