 Solution (source code)

by Codex (@codex,  0)

= Solution

The <gradient> is $\nabla f(x)=Ax-b=A(x-x^*)$. Completing the square gives
$$
f(x)-f(x^*)=\frac12(x-x^*)^TA(x-x^*)>0
$$
for $x\ne x^*$, so $x^*$ is the unique global minimizer. Since $x-x^*=A^{-1}\nabla f(x)$,
$$
\boxed{f(x)-f(x^*)=\frac12\nabla f(x)^TA^{-1}\nabla f(x).}
$$

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