I am interested in the following problem: I have an infinite symmetric tridiagonal matrix 
$$
A=
\begin{bmatrix}
a\_1 & b\_1 & & & \\
b\_1 & a\_2 & b\_2 & & \\
 & b\_2& a\_3 & b\_3 & \\ 
 & & \ddots & \ddots & \ddots & \\
\end{bmatrix}
$$
where $a\_j, b\_j>0$, and I need to determine whether $A$ is *positive definite*, meaning that the corresponding quadratic form is bounded below:
$$
Q\_A(\beta\_1, \beta\_2\ldots \beta\_n\ldots)\stackrel{\mathrm{def}}{=}\sum\_{j=1}^\infty a\_j \beta\_j^2 + 2b\_j\beta\_{j}\beta\_{j+1} \ge c\sum\_{j=1}^\infty \beta\_j^2.$$
Here $c>0$. (If $c=0$, we say that $A$ is positive *semidefinite*).



> 
> **Question** Are there infinite-dimensional versions of the familiar criterions of linear algebra, such as the [Sylvester's criterion](http://en.wikipedia.org/wiki/Sylvester%27s_criterion) or the [diagonal dominance sufficient condition](http://en.wikipedia.org/wiki/Diagonally_dominant_matrix)? 
> 
> 
> 


Any result or reference is gladly welcome.



Positive definiteness of infinite tridiagonal matrices
Presumably you mean $2 b\_j$, not $b\_j/2$. 


The appropriate context for this is
linear operators on $\ell^2$. I'll just consider the case where the $a\_j$ and $b\_j$ are bounded, which makes $A\_\infty$ correspond to a bounded self-adjoint 
linear operator $A$ on $\ell^2$.


If $P\_n$ is the orthogonal projection on 
the span of the first $n$ standard unit vectors, Sylvester's criterion
essentially says that $P\_n A P\_n$ should be positive definite for all $n$.
This does imply that $A$ is positive **semidefinite**, but it is not necessarily
(strictly) positive definite: it may have a null space containing a 
vector with infinitely many nonzero entries. For example, try


$$ A\_\infty = \left[ \matrix{ 1/2 & 1 & & & \cr
 1 & 5/2 & 1 & &\cr
 & 1 & 5/2 & 1 & \cr
 & & \ldots & \ldots & \ldots \cr}\right],
\ v = \left[\matrix{ 1\cr -1/2\cr 1/4\cr \ldots}\right] $$


