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 $\le 1$?

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    $\begingroup$ When $I-A$ is non-negative definite. $\endgroup$ – Kavi Rama Murthy Aug 13 at 8:39

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.

  • $\begingroup$ Thanks, Klaus. I had thought of Gershgorin's Theorem, but in fact for the matrix at hand the summed absolute eigenvalues may be $> 1$. $\endgroup$ – Lionel Barnett Aug 14 at 9:18
  • $\begingroup$ Sorry, I meant "summed absolute values" (in some rows). $\endgroup$ – Lionel Barnett Aug 16 at 9:49

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.

  • $\begingroup$ Thanks (and to Kavi Rama Murphy). Unfortunately, in my case (a rather arcane matrix arising in a statistical problem), showing $I-A$ PSD seems to be no easier than the original problem :-/ $\endgroup$ – Lionel Barnett Aug 14 at 9:22

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