TITLE: Diagonalization of Infinite Hermitian matrices
QUESTION [6 upvotes]: We know that $n\times n$ square Hermitian matrices can be diagonalized and have real eigenvalues. 
Suppose I have a countable sized Hermitian matrix $A=(a_{ij})$ where the indices $i$ and $j$ run over the natural numbers and the complex entries satisfy $\bar{a_{ij}}=a_{ji}$. 
Can such a matrix $A$ be diagonalized? Otherwise, what conditions can I impose that will make it diagonalizable? I'm interested in the case where only finitely many of the eigenvalues are nonzero.

REPLY [5 votes]: Such a matrix defines an unbounded operator $\mathcal A$ over the Hilbert space $\ell^2(\mathbb N)$, with domain $D(\mathcal A)$, formed of vectors $x$ such that $\mathcal Ax\in H$. It may happen of course that $\mathcal A$ be a bounded operator. The property $\overline{a_{ij}}=a_{ji}$ tells you that $\langle\mathcal Ax,y\rangle=\langle x,\mathcal Ay\rangle$ whenever $x,y\in D(\mathcal A)$. 
For an unbounded operator $\mathcal A$, we define the domain of the adjoint operator $\mathcal A^*$ to be the set of $y\in H$ such that the linear form $x\mapsto\langle\mathcal Ax,y\rangle$ is bounded.

Definition. The operator $\mathcal A$ is self-adjoint if $\langle\mathcal Ax,y\rangle=\langle x,\mathcal Ay\rangle$ whenever $x,y\in D(\mathcal A)$, and in addition $D(\mathcal A^*)=D(\mathcal A)$. 

It turns out that the symmetry property does not imply the equality of the domains. 
The theory of unitary diagonalisation of Hermitian matrices basically extends naturally to self-adjoint operators. See the book by Reed and Simon.

REPLY [2 votes]: As a comment on Denis's answer... You have to be a little bit careful about what you mean by "diagonalisation" though.  Consider $[0,1]$ with Lebesgue measure, and define $T:L^2[0,1]\rightarrow L^2[0,1]$ by $$ T(f)(x) = x f(x). $$
So $T$ is ``multiplication by $x$''.  This is easily seen to be self-adjoint.  So if we pick an orthonormal basis for $L^2[0,1]$, we get a self-adjoint matrix $A$ of the type you describe (and it induces a bounded map on $\ell^2$ of course).
But notice that $T$ has no eigenvectors or eigenvalues, and thus the same is true of $A$ (assuming we ask for our eigenvectors to be in $\ell^2$, but otherwise I don't see how we can talk about ``eigenvalue'').
Moral: you have to give up "diagonalisation" in favour of the more general idea of being "unitarily equivalent to a (real-valued) multiplication operator".