Scalar eigenvalue stability does not control nonnormal matrix growth

The matrix is nonnormal: $A{A}^{\mathsf{T}}=\left(\begin{array}{cc}17& -4\\ -4& 1\end{array}\right)$ while $A^{\mathsf{T}}A=\left(\begin{array}{cc}1& -4\\ -4& 17\end{array}\right)$. Its two eigenvalues are $-1$, so those of $hA$ are both $-1\mathrm{/}2\in [-2,0]$ and the scalar stability factor is $1\mathrm{/}2$.

The one-stage update is $I+hA=\left(\begin{array}{cc}1\mathrm{/}2& 2\\ 0& 1\mathrm{/}2\end{array}\right)\mathrm{.}$ It sends $v=(0,1)$ to $(2,1\mathrm{/}2)$, whose Euclidean norm is $\sqrt{17}\mathrm{/}2>1=\mathrm{\parallel }v{\mathrm{\parallel }}_2$. Therefore eigenvalue-wise scalar stability does not imply Euclidean nonexpansion here. However, for $u=(1,1)$, $⟨Au,u⟩=2>0$, so this matrix also violates the nonpositivity hypothesis in [F2].

To test omission of self-adjointness while retaining nonpositivity, take $B=\left(\begin{array}{cc}-1& 2\\ 0& -1\end{array}\right)$ and $\tilde{h}=1$, still with $s=1$. For every $u=(x,y)$, $⟨Bu,u⟩=-x^2+2xy-y^2=-(x-y{)}^2\le 0$. Both eigenvalues of $B$ are $-1$, so $\rho =1$, $\tilde{h}\rho =1\le 2$, and $\tilde{h}{\lambda }_i=-1\in [-2,0]$. Yet $B^{\mathsf{T}}\mathrm{\ne }B$ and $I+\tilde{h}B=\left(\begin{array}{cc}0& 2\\ 0& 0\end{array}\right)$ sends $(0,1)$ to $(2,0)$, increasing its Euclidean norm from $1$ to $2$. Thus the self-adjoint hypothesis in [F2] cannot be omitted even when nonpositivity and the spectral step restriction are retained.