From the Spectral Theorem in Operator Theory to the Spectral Theorem in Linear Algebra

Question

I'm still a bit confused about the very general Spectral Theorem in Operator Theory, since it's very abstract. So I thought it might be a good idea to apply the general theorem to the finite-dimensional case. Here's the theorem I got in my notes:

Spectral Theorem: Let $H$ be a complex Hilbert-space and $A: H \to H$ a normal operator. There is a unique spectral measure $\Phi$ on the spectrum $\sigma(A)$ such that $$f(A) = \int_{\sigma(A)} f d \Phi$$ for any continuous function $f$ on the spectrum $\sigma(A)$. Moreover $\Phi(U) \neq 0$ for all non-empty open subsets $U \subset \sigma(A)$, and for every bounded operator $B : H \to H$ we have $BA = AB$ if and only if $B \Phi(U) = \Phi(U) B$ for all $U$.

Let's now assume that $H := \mathbf{C}^n$ and that $A$ is self-adjoint. The spectrum of $A$ is real and finite, hence discrete. I want to show that there exists a orthonormal basis of eigenvectors of $A$.

So I figured to take the function $f := \mathbf{1}_{\{\lambda\}}$ for a $\lambda \in \sigma(A)$. We get $f(A) = \Phi(\{\lambda\})$. The spectral theorem then gives me a finite family of pairwise orthogonal projections $\{ \Phi(\{\lambda \})\}_{\lambda \in \sigma(A)}$, such that $\sum_\lambda \Phi(\{\lambda\})= \text{Id}$.

But I don't see how this does lead anywhere close to the Spectral Theorem in Linear Algebra. Can anyone help?

Thanks!

Answer 1

Because $A$ has finite spectrum, say $\lambda_1,\cdots,\lambda_k$, then $$ E_l = E\{\lambda_l\} $$ are orthogonal projections such that $$ E_l E_m = E_m E_l = 0,\;\;\; m \ne l \\ E_l E_l = E_l = E_l^* \\ E_1 + E_2 + \cdots + E_k = I, $$ and the matrix $A$ may be written as $$ A = \lambda_1 E_1 + \lambda_2 E_2 + \cdots + \lambda_k E_k. $$ Because of the properties of the projections, $AE_l = \lambda_l E_l$, which mean that every non-zero vector in the range of the projection $E_l$ is an eiegenvector with eigenvalue $\lambda_l$. The above formulation of the spectral theorem for normal matrices is quite standard, and follows directly from the general spectral theorem. If you want the eigenvector-only version of the spectrum theorem, then perform Gram-Schmidt on the columns of the projection matrices $E_l$ in order to obtain a full orthonormal basis of eigenvectors for $A$.

Answer 2

Each of the orthogonal projections preserves some nontrivial subspace. Take an orthonormal basis of each of these subspaces (via Gram-Schmidt, say). As these are pairwise orthogonal projections, the union of the orthonormal bases is an orthonormal basis of $\Bbb{C}^n$.