= Solution
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> $G(t)=\|e^{tA}\|_2$, equivalently the largest <singular value> of the <matrix exponential>. Let $s(A)$ be the <spectral abscissa> and let $w(A)=\lambda_{\max}((A+A^*)/2)$ be the <numerical abscissa>. The <Euclidean semigroup growth bounds> are
$$
\boxed{e^{ts(A)}\leq G(t)\leq e^{tw(A)},\qquad t\geq0.}
$$
For 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 $\|\mathbf q\|_2^2$ gives $2\operatorname{Re}(\mathbf q^*A\mathbf q)\leq2w(A)\|\mathbf q\|_2^2$, 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 $-3/(2R)\pm\tfrac12\sqrt{1+R^{-2}}$, so
$$
\boxed{e^{-t/R}\leq G(t)\leq\exp\left[\left(-\frac{3}{2R}+\frac12\sqrt{1+\frac1{R^2}}\right)t\right].}
$$
The upper exponent becomes positive when $R>\sqrt8$, even though the long-time spectrum is stable.
The two equations are $q_1'=-q_1/R$ and $q_2'=q_1-2q_2/R$. Solve the first, then use an <integrating factor> in the second. Writing $s=t/R$, $a=e^{-s}$, $b=e^{-2s}$ and $c=R(a-b)$ gives
$$
\boxed{B(t)=e^{tM}=\begin{pmatrix}a&0\\c&b\end{pmatrix}.}
$$
The maximizing initial state is the unit <eigenvector> of $B^*B$ belonging to its largest <eigenvalue>. Here
$$
B^*B=\begin{pmatrix}a^2+c^2&cb\\cb&b^2\end{pmatrix},\qquad G(t)^2=\frac{T+\sqrt{T^2-4a^2b^2}}2,\qquad T=a^2+b^2+c^2.
$$
For $t>0$, define
$$
d=\frac{a^2+c^2-b^2}{2cb}=\frac{e^s+1}{2R}+\frac R2(e^s-1),\qquad r=\sqrt{1+d^2}-d.
$$
The <optimal initial state for triangular stable shear> is therefore
$$
\boxed{\mathbf q_{\rm opt}(0)=\frac1{\sqrt{1+r^2}}\begin{pmatrix}1\\r\end{pmatrix}.}
$$
Multiplication 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 $r\to0$. Thus \b[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> uses
$$
\boxed{r\sim\sqrt{1+\left(\frac1R+\frac t2\right)^2}-\left(\frac1R+\frac t2\right).}
$$
This retains both the unequal damping rates and the shear. If $R\gg1$, it reduces to $r\sim(\sqrt{t^2+4}-t)/2$, the maximizing direction for the shear matrix $\begin{pmatrix}1&0\\t&1\end{pmatrix}$. 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.
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