# When are the eigenvalues of a positive-definite matrix $\le 1$?

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 $$\le 1$$?

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

## 2 Answers

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.

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

• 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 :-/ – Lionel Barnett Aug 14 at 9:22