= Solution
For the <instantaneous Lorenz error growth at a zero-y crossing>, direct differentiation of the <Lorenz system> gives
$$
J=\begin{pmatrix}-\sigma&\sigma&0\\r-z&-1&-x\\y&x&-b\end{pmatrix}.
$$
At a crossing of $y=0$ its <Symmetric matrix>[symmetric part] is particularly simple:
$$
\boxed{S=J+J^T=\begin{pmatrix}-2\sigma&\sigma+r-z&0\\
\sigma+r-z&-2&0\\0&0&-2b\end{pmatrix}.}
$$
Thus a perturbation initially in the $z$ direction has norm-squared rate $-2b$ and initially decays in <norm> at rate $b$. The $x$-dependent couplings between $y$ and $z$ are <Skew-symmetric matrix>[skew-symmetric] at the crossing and do not contribute to the instantaneous total <norm> derivative. They can rotate the perturbation later: a pure $\delta z$ immediately induces $\dot{\delta y}=-x\delta z$. It is therefore incorrect to claim that such an error remains an independently decaying $z$ mode along the complete trajectory.
For the $(x,y)$ block, the <trace> is $-2(\sigma+1)<0$ and the <determinant> is $4\sigma-(\sigma+r-z)^2$. A negative <determinant> gives one positive and one negative <eigenvalue>. Equivalently,
$$
\lambda_\pm=-(\sigma+1)\pm\sqrt{(\sigma-1)^2+(\sigma+r-z)^2},
\qquad \lambda_+>0\iff|\sigma+r-z|>2\sqrt\sigma.
$$
In particular the requested sufficient lower-height condition is
$$
\boxed{z<F(\sigma,r),\qquad F(\sigma,r)=\sigma+r-2\sqrt\sigma.}
$$
There is also a second growth range $z>\sigma+r+2\sqrt\sigma$; it must not be lost by removing the absolute value. At either endpoint the largest rate is zero. Between them both block <Eigenvalue>[eigenvalues] are negative. At a crossing in either growing range, some planar error directions have positive instantaneous exponential rate. This statement is local in time: a single instantaneous positive <eigenvalue> of $S$ does not prove sustained exponential error growth over an evolving Lorenz trajectory.
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