TITLE: Naive questions about "matrices" representing endomorphisms of Hilbert spaces.
QUESTION [15 upvotes]: This is a very basic question and might be way too easy for MO. I am learning analysis in a very backwards way. This is a question about complex Hilbert spaces but here's how I came to it: I have in the past written a paper about (amongst other things) compact endomorphisms of $p$-adic Banach spaces (and indeed of Banach modules over a $p$-adic Banach algebra), and in this paper I continually used the notion of a "matrix" of an endomorphism as an essential crutch when doing calculations and proofs. I wondered at the time where more "conceptual" proofs existed, and probably they do, but I was too lazy to find them.
Now I find myself learning the basic theory of certain endomorphisms of complex separable Hilbert spaces (continuous, compact, Hilbert-Schmidt and trace class operators) and my instinct, probably wrong, is to learn the theory in precisely the same way. So this is the sort of question I find myself asking.
Say $H$ is a separable Hilbert space with orthonomal basis $(e_i)_{i\in\mathbf{Z}_{\geq1}}$. Say $T$ is a continuous linear map $H\to H$. Then $T$ is completely determined by its "matrix" $(a_{ij})$ with $Te_i=\sum_ja_{ji}e_j$. But are there completely "elementary" conditions  which completely classify which collections of complex numbers $(a_{ij})$ arise as "matrices" of continuous operators?
I will ask a more precise question at the end, but let me, for the sake of exposition, tell you what the the answer is in the $p$-adic world.
In the $p$-adic world, $\sum_na_n$ converges iff $a_n\to 0$, and life is easy: the answer to the question in the $p$-adic world is that $(a_{ij})$ represents a continuous operator iff 
(1) For all $i$, $\sum_j|a_{ji}|^2<\infty$ (equivalently, $a_{ji}\to 0$ as $j\to\infty$), and 
(2) there's a universal bound $B$ such that $|a_{ij}|\leq B$ for all $i,j$.
[there is no inner product in the $p$-adic case, so no adjoint, and the conditions come out being asymmetric in $i$ and $j$].
See for example pages 8--9 of this paper of mine, although of course this isn't due to me---it's in Serre's paper on compact operators on $p$-adic Banach spaces from the 60s---see Proposition 3 of Serre's paper. In particular, in the $p$-adic world, one can identify the continuous maps $H\to H$ (here $H$ is a $p$-adic Banach space with countable ON basis $(e_i)$) with the collection of bounded sequences in $H$, the identification sending $T$ to $(Te_i)$.
In the real/complex world though, the analogue of this result fails: the sequence $(e_1,e_1,e_1,\ldots)$ is a perfectly good bounded sequence, but there is no continuous linear map $H\to H$ sending $e_i$ to $e_1$ for all $i$ (where would $\sum_n(1/n)e_n$ go?). 
Let's consider the finite rank case, so $T$ is a continuous linear map $H\to H$ with image landing in $\mathbf{C}e_1$. Then by Riesz's theorem, $T$ is just "inner product with an element of $H$ and then multiply by $e_1$". Hence we have an additional condition on the $a_{ij}$, namely that $\sum_j|a_{ij}|^2<\infty$. Furthermore a continuous linear map is bounded, as is its adjoint.
This makes me wonder whether the following is true, or whether this is still too naive:
Q) Say $(a_{ij})$ $(i,j\in\mathbf{Z}_{\geq1})$ is a collection of complex numbers satisfying the following:
There is a real number $B$ such that
1) For all $i$, $\sum_j|a_{ij}|^2\leq B$
2) For all $j$, $\sum_j|a_{ij}|^2\leq B$
Then is there a unique continuous linear map $T:H\to H$ with $Te_i=\sum_ja_{ji}e_i$?
My guess is that this is still too naive. Can someone give me an explicit counterexample? Or, even better, a correct "elementary" list of conditions characterising the continuous endomorphisms of a Hilbert space?
On the other hand, it clearly isn't a complete waste of time to think about matrix coefficients. For example there's a bijection between Hilbert-Schmidt operators $T:H\to H$ and collections $(a_{ij})$ of complexes with $\sum_{i,j}|a_{ij}|^2<\infty$, something which perhaps the experts don't use but which I find incredibly psychologically useful.

REPLY [5 votes]: So maybe the following is a counterexample to Kevin's original post. (It was created by computer scientist and friend Erik Vee as a "counterexample" to that exercise 3.14 in Zimmer (which says that a bounded operator is compact iff $a_{ij}$ goes to 0 as i and j go to $\infty$); and it was Robert Pollack who figured out that the reason it's not a counterexample is that it doesn't represent a continuous operator. So my role here is transcriber only.)
Define the matrix as follows. The first column is (1, 0, 0, ...), the second is (1/2, 1/2, 1/2, 1/2, 0, 0, ...), the nth has $\frac 1 n$ appearing $n^2$ times, followed by all zeros. Then $\ell_2$-norm of each column is exactly 1, and the $\ell_2$-norm of each row is bounded by $\sqrt{\frac{\pi^2}{6}}$. So this matrix has bounded $\ell_2$ rows and columns as necessary. 
But this matrix cannot represent a continuous operator. If it did, then, since it satisfies Kevin's/Zimmer's criterion, this operator -- call it $A$ -- would be compact, and hence a uniform limit of the operators $A_n$ given by the first n rows of $A$. But the operator $A - A_n$ has, in its nth column, $\frac 1 n$ appearing $n^2 - n$ times, which means that that column's $\ell_2$ norm is $1 - \frac 1 n$, which is bounded away from zero.
It's still unclear to me if this matrix represents an unbounded linear map, or if it doesn't represent a well-defined map at all.