$H(\lambda ^*)$ must be positive definite.

The solution we are seeking lies either interior to the trust region ( $\Vert s \Vert _2 < \Delta$) or on the boundary.
If it lies on the interior, the trust region may as well not have been there and therefore $s^*$ is the unconstrained minimizer of $q(s)$. We have seen in Equation 2.4 ($H s = -g$) how to find it. We have seen in Equation 2.6 that

$$H \quad ( s^T H s > 0 \quad \forall s)$$ (4.2)

in order to be able to apply 2.4.
If we found a value of $s^*$ using 2.4 which lies outside the trust region, it means that $s^*$ lies on the trust region boundary. Let's take a closer look to this case:
Theorem 1

First, we rewrite the constraints $\Vert s \Vert _2 = \Delta$ as $$c(s)=\frac{1}{2}\Delta-\frac{1}{2}\Vert s \Vert _2=0$$. Now, we introduce a Lagrange multiplier $\lambda$ for the constraint and use first-order optimality conditions (see Annexe, section 13.3 ). This gives:

$$\L (s,\lambda)=q(s)-\lambda c(s)$$ (4.3)

Using first part of Equation 13.22,we have

$$\nabla_s \L (s^*,\lambda^*)= \nabla q(s^*)-\lambda^* \nabla_s c(s^*)= H s^* + g + \lambda^* s^* = 0$$ (4.4)

which is 4.3.
We will now proof that $H(\lambda ^*)$ must be positive (semi)definite.
Suppose $s^F$ is a feasible point ( $\Vert s^F \Vert = \Delta$), we obtain:

$$q(s^F)=q(s^*)+\langle s^F-s^*, g(s^*)\rangle+ \frac{1}{2} \langle s^F-s^*, H (s^F-s^*)\rangle$$ (4.5)

Using the secant equation (see Annexes, Section 13.4), $g''-g'=H(x''-x')= H s \quad (s=x''-x')$, we can rewrite 4.5 into $g(s^*)= -\lambda^* s^*$. This and the restriction that $(\Vert s^F \Vert = \Vert s^* \Vert = \Delta)$ implies that:

$$\langle s^F-s^*, g(s^*)\rangle = \langle s^*-s^F, s^*\rangle \lambda^*$$

$$= (\Delta^2- \langle s^F,s^*\rangle) \lambda^*$$

$$= [\frac{1}{2}(\langle s^F,s^F\rangle+\langle s^*,s^*\rangle)-\langle s^F,s^*\rangle] \lambda^*$$

$$= [\frac{1}{2}(\langle s^F,s^F\rangle+\langle s^*,s^*\rangle)- \frac{1}{2}\langle s^F,s^*\rangle-\frac{1}{2}\langle s^F,s^*\rangle] \lambda^*$$

$$= \frac{1}{2}[\langle s^F,s^F\rangle-\langle s^F,s^*\rangle+\langle s^*,s^*\rangle-\langle s^F,s^*\rangle] \lambda^*$$

$$= \frac{1}{2}[\langle s^F,s^F-s^*\rangle+\langle s^*-s^F,s^*\rangle] \lambda^*$$

$$= \frac{1}{2}[\langle s^F,s^F-s^*\rangle-\langle s^*,s^F-s^*\rangle] \lambda^*$$

$$\langle s^F-s^*, g(s^*)\rangle = \frac{1}{2}\langle s^F-s^*,s^F-s^*\rangle \lambda^*$$ (4.6)

Combining 4.6 and 4.7

$$q(s^F)= q(s^*)+\frac{1}{2}\langle s^F-s^*,s^F-s^*\rangle\lambda^*+\frac{1}{2}\langle s^F-s^*, H (s^F-s^*)\rangle$$

$$= q(s^*)+\frac{1}{2}\langle s^F-s^*,( H+\lambda^* I )(s^F-s^*)\rangle$$

$$= q(s^*)+\frac{1}{2}\langle s^F-s^*, H(\lambda^*) (s^F-s^*)\rangle$$ (4.7)

Let's define a line $s^*+ \alpha v$ as a function of the scalar $\alpha$. This line intersect the constraints $\Vert s\Vert=\Delta$ for two values of $\alpha$: $\alpha=0$ and $\alpha=\alpha^F\neq 0$ at which $s=s^F$. So $s^F-s^*=\alpha^F v$, and therefore, using 4.8, we have that

$$q(s^F)= q(s^*)+\frac{1}{2}(\alpha^F)^2 \langle v,H(\lambda^*) v \rangle$$

Finally, as we are assuming that $s^*$ is a global minimizer, we must have that $s^F\geq s^*$, and thus that $\langle v,H(\lambda^*) v \rangle \geq 0 \quad \forall v$, which is the same as saying that $H(\lambda)$ is positive semidefinite.
If $H(\lambda ^*)$ is positive definite, then $\langle s^F-s^*, H(\lambda^*) (s^F-s^*)\rangle >0$ for any $s^F \neq s^*$, and therefore 4.8 shows that $q(s^F)>q(s^*)$ whenever $s^F$ is feasible. Thus $s^*$ is the unique global minimizer.
Using 4.2 (which is concerned about an interior minimizer) and the previous paragraph (which is concerned about a minimizer on the boundary of the trust region), we can state:
Theorem 2:

The justification of $\lambda^* ( \Vert s^* \Vert - \Delta ) =0$ is simply the complementarity condition (see Section 13.3 for explanation, Equation 13.22).
The parameter $\lambda$ is said to ``regularized'' or ``modify'' the model such that the modified model is convex and so that its minimizer lies on or within the trust region boundary.