Let $X$ be a real Banach space and $T:X\to X$ be a bounded operator. The spectral radius of $T$ is defined as $r(T)=\lim_n \|T^n\|^{1/n}$.
It is not hard to show that the spectrum of $T$ (i.e, the set $\{\lambda\in \mathbb R:T-\lambda Id \text{ is not invertible} \}$) is a subset of the interval $[-r(T),r(T)]$. When the Banach is complex, it is well-known that the spectral radius yields an "optimal bound", in the sense that there exists $\lambda$ in the complex spectrum such that $|\lambda| = r(T)$.
In the real case I would like to see an explicit example where every $\lambda$ in the real spectrum satisfies $-r(T)<\lambda<r(T)$. I only know examples where the spectrum is empty (e.g., rotation matrix by $\pi/2$).
For $-1<\lambda<1$ consider the matrix $$\begin{pmatrix} 0& 1& 0\\ -1& 0& 0\\ 0&0& \lambda\end{pmatrix}$$ The spectrum consists of $\pm i$ and $\lambda.$