 Solution

For real $\alpha ,\beta$, the equilibrium equations are $-{\beta }^{2}x-\alpha z=0$, $y=0$ and $\alpha x-({\beta }^2+1)z=\beta$. The determinant of the coefficient matrix is $-D$, where $D={\alpha }^2+{\beta }^2({\beta }^2+1)>0$ whenever $(\alpha ,\beta )\ne (0,0)$. Solving the two-by-two system, including cases with one parameter zero, givesSubstitution verifies the first equation by cancellation of $-\alpha {\beta }^3$ and $+\alpha {\beta }^3$, and the third because ${\alpha }^2\beta +{\beta }^3({\beta }^2+1)=\beta D$. The nonzero determinant proves uniqueness. Its squared length is ${\beta }^2({\alpha }^2+{\beta }^4)/D^2$, andThus $\Vert s_{\ast }\Vert \le 1/2$, which is inside the qubit physical Bloch vector body; in the Hilbert-Schmidt normalization of part (b), that body has radius $1/\sqrt{2}$.

The $y$ mode has eigenvalue $-1$. The $xz$ block has characteristic polynomial $(\lambda +{\beta }^2)(\lambda +{\beta }^2+1)+{\alpha }^2$, so the other eigenvalues areIf $|\alpha |>1/2$, they have real part $-{\beta }^2-1/2<0$. If $|\alpha |\le 1/2$, the square root is at most $1/2$, with equality only when $\alpha =0$; the larger eigenvalue is still negative unless $\alpha =\beta =0$. At $|\alpha |=1/2$ an eventual Jordan factor does not change decay because its eigenvalue is negative. Therefore for every allowed parameter pair all eigenvalues have negative real part, and $\boxed{s_{\ast }\text{ is globally attractive}}$.