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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!

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  • $\begingroup$ What is $C(\sigma(A))$? $\endgroup$ – littleO Mar 22 '17 at 22:35
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    $\begingroup$ The continuous functions on the spectrum of $A$. The expression $f(A)$ is understood as the value of $f$ in Continuous Functional Calculus. $\endgroup$ – Steven Mar 22 '17 at 22:36
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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 $A$.

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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 $\Bbb{C}^n$.

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  • $\begingroup$ How do I know that this orthonormal basis will consist of eigenvectors of $A$? $\endgroup$ – Steven Mar 22 '17 at 22:40
  • $\begingroup$ Recover $A$ using $f:=\operatorname{id}$ and show that these spaces are $A$-invariant, don't have time to work through computation at the moment but intuition is should follow from commuting condition on $\Phi$ $\endgroup$ – Neal Mar 22 '17 at 22:45
  • $\begingroup$ Alright, let me try: We have $\text{Id} = \sum_\lambda \lambda \mathbf{1}_{\{\lambda\}}$, so we get $A = \sum_\lambda \lambda \Phi(\{\lambda\})$ by Continuous Functional Calculus and the Spectral Theorem. So the ranges of $\Phi(\{\lambda\})$ consists only of eigenvectors - this should do, right? $\endgroup$ – Steven Mar 22 '17 at 22:49

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