The method of steepest descent is a gradient algorithm where the step size $\alpha_k $ is chosen to achieve the maximum amount of decrease of the objective function at each individual step. Specifically, $\alpha_k$ is chosen to minimize $\phi_k (\alpha) \triangleq f(\b{x}^{(k)} - \alpha \nabla f(\b{x}^{(k)}))$. In other words,
$$\alpha_k = \argmin{ \alpha \geq 0} f(\b{x}^{(k)} - \alpha \nabla f(\b{x}^{k}))$$
To summarize, the steepest descent algorithm proceeds as follows:
At each step starting from the point $\b{x}^{(k)} $, we conduct a line search in the direction $- \nabla f(\b{x}^{(k)}) $ until a minimizer $\b{x}^{(k+1)}$ is found.
If $\set{ \b{x}^{(k)}}^ \infty _ {k=0} $ is a steepest descent sequence for a given function $f: \R^n \to \R $, then for each $k$ the vector $\b{x}^{(k+1)} - \b{x}^{(k)}$ is orthogonal to the vector $\b{x}^{(k + 2)} - \b{x}^{(k+1)}$.
If $\set{ x^{(k)}}^ \infty _ {k = 0} $ is the steepest descent sequence for $f: \R^n \to \R $ and if $\nabla f(\b{x}^{(k)}) \neq \b{0} $, then $f(\b{x}^{(k+1)}) < f (\b{x}^{(k)})$.
In the case for some $k $ we have $\nabla f(\b{x}^{(k)}) = 0 $, then the point $\b{x}^{(k)} $ satisfies the FONC. In this case, $\b{x}^{(k+1)} = \b{x}^{(k)}$ can be used as a stopping criterion for the algorithm. However, this criterion can rarely be met.
A practical stopping criterion is to check if $\norm{ \nabla f(\b{x}^{(k)})}$ is less than a prespecified threshold, in which case we stop. Alternatively, we may use the following criterion
$$\abs{ f(\b{x}^{(k+1)}) - f(\b{x}^{(k)})} < \epsilon$$
where $\epsilon > 0 $ is a prespecified threshold.
Yet another alternative is to compute the norm $\norm{ \b{x}^{(k+1)} - \b{x}^{(k)}}$ of the difference between two successive iterates, and we stop if the norm is less than a prespecified threshold:
$$\norm{ \b{x}^{k+1} - \b{x}^{(k)}} < \epsilon $$
$\b{Q} \in \R^{n \times n}, \b{Q} = \transpose{\b{Q}}, \b{Q} > 0$, $\b{b} \in \R ^n$, and $\b{x} \in \R^n $.
The unique minimizer of $\b{f} $ can be found by setting the $\nabla f(\b{x}) $ to zero, where
$$\nabla f(\b{x}) = \b{Q} \b{x} - \b{b}$$
since $$D (\transpose{ \b{x}} \b{Q} \b{x}) = \transpose{ \b{x}} (\b{Q} + \transpose{ \b{Q}}) = 2 \transpose{ \b{x}} \b{Q}\br D (\transpose{ b } \b{x}) = \transpose{ \b{b}}$$
Also, we have $\b{F}(\b{x}) = \b{Q} = \transpose{ \b{Q}} > 0 $.
In fact, we can generalize the matrix $\b{Q} $ to be a non-symmetric matrix $\b{A} $. In that case, we just need to replace $\b{A}$ with
$$\b{Q} = \pare{\b{A} + \transpose{ \b{A}}} $$
Then it can be covered under above case.
To simplify the notation we write $g^{(k)} = \nabla f(\b{x}^{(k)})$. Then, the steepest descent algorithm for the quadratic function can be represented as
$$\b{x}^{(k+1)} = \b{x}^{(k)} - \alpha_k g^{(k)}$$
where
$$\begin{align*} \alpha_k &= \argmin{ \alpha \geq 0 } f(\b{x}^{(k)} - \alpha \b{g}^{(k)}) \br &= \argmin{ \alpha \geq 0 } (\frac{ 1 }{ 2 } \transpose{(\b{x}^{(k)} - \alpha \b{g}^{(k)})} \b{Q} (\b{x}^{(k)} - \alpha \b{g}^{(k)}) - \transpose{(\b{x}^{(k)} - \alpha \b{g}^{(k)})}\b{b}) \end{align*}$$
In the quadratic case, we can find an explicit formula for $\alpha_k $.
We want to find a specific formula for $\alpha_k $.
Assume $\b{g}^{(k)} \neq 0 $, for if $\b{g}^{(k)} = 0 $ is a stop criterion. Because $\alpha_k \geq 0 $ is a minimizer of $\phi_k (\alpha) = f(\b{x}^{(k)} - \alpha \b{x}^{(k)}) $, we apply the FONC to $\phi_k(\alpha)$ to obtain
$$\phi'_k(\alpha) = \transpose{(\b{x}^{(k)} - \alpha \b{g}^{(k)})} \b{Q} (- g^{(k)}) - \transpose{ \b{b}}(-\b{g}^{(k)}) = 0 $$
Hence,
$$$alpha_k= \frac{ \transpose{ \b{g}^{(k)}} \b{g}^{(k)}}{ \transpose{ \b{g}^{(k)}} \b{Q} \b{g}^{(k)}}$
The method of steepest descent for the quadratic takes the form
$$\b{x}^{(k+1)} = \b{x}^{(k)} - \pare{\frac{ \transpose{ \b{g}^{(k)}} \b{g}^{(k)}}{ \transpose{ \b{g}^{(k)}} \b{Q} \b{g}^{(k)}}} \b{g}^{(k)} $$