Consider $M$, a positive definite matrix. Let $M^{(1)}$ be the diagonal matrix which agrees with $M$ on the diagonal ($M_{ii}=M^{(1)}_{ii}$). We have that $M^{(1)}$ is positive definite because it is diagonalizable and it has non-negative eigenvalues.
What about general bands? Let $M^{(b)}$ be the restriction of $M$ on a band: $M_{ij}^{(b)}=M_{ij}$ when $i$ and $j$ differ by less than $b$ in absolute value and $M^{(b)}_{ij}=0$ for otherwise. Is $M^{(b)}$ positive definite for all $b$?
Best Answer
No. The matrix
$M = \begin{bmatrix}5 & 4 & 4 \\\\ 4 & 5 & 4 \\\\ 4 & 4 & 5\end{bmatrix} = \begin{bmatrix}2 & 2 & 2\end{bmatrix}\begin{bmatrix}2 \\\\ 2 \\\\ 2\end{bmatrix} + I$
is positive definite, but
$\begin{bmatrix}1 & -\sqrt{2} & 1\end{bmatrix}M^{(2)}\begin{bmatrix}1 \\\\ -\sqrt{2} \\\\ 1\end{bmatrix} = 20 - 16\sqrt{2}<20 - 22.4 < 0$,
so $M^{(2)}$ is not.