The eigenvalues of a positive-definite matrix are guaranteed to be $> 0$; but does anyone know of sufficient conditions when they will also all be $\le 1$?
There are certainly many possible answers to this question, including the obvious one given by @Kavi Rama Murthy. Here is a slightly less obvious one: Gershgorin's circle theorem implies that if the sum of the absolute values of the entries in each row does not exceed $1$, then the eigenvalues are all below $1$.
Note that this estimate is usually rather rough, though.
I just want to expand on @ Kavi Rama Murphy's comment, to show it's iff.
If $A$ is a Positive Definite matrix then let $T$ be the associated linear operator, and $V$ an inner product space, with an inner product induced norm $||.||$.
Since $T$ is Positive Definite, all of its eigenvalues are $> 0$. In addition, since it is positive definite, it is self-adjoint. SVD shows that $\forall v\in V, ||Tv|| \leq s^*||v||$, where $s^*$ is the maximum singular value. Since $T$ is self-adjoint AND positive definite, the singular values equal the eigenvalues. If by hypothesis all eigenvalues are $\leq 1$. Thus so $I - T$ or $I - A$ is PSD. The other direction (in the comment), should be clear.