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For symmetric negative-definite matrices, put $a={\lambda }_{\max }(A)$, $b={\lambda }_{\max }(B)$ and $c={\lambda }_{\max }(A+B)\le a+b$. The spectral theorem gives exact exponential norm bounds, and $\Vert [e^{sA},B]{\Vert }_2\le 2\Vert B{\Vert }_2{e}^{sa}$. Integration in the exponential product defect identity proves the displayed estimate if $c<a+b$. At equality its continuous limit is $2t\Vert B{\Vert }_2{e}^{tc}$; the quotient must not be interpreted as division by zero.

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.