Here we present a non-inductive proof of the spectral theorem for normal matrices (which doesn’t use, for instance, proposition 8.6.4 in Artin’s *Algebra*). (But it does seem to be the same as the proof in Herstein’s *Topics in algebra*.) It is motivated by a similar direct proof (presented in my class) for Hermitian operators with distinct eigenvalues.

We work in matrix form, so we need to prove that there exists an orthonormal basis of consisting of eigenvectors of a normal matrix .

First we show that has no generalized eigenvectors of order (or higher). Fix , and define . Since commutes with every matrix, is normal. It suffices to show that whenever (here is a vector). Suppose the latter holds. Since commute,

from which it similarly follows that

as desired.

Hence has all Jordan blocks of size ( is diagonalizable). Let be the distinct eigenvalues of , with corresponding eigenspaces , so that .

We now prove that the are pairwise orthogonal. Let be two distinct indices and take , . Let , so is normal as before. Then (by definition), so

Therefore

so forces , establishing the desired orthogonality.

Finally, we can just choose arbitrary orthonormal bases for the (nondegenerate) individual , and merge the resulting bases into a single orthonormal basis of eigenvectors for , so we’re done.

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Compare with http://sbseminar.wordpress.com/2012/12/03/a-calculus-free-proof-of-the-spectral-theorem/

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It would be interesting to see if there’s a direct proof that the minimal polynomial of a normal matrix is squarefree.

Also, there’s a nice perturbation approach to the spectral theorem (due to von Neumann-Wigner; see http://www.math.harvard.edu/~knill/teaching/math19b_2011/handouts/lecture35.pdf).

For the operator version, need to look at more carefully, but see https://terrytao.wordpress.com/2011/12/20/the-spectral-theorem-and-its-converses-for-unbounded-symmetric-operators/

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