In their book An Introduction to Optimization, on the chapter on gradient algorithms, to prepare for discussing convergence properties of the descent methods, authors Chong and Zak have following:
$f(x) = \frac{1}{2}x^TQx-b^Tx$, where $Q$ is symmetric and $Q>0$.
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$V(x)=f(x)+\frac{1}{2}(x^*)^TQx^*=\frac{1}{2}(x-x^*)^T Q (x-x^*)$, where $x^*$ is the solution point obtained by solving $Qx=b$, that is, $x^*=Q^{-1}b$
I could not follow the equation $f(x)+\frac{1}{2}(x^*)^TQx^*=\frac{1}{2}(x-x^*)^T Q (x-x^*)$. Can you prove it?
\begin{align} V(x)=&f(x)+\frac{1}{2}(x^*)^TQx^* \\ =&\frac12 x^TQx-b^Tx+ \frac{1}{2}(x^*)^TQx^* \\ =&\frac12 x^TQx-(Qx^*)^Tx+ \frac{1}{2}(x^*)^TQx^* \\ =&\frac12 x^TQx-(x^*)^TQx+ \frac{1}{2}(x^*)^TQx^* \quad(\because Q^T=Q)\\ =&\frac12 x^TQx-\frac12(x^*)^TQx-\frac12(x^*)^TQx+ \frac{1}{2}(x^*)^TQx^*\\ =&\frac12 x^TQx-\frac12x^TQx^*-\frac12(x^*)^TQx+ \frac{1}{2}(x^*)^TQx^* \quad(\because Q^T=Q)\\ =& \frac{1}{2}x^T Q (x-x^*) - \frac{1}{2}(x^*)^T Q (x-x^*) \\ =&\frac{1}{2}(x-x^*)^T Q (x-x^*) \end{align}