Let $M$ be a positive definite and symmetric matrix:
$$M = \left(\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1 n}\\ a_{21} & a_{22} & \cdots & a_{2 n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \end{array}\right)$$
where $a_{ij} = a_{ji}$. Consider the band matrices $B^{(d)}$ with components $b_{i j}^{(d)} = a_{i j}$ if $|i-j| < d$ and $b_{i j}^{(d)}=0$ otherwise. Thus $B^{(d)}$ is a band matrix of "width" $d$.
For example
$$B^{(2)} = \left(\begin{array}{ccccccc} a_{11} & a_{12} & 0 & 0 & \cdots & 0 & 0\\ a_{21} & a_{22} & a_{32} & 0 & \cdots & 0 & 0\\ 0 & a_{32} & a_{33} & a_{43} & \cdots & 0 & 0\\ 0 & 0 & a_{34} & a_{44} & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & 0 & \cdots & a_{n - 1, n - 1} & a_{n - 1, n}\\ 0 & 0 & 0 & 0 & \cdots & a_{n, n - 1} & a_{n n} \end{array}\right)$$
Is $B^{(d)}$ positive definite?
Motivation: I am trying to construct a preconditioner for an optimization problem. Computing the full Hessian is computationally expensive, but I can compute a few diagonals without problems. I want to use a band approximation like this as a preconditioner, but I need to be sure it will be posdef.
The answer is NO. See https://mathoverflow.net/questions/29921/are-the-banded-versions-of-a-positive-definite-matrix-positive-definite for a countereexample.
Only $B^{(1)}$ is guaranteed to be positive definite in general.