Consider the unconstrained optimization problem $$\underset{\mathbb{R}^n}{\text{min}} f(x) $$ with $$ f (x) = \frac12 x^T Qx - c^T x + \frac{1}{2\mu}(a^T x - \beta)^2 , $$
in which $c$ and $a$ are $n$ vectors, Q an $n\times n$ symmetric and positive-definite matrix, $\beta \in \mathbb{R}$, and $\mu > 0$.
The steepest-descent direction for $f$ at a point $x$ can be written $p_{sd} = -Kx + d$.
a) Determine an expression for matrix K and vector d in terms of the objects defined above.
The steepest-descent direction is found by forming $$\phi(t)=f(x_0+td),$$ where $$d=\nabla f(x_0)$$ whereof $x_0$ is some starting point for minimization. Let us consider $$f(x)=\frac12x^TQx-c^Tx+\frac{1}{2\mu}(a^Tx-\beta)^2$$ Now put $x=x_0+td$ and obtain:
$$f(x_0+td)=\frac12(x_0+td)^TQ(x_0+td)-c^Tx_0+\frac{1}{2\mu}(a^Tx_0-\beta)^2$$
Since $$\phi(t)=f(x_0+td),$$ then we have
\begin{equation} \phi(t)= \frac12(x_0+td)^TQ(x_0+td)-c^Tx_0+\frac{1}{2\mu}(a^Tx_0-\beta)^2 \end{equation} The descent direction is a vector, and can be found by differentiation the latter with respect to $t$. Hence, we have
\begin{equation} \phi'(t)= \frac12(x_0+d)^TQ(x_0+d)-c^Tx_0+\frac{1}{2\mu}(a^Tx_0-\beta)^2 \end{equation}
Since the descent direction is given by $p_{sd} = -Kx + d$, then we can solve for $K$ by equating this to the latter, and obtain:
\begin{equation} -Kx + d= \frac12(x_0+d)^TQ(x_0+d)-c^Tx_0+\frac{1}{2\mu}(a^T-\beta)^2 \end{equation}
Solving for $K$, we obtain \begin{equation} K= \frac1x\bigg(c^Tx_0+d-\frac12(x_0+d)^TQ(x_0+d)-\frac{1}{2\mu}(a^T-\beta)^2\bigg) \end{equation}
However, I am not sure I differentiated that $\phi(t)$ expression correctly, given that we have matrices and vectors in the variables. However, t is still a n-vector variable, and must be differentiated accordingly?
UPDATE:
Now we expand the expression $$\phi(t)=f(x_0+td)=\frac12(x_0+td)^TQ(x_0+td)-c^Tx_0+\frac{1}{2\mu}(a^Tx_0-\beta)^2$$, set $d=\nabla f(x_0)$ and assume linearity $f(a+b)=f(a)+f(b)$: \begin{equation} \begin{split} &\phi(t)= \frac{1}{2}\bigg(x_0 - t\nabla f(x_0) \bigg)^T\bigg(Qx_0 - Q\nabla f(x_0)t \bigg)-c^Tx_0\\&+\frac{1}{2\mu}\bigg(a^T\nabla f(x_0)-\beta\bigg)\bigg(a^T\nabla f(x_0)-\beta\bigg) \end{split} \end{equation} which gives \begin{equation} \begin{split} &\phi(t)= \frac{1}{2}\bigg(x_0Q\nabla f(x_0)+x_0Qx_0-Qx_0\nabla f(x_0) t+Q\nabla^2f(x_0)t^2\bigg) -c^Tx_0\\&+\frac{1}{2\mu}\bigg((a^T)^2 \nabla^2f(x_0)-2\beta a^T\nabla f(x_0)+\beta^2\bigg) \end{split} \end{equation} Now we differentiate with respect to $t$ and obtain: \begin{equation} \begin{split} &\phi'(t)= \frac{1}{2}\bigg(2Q\nabla^2f(x_0)t-Qx_0\nabla f(x_0)\bigg) \end{split} \end{equation} To represent $K$, we have:
$$p_{sd} = -Kx + d=\phi'(t),$$ hence, \begin{equation} -Kx + d=\frac{1}{2}\bigg(2Q\nabla^2f(x_0)t-Qx_0\nabla f(x_0)\bigg) \end{equation} which gives the following:
\begin{equation} K=\frac{1}{2x}\bigg(Qx_0\nabla f(x_0)-2Q\nabla^2f(x_0)t+2d\bigg) \end{equation} Assuming $d=-\nabla f(x_0)$, and putting $x=x_0$ we obtain: \begin{equation} K=\frac{1}{2x_0}\bigg(\nabla f(x_0)\big(Qx_0-2)-2Q\nabla^2f(x_0)t\bigg) \end{equation}
