Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 331 / 3 / b / Solution

Direct multiplication shows $L{L}^t\ne L^{t}L$ for $\beta \ne 0$, so $L$ is a non-normal matrix. MeanwhileThe symmetric part has eigenvalues $-\epsilon \pm |\beta |$, so instantaneous growth is possible exactly when $|\beta |>\epsilon$. Under the intended $0\le \beta <1$ regime this is $\boxed{\beta >\epsilon }$.