Because $L$ is diagonalizable, write
$$
L=V\Lambda V^{-1},
$$
where every diagonal entry of $\Lambda$ has negative real part. Define the equivalent norm
$$
\boxed{\|x\|_V=\|V^{-1}x\|_2}.
$$
Then
$$
\|e^{tL}x\|_V
=\|e^{t\Lambda}V^{-1}x\|_2
\leq\|V^{-1}x\|_2
=\|x\|_V
$$
for every $t\geq0$. Equivalently, this norm comes from the positive-definite inner-product matrix $H=V^{-\dagger}V^{-1}$. Thus stable eigenvalues always admit a norm with no growth, even though the standard Euclidean norm may show transient amplification.


Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-331/3/ii/solution