A <normal matrix> satisfies $AA^\dagger=A^\dagger A$. Direct multiplication, with all parameters real, gives the <normality criterion for a two-mode shear model>:
$$
AA^\dagger-A^\dagger A=
\begin{pmatrix}a^2-c^2&-ib(a+c)\\ib(a+c)&c^2-a^2\end{pmatrix}.
$$
For $a=1$, $b=c=q=10/Re$, both $1-q^2=0$ and $q(1+q)=0$ are required. Their only common solution is $q=-1$. Hence
$$
\boxed{Re_N=-10\ \text{formally};\qquad\text{no normal case exists for }Re>0.}
$$
The original PDF really has these signs and $b=c=10/Re$. Thus parts (v) and (vi) have no physical positive-Reynolds-number case as printed. We give their algebraic continuation at $Re=-10$ below, without silently changing the matrix. The tempting value $Re=10$ makes the off-diagonal magnitudes equal but fails the off-diagonal commutator condition.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 331 / 4 / b / iv / Solution

{title2=$a^2=c^2,\quad b(a+c)=0$}

For $A=\begin{pmatrix}-i\omega_1&a\\c&-i(\omega_1-b)\end{pmatrix}$ with real parameters,
$$
AA^\dagger-A^\dagger A=\begin{pmatrix}a^2-c^2&-ib(a+c)\\ib(a+c)&c^2-a^2\end{pmatrix}.
$$
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.

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