I have this exercise that I really don't know how to solve it. Can someone help me?
Let $n ∈ \mathbb{N}$, $Q ∈ \mathbb{R}^{n×n}$ symmetric and positive definite, $b ∈ \mathbb{R}^n$, $c ∈ \mathbb{R}$ and let $F : \mathbb{R}^n → \mathbb{R}$ be given by
$F(x) = x^TQx + b^T x+c.$
Assume that d$ ∈ \mathbb{R}^n$ is descent direction of $F$ at $x ∈ \mathbb{R}^n$ and prove that
$α^∗:=\frac{∇F(x)d}{d^TQd}$
satisfies $F(x + α^∗d) = \min_α F(x + αd)$.
I think you're missing a minus sign in the expression of $\alpha$.
Assuming that your functional is $F(x)=x^T Qx+b^T x+c$
Let $\alpha^\ast$ satisfy $F(x+\alpha^\ast d)=\min_\alpha F(x+\alpha d)$, this means that the function of 1 variable $$\varphi(\alpha):=F(x+\alpha d)$$
has a minimim at $\alpha=\alpha^\ast$, since $\varphi$ is differentiable it follows that
$$\label{eq1}\tag{1}0=\varphi'(\alpha^\ast)=\frac{d}{d\alpha}F(x+\alpha d)\Big|_{\alpha=\alpha^\ast}=\nabla F(x+\alpha^\ast d)\cdot d$$
The gradient of $F$ equals $\nabla F(x)=2Qx+b$, therefore, substituting in \eqref{eq1}:
$$0=\left[2Q(x+\alpha^\ast d)+b\right]\cdot d=\nabla F(x)\cdot d+2\alpha^\ast Q(d)\cdot d,$$
getting
$$\alpha^\ast =-\frac{1}{2}\frac{\nabla F(x)\cdot d}{d\cdot Qd}$$
I think that your functional should look like $F(x)=\frac{1}{2}x^T Qx-b^Tx+c$, since in this way minimizing $F$ is equivalent to solving the linear system $Qx=b$, infact in this case
$$\nabla F(x)=Qx-b,$$
and, following the same steps as before, you get
$$\left[Q(x+\alpha^\ast d)-b\right]\cdot d=0\implies \nabla F(x)\cdot d+\alpha^\ast Qd\cdot d=0\implies \alpha^\ast=\frac{-\nabla F(x)\cdot d}{d\cdot Qd}$$