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Take the norm to be the Euclidean norm. The solution is $\mathbf{q}(t)=e^{tA}\mathbf{q}(0)$, so optimal growth is the operator norm $G(t)=\Vert e^{tA}{\Vert }_2$, equivalently the largest singular value of the matrix exponential. Let $s(A)$ be the spectral abscissa and let $w(A)={\lambda }_{\max }((A+A^{\ast })/2)$ be the numerical abscissa. The Euclidean semigroup growth bounds areFor the lower bound, the spectral radius of $e^{tA}$ is $e^{ts(A)}$ and is bounded by its operator norm. For the upper bound, differentiating $\Vert \mathbf{q}{\Vert }_2^2$ gives $2\mathrm{Re}({\mathbf{q}}^{\ast }A\mathbf{q})\le 2w(A)\Vert \mathbf{q}{\Vert }_2^2$, and integration gives the result. A non-normal matrix can have $w(A)>s(A)$, allowing transient amplification even when all its eigenvalues have negative real parts.

For the triangular matrix, $s(M)=-1/R$. The eigenvalues of its Hermitian part are $-3/(2R)\pm \frac{1}{2}\sqrt{1+R^{-2}}$, soThe upper exponent becomes positive when $R>\sqrt{8}$, even though the long-time spectrum is stable.

The two equations are $q_1^{\prime }=-q_1/R$ and $q_2^{\prime }=q_1-2q_2/R$. Solve the first, then use an integrating factor in the second. Writing $s=t/R$, $a=e^{-s}$, $b=e^{-2s}$ and $c=R(a-b)$ givesThe maximizing initial state is the unit eigenvector of $B^{\ast }B$ belonging to its largest eigenvalue. HereFor $t>0$, defineThe optimal initial state for triangular stable shear is thereforeMultiplication by any nonzero complex scalar gives the same growth ratio; at $t=0$, every nonzero initial state is optimal.

For $t\gg R$, $d\to \infty$ and $r\to 0$. Thus the optimal initial state tends to $(1,0{)}^T$, and $G(t)\sim e^{-t/R}\sqrt{1+R^2}$. The initial first component excites the slower eigenmode and its amplified second component.

For $t\ll R$, the exact expression gives $d=(R^{-1}+t/2)[1+O(t/R)]$. Accordingly the short-relative-time optimal state for triangular shear usesThis retains both the unequal damping rates and the shear. If $R\gg 1$, it reduces to $r\sim (\sqrt{t^2+4}-t)/2$, the maximizing direction for the shear matrix $\left(\begin{array}{cc}1& 0\\ t& 1\end{array}\right)$. If time also tends to zero at fixed $R$, it instead tends to $r=\sqrt{1+R^{-2}}-R^{-1}$, the largest-eigenvalue direction of the Hermitian part of $M$. Merely setting $t/R$ small does not justify setting $t$ small; the exact formula resolves both possibilities.

In the optimal initial state for triangular stable shear, expansion of $e^{t/R}$ gives the displayed leading parameter with relative error $O(t/R)$. Keeping both terms distinguishes a large-$R$ shear regime from the limit $t\to 0$ at fixed $R$. The latter direction is the largest-eigenvalue direction of the Hermitian part of $M$; the former can instead have large absolute time and substantial shear amplification even though $t/R$ is small.