Symmetric nonnegative matrix rank-one perturbation D-stability conjecture

Counterexample in the smallest possible dimension

The conjecture is false. In fact, the asserted matrix need not even be positive stable, so diagonal scaling is unnecessary. The counterexample uses a symmetric, irreducible, entrywise nonnegative matrix of order three.

Let

$$H=\begin{pmatrix} 16&16&0\\ 16&0&4\\ 0&4&31 \end{pmatrix}, \qquad z=\begin{pmatrix}1\\1\\4\end{pmatrix}.$$

The matrix $H$ is symmetric, entrywise nonnegative, and irreducible. Direct multiplication gives

$$Hz=32z.$$

Because $z>0$, the Perron--Frobenius theorem shows that

$$\rho(H)=32,$$

and that this eigenvalue is geometrically simple, exactly as required in the conjecture.

Choose the strictly positive vectors

$$v=\begin{pmatrix}160\\ \tfrac{1}{10}\\1\end{pmatrix}, \qquad w=\begin{pmatrix}\tfrac{1}{10}\\ \tfrac{1}{10}\\40\end{pmatrix}.$$

The matrix in the conjecture is then

$$B=\rho(H)I-H+vw^{\mathsf T} =\begin{pmatrix} 32&0&6400\\ -\tfrac{1599}{100}&\tfrac{3201}{100}&0\\ \tfrac{1}{10}&-\tfrac{39}{10}&41 \end{pmatrix}.$$

Its characteristic polynomial is

$$f(\lambda)=\det(\lambda I-B) =\lambda^3-\frac{10501}{100}\lambda^2 +\frac{300873}{100}\lambda -\frac{10515528}{25}.$$

Write

$$a_1=\frac{10501}{100}, \qquad a_2=\frac{300873}{100}, \qquad a_3=\frac{10515528}{25}.$$

All three numbers are positive, but

$$a_1a_2-a_3 =-\frac{1046743827}{10000}<0.$$

Consequently,

$$f(a_1)=a_1a_2-a_3<0, \qquad \lim_{\lambda\to+\infty}f(\lambda)=+\infty,$$

so $f$ has a real root $r>a_1$. On the other hand,

$$f(\lambda)=\lambda^3-a_1\lambda^2+a_2\lambda-a_3<0 \qquad (\lambda\leq 0),$$

and hence $f$ has no nonpositive real root. If its other two roots were real, they would therefore both be positive, contradicting Vieta's identity

$$\lambda_2+\lambda_3=a_1-r<0.$$

Thus the remaining roots form a complex-conjugate pair satisfying

$$\operatorname{Re}(\lambda_2) =\operatorname{Re}(\lambda_3) =\frac{a_1-r}{2}<0.$$

Therefore $B$ is not positive stable. Taking the positive diagonal matrix $D=I$ already shows that $B$ is not $D$-stable.

An infinite family with the same symmetric matrix

The failure is not isolated. Keep the same matrix $H$ and, for every real $t\geq31$, define

$$v_t=\begin{pmatrix}4t\\4/t\\1\end{pmatrix}, \qquad w_t=\begin{pmatrix}4/t\\4/t\\t\end{pmatrix}.$$

Both vectors are strictly positive, and

$$B_t=32I-H+v_tw_t^{\mathsf T} =\begin{pmatrix} 32&0&4t^2\\ -16+16/t^2&32+16/t^2&0\\ 4/t&-4+4/t&t+1 \end{pmatrix}.$$

Writing

$$\det(\lambda I-B_t) =\lambda^3-a_1(t)\lambda^2+a_2(t)\lambda-a_3(t),$$

direct expansion gives

$$a_1(t)=\frac{t^3+65t^2+16}{t^2}, \qquad a_2(t)=\frac{16(3t^3+68t^2+t+33)}{t^2}, \qquad a_3(t)=\frac{256(t^2+2)(t^2+t+1)}{t^2}.$$

These coefficients are positive, whereas

$$a_1(t)a_2(t)-a_3(t) =-\frac{16}{t^4}Q(t),$$

where

$$Q(t) =13t^6-247t^5-4373t^4-114t^3-3201t^2-16t-528.$$

For $u\geq0$, expansion about $t=31$ gives

$$Q(31+u) =13u^6+2171u^5+144737u^4+4829624u^3 +81274304u^2+570894031u+421116864>0.$$

The same elementary characteristic-polynomial argument therefore shows that $B_t$ has two eigenvalues with strictly negative real part for every $t\geq31$. The concrete counterexample above is the instance $t=40$.

The original conjecture is Conjecture 2.16 of J. Bierkens and A. C. M. Ran, A singular M-matrix perturbed by a nonnegative rank one matrix has positive principal minors; is it D-stable?, Linear Algebra and its Applications 457 (2014), 191--208, https://doi.org/10.1016/j.laa.2014.05.022. Their Corollary 2.14 establishes the assertion in dimension two, so the dimension-three counterexample is minimal. The later article by B. Anehila and A. C. M. Ran, https://doi.org/10.2989/16073606.2021.1951871, concerns the distinct Conjecture 2.17 about large rank-one perturbations and does not settle the symmetric Conjecture 2.16 considered here.