Is the Spectral Radius of a Primitive Matrix Bounded?

Question

Let $A$ be a primitive matrix. If I add a nonnegative matrix $\epsilon E$ ($\epsilon > 0$), does the following limit for the spectral radius exist $$\lim_{\epsilon \to \infty} \rho(A+\epsilon E)$$ or does it grow without bound? Numerical experiments for my specific matrices indicate that the spectral radius grows without bound but I want to see if there are general results. I make no assumptions on the form of matrix $E$.

Answer 1

No, it grows without a bound. To see that, first note that for nonnegative matrices Gelfand's formula implies that the spectral radius function is monotonic. That is, if $X\ge Y\ge0$, then $\rho(X)=\lim_{k\to\infty}\|X^k\|^{1/k}\ge\lim_{k\to\infty}\|Y^k\|^{1/k}=\rho(Y)$.

Suppose $E\ne0$. Since $A$ is nonnegative and primitive, $A^m$ and $A^{m+1}$ are positive for some positive integer $m$. Therefore $$ (A+\epsilon E)^{m+1}\ge P+\epsilon Q:=A^{m+1}+\epsilon A^mE\ge0. $$ $P=A^{m+1}$ is positive. Since $Q=A^mE$ is nonzero and nonnegative, it has a positive entry. If this is a diagonal entry $q_{ii}$, we look at the diagonal entry $p_{ii}+\epsilon q_{ii}$ of $P+\epsilon Q$. If this is an off-diagonal entry $q_{ij}>0$, we look at the principal $2\times2$ submatrix $(P+\epsilon Q)(\mathcal I,\mathcal I)$, where $\mathcal I=\{i,j\}$ and $$ (P+\epsilon Q)(\mathcal I,\mathcal I) \ge \pmatrix{p_{ii}&p_{ij}+\epsilon q_{ij}\\ p_{ji}&p_{jj}} \ge C := p\pmatrix{1&r\epsilon\\ 1&1} $$ for some $p,r>0$. So, when $\epsilon\to\infty$, we have $p_{ii}+\epsilon q_{ii}\to\infty$ in the first case and $\rho(C)=p(1+\sqrt{r\epsilon})\to\infty$ in the second case. The monotonicity of spectral radius thus implies that $\lim_{\epsilon\to\infty}\rho(P+\epsilon Q)=\infty$. Hence $\rho(A+\epsilon E)\ge\rho(P+\epsilon Q)^{1/(m+1)}$ is unbounded.