3
$\begingroup$

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.

$\endgroup$
1
  • 3
    $\begingroup$ You could try using the following: mathoverflow.net/a/89500/8430 --- in summary, if $0 < \frac{b_k^2}{a_k a_{k+1}} < \frac{1}{4}$ will ensure positive definiteness. $\endgroup$
    – Suvrit
    Feb 12, 2015 at 20:31

2 Answers 2

4
$\begingroup$

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

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

$\endgroup$
1
  • $\begingroup$ Of course I mean $2b_j$, thank you. $\endgroup$ Feb 12, 2015 at 20:45
3
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.