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For $A=\left(\begin{array}{cc}-i{\omega }_1& a\\ c& -i({\omega }_1-b)\end{array}\right)$ with real parameters,Thus the displayed conditions are necessary and sufficient for a normal matrix. In the family $a=1$, $b=c=q$, normality requires $q=-1$, not $q=1$. If $q=10/Re$, this is the formal value $Re=-10$ and no positive Reynolds number qualifies. At that value $A$ is a skew-Hermitian matrix and its exponential is unitary, so every initial direction has optimal energy amplification of a linear system equal to one.

For $L=\left(\begin{array}{cc}-1/R& 1\\ 0& -1/(4R)\end{array}\right)$ with $R>1$, put $s=\sqrt{1-R^{-2}}$ and $q=(5-3s)/(5+3s)$. The optimal energy amplification of a linear system reaches its maximum atThe maximizing initial orientation is ${\rho }^{+}$ and the final orientation is ${\rho }^{-}$ from the orientation interval for transient energy growth. Taking $R\to \infty$ gives the displayed viscous-time and Reynolds-squared asymptotics. These constants belong to this specified triangular model, not to every non-normal matrix.

For the linear evolution from part (iii), maximize the energy ratio at each time. The optimal energy amplification of a linear system isThus the optimizing initial state is a right singular vector, not generally an eigenvector of $L$.

There is a convenient exact way to find the maximizing time. Let $x_0$ be a unit optimizing initial vector and $y=M(t)x_0$. The singular value decomposition gives $M^{T}y=G_{\mathrm{opt}}{x}_0$. For $t>0$, the top singular value is simple. At an interior time maximum,Since $LM=ML$, this also equals $G_{\mathrm{opt}}{x}_0^{T}L{x}_0$. Thus both initial and final orientations are neutral energy-growth directions. The positive entries of $M^{T}M$ let the top right singular vector be chosen with positive components. Its orientation decreases through the growth sector, so it must start at ${\rho }^{+}$ and end at ${\rho }^{-}$.

Put $q=a/d=e^{-3t/(4R)}$ and $K=4R/3$. The solution givesSetting this equal to ${\rho }^{-}$ yieldsTherefore the exact optimal time for $R>1$ isHere and below $\log$ is the natural logarithm. This is the unique positive stationary time: $q$ decreases monotonically from one to zero, and the neutral-direction condition determines one $q_{\ast }\in (0,1)$. Also $G_{\mathrm{opt}}(0)=1$, $G_{\mathrm{opt}}(\infty )=0$, and $R>1$ admits initial energy growth, so this stationary point is the global maximum. For $R\le 1$, the instantaneous energy-growth criterion for a linear system gives no amplification, and the maximum is one at time zero.

A normal matrix admits unitary diagonalization of a normal matrix: $L=V\mathrm{diag}({\lambda }_j)V^{†}$ with orthonormal eigenvectors. ThusFor $t\ge 0$, the largest weight belongs to an eigenvalue with maximal real part. The weighted average is bounded by that weight and attains it on the corresponding eigenspace. Therefore the optimal energy amplification of a linear system isIf the largest real part is repeated, any nonzero combination within that maximal eigenspace is optimal. For negative times the ordering reverses; the displayed formula concerns forward evolution.

At each fixed time, the maximum over nonzero initial conditions is the optimal energy amplification of a linear system:Write $a=e^{{\lambda }_{1}t}$, $d=e^{{\lambda }_{2}t}$ and $b=(a-d)/({\lambda }_1-{\lambda }_2)$. The symmetric matrix $A^{T}A$ has trace $\tau =a^2+b^2+d^2$ and determinant $a^2{d}^2$. Its largest eigenvalue isAn optimal initial condition is a right singular vector of $A(t)$ associated with its largest singular value.

If ${\lambda }_2>{\lambda }_1$, let $\alpha =1/({\lambda }_2-{\lambda }_1)$. Then $A(t)/e^{{\lambda }_{2}t}\to \left(\begin{array}{cc}0& 0\\ \alpha & 1\end{array}\right)$, so the precise long-time asymptotic statement isThe Cauchy-Schwarz inequality shows that the unit initial conditions achieving this leading factor areThis is an adjoint eigenvector, since $L^T(\alpha ,1{)}^T={\lambda }_2(\alpha ,1{)}^T$. It differs from the right eigenvector $v_2=(0,1{)}^T$: the initial condition maximizes the projection onto the slow mode, whereas the eventual state aligns with that right eigenvector. For ${\lambda }_2<0$, even the optimal energy ultimately decays to zero; the prefactor describes enhanced excitation of the slow mode.

For $L=\left(\begin{array}{cc}{\lambda }_1& 0\\ 1& {\lambda }_2\end{array}\right)$ with distinct real negative eigenvalues, the matrix exponential is $A=\left(\begin{array}{cc}a& 0\\ b& d\end{array}\right)$, where $a=e^{{\lambda }_{1}t}$, $d=e^{{\lambda }_{2}t}$ and $b=(a-d)/({\lambda }_1-{\lambda }_2)$. The optimal energy amplification of a linear system isIts eigenvectors are nonorthogonal, and immediate energy growth occurs exactly when ${\lambda }_1{\lambda }_2<1/4$. If ${\lambda }_2>{\lambda }_1$, putting $\alpha =({\lambda }_2-{\lambda }_1{)}^{-1}$ gives $G(t)\sim (1+{\alpha }^2)e^{2{\lambda }_{2}t}$. The optimal asymptotic initial direction $(\alpha ,1)$ is an adjoint eigenvector, while the eventual state aligns with the right eigenvector $(0,1)$.