Let $A$ be a positive matrix. $B$ is a small perturbation of $A$, and $B$ is still a positive matrix.
By Perron-Frobenius Theorem, it is known that $r(A)$ and $r(B)$ are algebracially simple eigenvalue of $A$ and $B$.
Here, $r(A)$ is the spectral radius of $A$.
Is there an estimate between $r(A)$ and $r(B)$ having the following type?
$$\vert r(A) – r(B)\vert \leq C \Vert A-B\Vert $$
for some $C$.
Best Answer
Take for example $$ A = \begin{pmatrix} 1&1&1\\ 1&1&1\\ 1&3&2\\ \end{pmatrix} \quad E = \begin{pmatrix} 0&1&0\\ 0&0&0\\ 0&0&0\\ \end{pmatrix} $$ and $B_\delta = A+\delta E$.
You get (after lot of computations) that $r(A)=4$ and $r(B_\delta) = 4 + O(\sqrt{\delta})$ for $\delta$ converging to zero, so $$|r(A)-r(B_\delta)|= O(\sqrt{\delta}) > C\delta = C\|B_\delta-A\|$$ for every constant $C$ for $\delta$ small enough.