= Solution
At a fixed point the first equation gives $\alpha S_*I_*=B$. The second gives $I_* =\alpha S_*I_*$, so $I_*=B>0$ and
$$
\boxed{(S_*,I_*)=(1/\alpha,B).}
$$
There is no disease-free fixed point: with $I=0$, susceptibles still increase by $B$ each step.
Set $r=\alpha B>0$. The <Jacobian matrix> of the update at the fixed point is
$$
M=\begin{pmatrix}1-r&-1\\r&1\end{pmatrix},\qquad
\det M=1,\quad\operatorname{tr}M=2-r.
$$
Its characteristic equation and multipliers are
$$
\lambda^2-(2-r)\lambda+1=0,\qquad
\lambda_\pm=1-\frac r2\pm\frac12\sqrt{r(r-4)}.
$$
For $0<r<4$, these are distinct complex conjugates with product one, so both have modulus one. The linearized iterates are bounded and oscillatory: \b[neutral linear stability holds for $0<\alpha B<4$], with no linear asymptotic attraction.
At $r=4$, the repeated multiplier is $-1$, but $M\ne-I$. Writing $M=-I+N$ gives $N\ne0$, $N^2=0$ and $M^n=(-1)^n(I-nN)$, so generic linearized perturbations grow linearly. For $r>4$, a real multiplier has modulus greater than one. Thus \b[the equilibrium is linearly unstable for $\alpha B\geq4$].
For the unit-modulus range, the nonlinear behavior is not determined by this calculation: <unit-modulus multipliers do not determine nonlinear stability>.
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