 Solution

The matrix $C=A-b{b}^T$ 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 givesThis 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 $d\ge 2$ 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.