# Axler Proof of Real vs. Complex Spectral Theorem

In Professor Axler's Linear Algebra Done Right, he separates the proof of the spectral theorem for normal operators on a complex v.s. from the spectral theorem for self-adjoint operators over a real v.s., and I am wondering if each proof also works for the other case.

To my understanding, sketches of the proofs are as follows:

1. Normal operator $$T$$ on a complex v.s.: $$T$$ has an eigenvalue $$\lambda$$. Schur's theorem says we can express $$T$$ as an upper triangular matrix (apply the inductive hypothesis of Schur's theorem to the range of $$T - \lambda I$$ and complete the basis). Result that $$||Tx|| = ||T^*x||$$ for normal operators gives the result.

2. Self-adjoint operator $$T$$ on a real v.s.: $$T$$ has an eigenvalue $$\lambda$$. Apply the inductive hypothesis of the spectral theorem to the orthogonal complement of the span of an eigenvector (allowed to do this because this is a $$T$$-invariant subspace) to get result.

It seems to me like the key result is showing the existence of eigenvalues, and each proof would work for the other case (to apply (2) to normal operators, we need to verify that the complement of an eigenspace is $$T$$-invariant if $$T$$ is normal). I just want to make sure I'm not missing some subtlety that explains why Professor Axler separated these 2 cases.

Thank you!

• You are correct. In finite dimensional vector space, we want to know what operators can be diagonalised under the standard orthonormal basis. If the field is $\mathbb{R}$, the operators are self-adjoint. If the field is $\mathbb{C}$, the operators are normal. These proofs are similair. Dec 8, 2023 at 5:19

For example, the result that if $$T$$ in normal then $$\|Tx\| = \|T^*x\|$$, which is crucially used in the proof of the complex spectral theorem, plays no role in the proof of the real spectral theorem.