Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 66 / 4

By orthogonal diagonalization of a real symmetric matrix, write $A=Q\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_d)Q^T$, with orthogonal $Q$. Its matrix exponential has the same eigenvectors and positive eigenvalues $e^{t{\lambda }_j}$. Orthogonal invariance of the induced Euclidean norm gives the exact identityThis proves the requested inequality with equality. If a real number $\beta$ gave the bound for every $t\ge 0$, evaluating on a unit eigenvector for ${\lambda }_{\max }(A)$ at any $t>0$ would give $e^{t{\lambda }_{\max }}\le e^{t\beta }$, so $\beta \ge {\lambda }_{\max }$. Thus the stated exponent is the smallest possible. For a symmetric matrix the spectral abscissa and Euclidean logarithmic norm coincide, unlike the general nonsymmetric case in Question 1.

Set $S=A+B$ and $E(t)=F(t)-e^{tS}$. Since $F(0)=I$, $E(0)=0$. Differentiate the ordered exponential products, using that each matrix commutes with its own exponential:Here the commutator convention is $[X,Y]=XY-YX$; this fixes both signs. The error satisfies $E^{\prime }=SE+D$, so the variation-of-constants formula givesThis is the symmetrized exponential-splitting defect identity. Symmetry of $A,B$ was not needed for the identity itself; it will be used to bound their exponentials in part (c).

Let $\beta =\mu [A]+\mu [B]$ and $\gamma =\mu [A+B]$, where all three matrices are symmetric. Submultiplicativity and the triangle inequality giveand similarly the other commutator term is bounded by $2\Vert B{\Vert }_2{e}^{x\beta }$. Apply part (a) also to $A+B$ in the integral from part (b). The outer factor one-half cancels these twos, leavingThe exponential divided difference evaluates this integral. Thuswhen $\beta \ne \gamma$, andwhen they coincide. The second expression is both the direct equal-exponent integral and the continuous limit of the first. The Rayleigh-Ritz variational principle also gives $\gamma \le \beta$, although the integral computation does not require a strict inequality. These are valid coarse norm bounds; the cancellation between the two products can make the actual small-step error substantially smaller.