3
$\begingroup$

A matrix $A \in \mathbb{R}^{n \times n}$ is called completely positive if there exists an entrywise nonnegative matrix $B \in \mathbb{R}^{n \times r}$ such that $A = BB^{T}$. All eigenvalues of $A$ are real and nonnegative.

My question is when will a completely positive matrix have all positive eigenvalues?

The only completely positive matrix I know so far have zero eigenvalues is \begin{equation} A = \begin{pmatrix} 41 & 43 & 80 & 56 & 50 \\ 43 & 62 & 89 & 78 & 51 \\ 80 & 89 & 162 & 120 & 93 \\ 56 & 78 & 120 & 104 & 62 \\ 50 & 51 & 93 & 62 & 65 \end{pmatrix} . \end{equation} So probably that the case completely positive matrix has zero eigevalues are rare. But I could not find any documment on this.

Update: due to @Robert Israel answer.

Usually, we do not know $B$ in general and indeed the composition may not unique. Therefore the condition depends only in $A$ would be easier to verify. For example with \begin{equation} A = \begin{pmatrix} 18 & 9 & 9 \\ 9 & 18 & 9 \\ 9 & 9 & 18 \end{pmatrix} \end{equation} there are at least three decompose satisfies $B \geq 0$. In particular \begin{equation} B_{1} = \begin{pmatrix} 4 & 1 & 1 \\ 1 & 4 & 1 \\ 1 & 1 & 4 \end{pmatrix} , \quad B_{2} = \begin{pmatrix} 3 & 3 & 0 & 0 \\ 3 & 0 & 3 & 0 \\ 3 & 0 & 0 & 3 \end{pmatrix} , \quad B_{3} = \begin{pmatrix} 3 & 3 & 0 \\ 3 & 0 & 3 \\ 0 & 3 & 3 \end{pmatrix} . \end{equation}

$\endgroup$
  • 1
    $\begingroup$ How about examples similar to $\begin{bmatrix}2 & 2 \\ 2 & 2\end{bmatrix} = \begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}\begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}^T$? $\endgroup$ – Neal Feb 11 at 14:49
  • $\begingroup$ Oh I see.. Thanks @Neal! So it seems that the case completely positive matrix has zero eigevalues are not rare $\endgroup$ – mortal Feb 11 at 14:51
  • 2
    $\begingroup$ I think it depends what you mean by "rare". It is not difficult to create examples, but I would not be surprised at all if Robert Israel is correct and the set of such matrices is closed nowhere dense. $\endgroup$ – Neal Feb 11 at 15:51
  • $\begingroup$ @Neal "rare" means that if I assume the matrices I consider has all positive eigenvalues is not a too restricted assumption. Like as Robert Israel said, if the set of completely positive matrices with 0 as an eigenvalue form a closed nowhere dense set in the completely positive matrices, then consider the class with all positive eigenvalues is not too restricted, I guess. $\endgroup$ – mortal Feb 12 at 11:27
4
$\begingroup$

It will have all positive eigenvalues iff $0$ is not an eigenvalue, i.e. iff it is nonsingular, and (if $B$ is also $n \times n$) this is equivalent to $B$ being nonsingular.

EDIT: You are right about the completely positive matrices with $0$ as an eigenvalue being "rare": they form a closed nowhere dense set in the completely positive matrices.

EDIT: Closed because determinant is a continuous function, so $\{A: \det(A)=0\}$ is closed.

Nowhere dense: Given $A = BB^T$ where $B$ is $n \times r$, consider $C(t) = [(1-t)B | tI]$ (i.e. the $n \times (r+n)$ matrix constructed by adjoining the columns of $(1-t) B$ and $tI$, where $I$ is the $n \times n$ identity matrix. Then for $0 \le t \le 1$, $A(t) = C(t) C(t)^T$ is a completely positive matrix; $A(0) = A$ and $A(1) = I$. Now $\det A(t)$ is a polynomial in $t$ and not identically $0$, so it is nonzero for almost all $t$. In particular, there are nonsingular completely positive matrices $A(t)$ arbitrarily close to $A$.

$\endgroup$
  • $\begingroup$ thanks for your point. However it seems not very useful since first, usually, we don't know $B$. And second, the decomposition may not unique therefore in my opinion it would be much more difficult to verify if you put some condition on $B$. I will modify my post $\endgroup$ – mortal Feb 11 at 12:43
  • 2
    $\begingroup$ If you want to test whether a particular $A$ is nonsingular, you can just take the determinant. $\endgroup$ – Robert Israel Feb 11 at 12:47
  • 1
    $\begingroup$ The hard part, AFAIK, it to find an entrywise nonnegative $B$ for a given $A$. For this I would try numerical optimization. $\endgroup$ – Robert Israel Feb 11 at 15:01
  • $\begingroup$ Yes indeed I found some algorithm to find $B$, however it require that $A$ is nonsingular. That's the reason why I asked if this situation usually happen $\endgroup$ – mortal Feb 11 at 15:18
  • $\begingroup$ "they form a closed nowhere dense set in the completely positive matrices" this a positive result for me. But could you please show me the reference about it or is it a new result? $\endgroup$ – mortal Feb 12 at 11:20
2
$\begingroup$

A nice result on the subject is given in:

Kogan, Natalia; Berman, Abraham, Characterization of completely positive graphs, Discrete Math. 114, No. 1-3, 297-304 (1993). ZBL0783.05071.

$\endgroup$
  • $\begingroup$ thank you! I will have a look on that $\endgroup$ – mortal Feb 12 at 11:28

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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