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Let $A=(a_{ij}) \in \mathbb{R}^{n \times n}$ be a matrix with $a_{ij} = a_{ji} \geq 0$ and $B=(b_{ij})$ with $b_{ij} = \sqrt{a_{ij}}$.

Is $B$ positive-definite whenever $A$ is?

In other words:

$\sum_{1 \leq i,j \leq n} c_i c_j a_{ij} > 0 \iff \sum_{1 \leq i,j \leq n} c_i c_j \sqrt{a_{ij}} > 0$ for every choice of $c_i \in \mathbb{R}$?

A counterexample or any hint to a proof would be appreciated, thanks!

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It would be very strange. It is known that $A$ is positive-definite whenever $B$ is (more generally, the componentwise product of two positive-definite matrices is positive definite, because it is a submatrix of the Kronecker product), so we would get an if-and-only-if assertion, which would (by real algebraic geometry or something like that) mean some identities between dets, which almost certainly don't hold. – darij grinberg Jan 20 '11 at 16:12
And as a corollary to darij's observation: Not only are there counterexamples, but almost every matrix A with the required properties gives a counterexample. – Johannes Hahn Jan 20 '11 at 17:12
Uhm, I don't think so. – darij grinberg Jan 20 '11 at 18:09
Hmm, I didn't realise (before doing the calculation) that this is true for $n=2$ – Yemon Choi Jan 20 '11 at 21:50
@darij: The set where your "some identity" does not hold is a null set, open and dense subset of $\lbrace sym.matrices\rbrace = \mathbb{R}^{n(n+1)/2}$. If you intersect it with the space of all matrices with the required properties it is still a null set which is open and dense in this subspace (positive definite + positive entries is an open set in $\mathbb{R}^{n(n+1)/2}$ so there are no problems with dimensions etc.). – Johannes Hahn Jan 21 '11 at 11:30
up vote 8 down vote accepted


$B=\left(\begin{array}{ccc} 1 & 0 & 1 \\\\ 0 & 1 & 1 \\\\ 1 & 1 & \sqrt3 \end{array}\right)$.

This is for $n=3$, and easily extends to all $n\geq 3$.

For $n=2$ it holds, though.

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ah, you beat me to typing it up. (btw, I think the pmatrix environment works here, so you don't have to type up the full array.) – Willie Wong Jan 20 '11 at 16:29
you probably should also specify that the one you wrote down is $B$, the "square root" matrix for clarity's sake. – Willie Wong Jan 20 '11 at 16:31
Thank you very much for your quick answer! This is just the counterexample I need. – Christian Stahlhut Jan 21 '11 at 14:12

As a slightly longish comment, however, I would like to point out that there is a rich class of matrices, for which component-wise square root remains positive definite. The simplest example is the "unit" of this class of matrices, namely the matrix of all ones.

In general, there are several wonderful matrices, for which not only componentwise squareroot, but any root, continues remaining positive definite (in other words, the matrix $a_{ij}^\epsilon$ is positive definite too).

Simple, but instructive examples include:

  1. $a_{ij} = \frac{1}{\lambda_i + \lambda_j}$, where the $\lambda$s are nonnegative.
  2. $a_{ij} = \min(i,j)$
  3. $a_{ij} = \mbox{gcd}(i,j)$

These matrices are called infinitely divisible matrices. Please see the linked PDF for a very nice introduction to such matrices. Additional references can also be found in my related answer to an old question

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Thanks for these hints! – Christian Stahlhut Jan 21 '11 at 14:19
The link does not work! – kjetil b halvorsen Apr 27 at 17:13

In the tradition (local to this question) of pointing out special cases or related results as answers, there is the amazing theorem of I. J. Schoenberg, of which I state the simplest form: Let $a_1, \dotsc, a_{n(n-1)/2}$ be the edge lengths of a simplex in $\mathbb{R}^n.$ Then, so are $a_1^\alpha, \dotsc, a_{n(n-1)/2}^\alpha,$ for any $0\leq \alpha \leq 1.$ This theorem is really about positive-semidefinite functions in the sense of Bochner (but has a direct spectral interpretation if you use the Cayley-Menger matrix).

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Hmm.. interesting. I don't know what to make of it yet, but thank you for your comment! – Christian Stahlhut Jan 21 '11 at 14:23

Whenever $A$ is negative semi-definite on the subspace of vectors $(a_1,\dots,a_n)$ with $\sum_i a_i=0$, then the same will be true for $B$. This is a result of Schoenberg.

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