= Solution
The <matrix> $C=A-bb^T$ is an <invertible matrix>. Indeed, on the line spanned by $b$ it has <eigenvalue> $-1/2$, and on $b^\perp$ it agrees with the positive-definite restriction of $A$. Therefore the full <Newton step> is well defined and gives
$$
\boxed{x_1=x_0-(2C)^{-1}(2Cx_0)=0.}
$$
This is the unique <stationary point> because $C$ is an <invertible matrix>. The algebra applies to every starting point, but it does not make the answer a minimum: for $d\geq2$ it is the <saddle point> just classified, and for $d=1$ it is a maximum. A safeguarded minimization algorithm that replaces an <indefinite matrix> used as its <Hessian matrix> or damps the step need not take this exact step; the assertion concerns the ordinary full <Newton step> formula.
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