= Solution
Direct differentiation gives
$$
A=\begin{pmatrix}0&a&0\\b&0&0\\0&0&0\end{pmatrix},\qquad p=0,\qquad s=\begin{pmatrix}0&(a+b)/2&0\\(a+b)/2&0&0\\0&0&0\end{pmatrix}.
$$
The <curl> is $(0,0,b-a)$, so the preceding convention gives
$$
\boxed{\omega=(0,0,(a-b)/2).}
$$
Put $h=(a+b)/2$. The normalized <principal axes> and <eigenvalues> of the <symmetric tensor> $s$ are
$$
\boxed{\begin{array}{c|c}
\text{axis}&\text{principal value}\\\hline
(1,1,0)/\sqrt2&h\\
(1,-1,0)/\sqrt2&-h\\
(0,0,1)&0
\end{array}.}
$$
The planar axes lie along the two angle bisectors of the coordinate axes. If $h\ne0$, one planar direction extends and the perpendicular direction contracts, with no change along the third axis. The labels need not be ordered by size when $h<0$. If $a+b=0$, then $s=0$, all principal values vanish and any orthonormal frame is a principal frame; the flow is pure local rotation in the plane.
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