Positive definiteness of infinite tridiagonal matrices

Question

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 or the diagonal dominance sufficient condition?

Any result or reference is gladly welcome.

Answer 1

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

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

Answer 2

Here's an easy answer based on the diagonal dominance criterion.

If the entries of $A$ satisfy the inequalities \begin{equation} \begin{cases} a_j\ge b_{j-1}+b_j + c, & j\ge 2 \\ a_1\ge b_1 + c \end{cases} \end{equation} where $c\ge 0$, then the quadratic form $Q_A$ satisfies \begin{equation} Q_A(\beta_1,\beta_2,\beta_3\ldots) \ge c\sum_{j=1}^\infty \beta_j^2. \end{equation} Proof: \begin{equation} \begin{split} \sum_{j=1}^\infty a_j\beta_j^2+2b_j\beta_j\beta_{j+1} &= a_1\beta_1^2+2b_1\beta_1\beta_2 + \sum_{j=2}^\infty a_j\beta_j^2+2b_j\beta_j\beta_{j+1}\\ &\ge b_1\beta_1^2+2b_1\beta_1\beta_2+b_1\beta_2^2+\sum_{j=2}^\infty b_j\beta_{j+1}^2+b_j\beta_j^2+2b_j\beta_j\beta_{j+1} + c\sum_{j=1}^\infty\beta_j^2 \\ &=\sum_{j=1}^\infty b_j\left(\beta_j+\beta_{j+1}\right)^2+c\sum_{j=1}^\infty \beta_j^2. \end{split} \end{equation}