Conditions for an operator to be diagonalisable in a Hilbert space

Question

I have a doubt that involves Hilbert spaces and diagonalisation. So if we say that a diagonal operator in a basis can be written as, (with Dirac's notation)

$$A=\sum a_i |i\rangle \langle i|$$

In any space I know that this diagonalisation condition can be more complicated. But in a Hilbert space when a operator is expressable as I written before? Can I always write an operator in this form?

It's known that if the operator is hermitian, is diagonalisable. But as fas as I know this condition is not necessary and sufficient, and I don't know what restrctions can be imposed to matrixes in a Hilbert space.

I have not specified the dimension, so there could be some difference bethen the finite and infinite dimensions.

Answer 1

Let $ \mathcal{H}$ being be a real or complex Hilbert space, with inner product $\langle \cdot,\cdot\rangle$, and $L:\mathcal{H}\to \mathcal{H}$ being a compact linear operator.

Let us suppose that $L$ is a selfadjoint operator, that is $$\langle Lu,v\rangle=\langle u,Lv\rangle,\qquad u,\,v\in \mathcal{H}.$$ Then there exists a discrete orthonormal set $\{\phi_1,\,\phi_2,,\,\ldots\}$, and a set of scalar $\{\lambda_1,\,\lambda_2,\ldots\}$ (both sets can be finite sets, when the dimesion of $\mathcal{H}$ is finite), being such that $$L(u)=\sum_i\lambda_i\langle \phi_i,u\rangle\phi_i.$$

This results also holds if $L$ is a normal operator.

Answer 2

I find it difficult to parse your question, but I suspect that you mean to ask the following: if $A$ is a diagonalizable over a (finite dimensional) Hilbert space, then it is necessarily possible to find an orthonormal basis $\{|i\rangle\}_{i=1,\dots,n}$ such that $A = \sum_{i} a_i |i\rangle\langle i|$ for coefficients $a_i \in \Bbb C$?

The answer to this question is no. In particular, every operator of this form is normal, i.e. satisfies $AA^\dagger = A^\dagger A$. In fact, an operator is normal if and only if it can be written in this fashion for some choice of basis $\{|i\rangle\}$ an associated coefficients. On the other hand, there exist operators that are diagonalizable not normal. For example, consider the operator over $\Bbb C^2$ associated with the matrix $$ \pmatrix{1&2\\0&2}. $$

Answer 3

The spectral theorem provides a "diagonalisation" of self-adjoint operators (and more generally also for normal operators). Let $A$ be a (not necessarily bounded) self-adjoint operator on a Hilbert space $\mathcal{H}$, and $\sigma(A)$ its spectrum. The spectral theorem guarantees that a projection-valued measure $E$ exists such that $$A = \int_{\sigma(A)} \lambda \ \mathrm{d}E(\lambda).$$ If the spectrum of $A$ is discrete and consists of eigenvalues only (e.g. if $\mathcal{H}$ is finite-dimensional), then this representation reduces to the familiar formula $$A = \sum_{\lambda \in \sigma(A)} \lambda P_\lambda,$$ where $P_\lambda$ is the projection operator to the eigenspace corresponding to the eigenvalue $\lambda$.