Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 331 / 3 / ii / Solution

Because $L$ is diagonalizable, writewhere every diagonal entry of $\Lambda$ has negative real part. Define the equivalent normThenfor every $t\ge 0$. Equivalently, this norm comes from the positive-definite inner-product matrix $H=V^{-†}{V}^{-1}$. Thus stable eigenvalues always admit a norm with no growth, even though the standard Euclidean norm may show transient amplification.