Completely positive matrix with positive eigenvalue

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 $$$$A = \begin{pmatrix} 41 & 43 & 80 & 56 & 50 \\ 43 & 62 & 89 & 78 & 51 \\ 80 & 89 & 162 & 120 & 93 \\ 56 & 78 & 120 & 104 & 62 \\ 50 & 51 & 93 & 62 & 65 \end{pmatrix} .$$$$ 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 $$$$A = \begin{pmatrix} 18 & 9 & 9 \\ 9 & 18 & 9 \\ 9 & 9 & 18 \end{pmatrix}$$$$ there are at least three decompose satisfies $$B \geq 0$$. In particular $$$$B_{1} = \begin{pmatrix} 4 & 1 & 1 \\ 1 & 4 & 1 \\ 1 & 1 & 4 \end{pmatrix} , \quad B_{2} = \begin{pmatrix} 3 & 3 & 0 & 0 \\ 3 & 0 & 3 & 0 \\ 3 & 0 & 0 & 3 \end{pmatrix} , \quad B_{3} = \begin{pmatrix} 3 & 3 & 0 \\ 3 & 0 & 3 \\ 0 & 3 & 3 \end{pmatrix} .$$$$

• How about examples similar to $\begin{bmatrix}2 & 2 \\ 2 & 2\end{bmatrix} = \begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}\begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}^T$? – Neal Feb 11 at 14:49
• Oh I see.. Thanks @Neal! So it seems that the case completely positive matrix has zero eigevalues are not rare – mortal Feb 11 at 14:51
• 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. – Neal Feb 11 at 15:51
• @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. – mortal Feb 12 at 11:27

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 $$A$$.

• 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 – mortal Feb 11 at 12:43
• If you want to test whether a particular $A$ is nonsingular, you can just take the determinant. – Robert Israel Feb 11 at 12:47
• The hard part, AFAIK, it to find an entrywise nonnegative $B$ for a given $A$. For this I would try numerical optimization. – Robert Israel Feb 11 at 15:01
• 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 – mortal Feb 11 at 15:18
• "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? – mortal Feb 12 at 11:20

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.

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