In the case where the Hessian, B, is indefinite, it is required that we find a $\lambda \in (-\lambda_1, \infty)$, s.t. $Q(\Lambda+\lambda \text{I})Q^T$ is definite, where $\lambda_1$ is the smallest eigenvalue. Since the Hessian is indefinite, the smallest value is negative, thus $(\Lambda+\lambda \text{I})$ is a positive diagonal matrix.
In the case where the Hessian is Positive Definite, the first choice for $\lambda$ for the subproblem is $0$, so that we take the full Newton step if possible, otherwise we find the minimum $\lambda$ such that the step remains within the region.
In the case of indefinite Hessian, is it possible to choose a first $\lambda$ that guarantees a solution for the optimization algorithm that doesn't push $\lambda$ outside $(-\lambda_1, \infty)$?
The function that I have under consideration is $x\mapsto \log(x^T \text{diag}(a)x +\epsilon)$, where $a_i = c^{\frac{i-1}{n-1}}$, where $c$ is a constant, $x\in \mathbb{R}^n$ with $n>1$, and $\epsilon=10^{-4}$.
I found that in some situations multiple attempts at finding an initial $\lambda$ are necessary.