Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 80 / 6

The Euclidean matrix norm is the induced operator norm $\Vert C{\Vert }_2={\sup }_{\Vert x{\Vert }_2=1}\Vert Cx{\Vert }_2$. By the spectral theorem, a real symmetric matrix has $A=Q\Lambda Q^T$ with $Q$ orthogonal and $\Lambda =\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_m)$. HenceOrthogonal transformations preserve the Euclidean norm, and the induced norm of a real diagonal matrix is the maximum absolute diagonal entry. Therefore, for $t\ge 0$,This gives the requested inequality with equality. If an exponent $\beta <\mu (A)$ were to work with prefactor one for every $t$, choose a unit eigenvector for $\mu (A)$. Its image has length $e^{t\mu (A)}>e^{t\beta }$ for every $t>0$, a contradiction. Thus $\mu (A)$ is the smallest possible exponent.

For symmetric matrices the spectral abscissa equals the Euclidean logarithmic norm. Symmetry is essential to this particular exact norm formula; a general non-normal matrix can have transient growth not controlled with prefactor one by its spectral abscissa.

For the ordered matrix exponential product $\Phi (t)=e^{tA}{e}^{tB}$, differentiation givesThe matrix commutator here is $[X,Y]=XY-YX$. Put $E(t)=\Phi (t)-e^{t(A+B)}$. Then $E(0)=0$ andMultiplying on the left by $e^{-t(A+B)}$ gives an exact product derivative. Integration from zero to $t$, followed by multiplication by $e^{t(A+B)}$, proves the exponential product defect identity:No commutation of $A$ and $B$ was assumed. The ordered product is the usual Lie-Trotter splitting; the PDF calls this product the Beam-Warming splitting. If $[A,B]=0$, the commutator term vanishes and the product is exact. In general its small-step expansion has local error $t^2[A,B]/2+O(t^3)$.

Let $a=\mu (A)$, $b=\mu (B)$ and $c=\mu (A+B)$. The sum is also symmetric negative definite. Part (a) gives the exponential norm bounds, while the triangle inequality and submultiplicativity giveApplying these estimates to the exponential product defect identity yieldsIf $c<a+b$, evaluate the scalar integral to obtain the decaying bound for symmetric negative matrix splitting:The sign condition is natural: for every unit vector $v$, $v^T(A+B)v\le a+b$, so the Rayleigh quotient gives $c\le a+b$.

When $c=a+b$, the integrand equals $e^{tc}$ independently of $x$, givingThis also follows by taking the continuous limit of the quotient as $a+b-c\to 0$. There is no division-by-zero exception in the original integral argument. All exponents are negative, so these bounds decay for large $t$; they are coarse bounds and need not reproduce the sharper quadratic small-$t$ error.