Here is a general continuity type result on perturbations of the spectral radius:

**Theorem.** Let $A,B \in \mathbb{C}^{n \times n}$ be matrices and let $r(A)$ denote the spectral radius of $A$. For each $r > r(A)$ we set
\begin{align*}
\alpha(r) := \sup_{\lvert\lambda\rvert \ge r} \lVert (\lambda - A)^{-1} \rVert
\end{align*}
(note that $\alpha(r) < \infty$). If $\lVert B \rVert < 1/\alpha(r)$, then $r(A+B) < r$.

*Proof.* This is a simple consequence of the Neumann series: Let $\mu$ be a complex number of moduls $\lvert \mu \rvert \ge r$. Then $\lVert (\mu - A)^{-1} B \rVert < 1$, so the matrix
\begin{align*}
\mu - (A+B) = (\mu - A) (I - (\mu - A)^{-1} B)
\end{align*}
is invertible since $(\mu - A)$ and $I - (\mu - A)^{-1}B$ are invertible (the latter due to the Neumann series).

The following formulation of the theorem is probably a bit easier to read:

**Corollary.** Let $A \in \mathbb{C}^{n \times n}$. For every $r > r(A)$ there exists $\delta(r) > 0$ such that $r(A+B) < r$ for every matrix $B \in \mathbb{C}^{n \times n}$ which fulfils $\lVert B \rVert < \delta(r)$.

*Proof.* Take $\delta := 1/\alpha(r)$.

**Remark 1.** If $r(A) = 1$ and if $A$ is in addition power-bounded, i.e. $M := \sup_{n \in \mathbb{N}_0} \lVert A^n \rVert < \infty$, then we have $\alpha(r) \le M/(r - 1)$ for each $r > 1$ (this is again a simple consequence of the Neumann series representation of the resolvent). Hence, in this special case we can choose $\delta(r) = (r-1)/M$ in the corollary.

**Remark 2.** The above results are very rough (as indicated by their elementary proofs). There exist more precise (and quantitative) results, as for instance indicated by Federico Poloni in the comments.