Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 331 / 4 / a

The matrix exponential is the absolutely convergent power seriesIt converges in any finite-dimensional operator norm, since $\Vert L^m\Vert \le \Vert L{\Vert }^m$. Termwise differentiation gives $B^{\prime }=LB$, $B(0)=I$, so the solution is $q(t)=B(t)q_0$. Also $e^{Lt}{e}^{-Lt}=I$, proving invertibility without requiring $L$ itself to be invertible.

The question's expansion in eigenvectors tacitly needs an eigenbasis: an invertible matrix alone need not be a diagonalizable matrix. The exponential solution and the singular-value arguments remain valid without that assumption.

The relevant norm is the operator norm induced by the complex Euclidean norm, also called the matrix 2-norm:The maximum exists because the unit sphere in finite dimensions is compact. It obeys $\Vert Bq{\Vert }_2\le \Vert B{\Vert }_2\Vert q{\Vert }_2$ and is the norm relevant to the specified perturbation energy. Other matrix norms do not in general give this gain.

A normal matrix admits unitary diagonalization of a normal matrix: $L=V\mathrm{diag}({\lambda }_j)V^{†}$ with orthonormal eigenvectors. ThusFor $t\ge 0$, the largest weight belongs to an eigenvalue with maximal real part. The weighted average is bounded by that weight and attains it on the corresponding eigenspace. Therefore the optimal energy amplification of a linear system isIf the largest real part is repeated, any nonzero combination within that maximal eigenspace is optimal. For negative times the ordering reverses; the displayed formula concerns forward evolution.

The singular value decomposition of $B$ is $B=U\Sigma V^{†}$, where $U,V$ are unitary and $\Sigma =\mathrm{diag}({\sigma }_1,\dots ,{\sigma }_n)$, ${\sigma }_1\ge \cdots \ge {\sigma }_n>0$ since $B$ is invertible. Its right singular vector $v_j$ and left singular vector $u_j$ are the corresponding unit columns of $V$ and $U$, withEquivalently $B^{†}B{v}_j={\sigma }_j^2{v}_j$ and $B{B}^{†}{u}_j={\sigma }_j^2{u}_j$. The positive numbers ${\sigma }_j$ are the singular values. Degenerate singular subspaces allow any orthonormal choice of paired vectors.

For $q(t)=B{q}_0$, the energy ratio is the Rayleigh quotient of the Hermitian positive-definite matrix $B^{†}B$:Expanding $q_0$ in the orthonormal right singular vectors makes this a weighted average of ${\sigma }_j^2$. HenceAny nonzero vector in the largest right-singular subspace is optimal if ${\sigma }_1$ is repeated. Its amplified state is proportional to the paired left singular vector. For a non-normal matrix $L$, this optimal initial direction need not be an eigenvector of $L$, explaining transient growth from non-normal modes.