0
$\begingroup$

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 $\le 1$?

$\endgroup$
  • 8
    $\begingroup$ When $I-A$ is non-negative definite. $\endgroup$ – Kavi Rama Murthy Aug 13 at 8:39
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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