I have a feeling this is a very easy question but I can't figure out what I'm missing so any help would be appreciated.
Let's say we have a quadratic form $$ f(x) = \frac{1}{2} x^T A x - b^T x$$ where $A$ is a $n \times n$ symmetric and positive definite matrix and $x,b$ are $n \times 1$ vectors.
If we introduce the weighted norm as $$\| x\|^2_A = x^T A x$$ and $x^*$ is the solution of the system $Ax=b$, then I want to prove that $$ \frac{1}{2} \| x-x^*\|^2_A = f(x) - f(x^*)$$
What I've done so far:
$$\frac{1}{2}\|x-x^*\|^2_A \\= \frac{1}{2}(x-x^*)^T A (x-x^*) \\=\frac{1}{2}(x^TAx -x^TAx^* - x^{*T}Ax+x^{*T}Ax^*) \\ = \frac{1}{2}(x^T Ax -b^Tx - b^Tx + x^{*T}Ax^*) \\= \frac{1}{2}x^TAx - b^Tx + \frac{1}{2}x^{*T}Ax = \\f(x) + \frac{1}{2} x^{*T} Ax^*$$
I'm missing a term $-b^Tx^*$ to complete the proof but no matter how I look at it I can't find where I've made a mistake, if there is one.
Thanks in advance.
$\frac{1}{2} x^{*T} Ax^* = -(\frac{1}{2} x^{*T} Ax^* - x^{*T} Ax^*) = -(\frac{1}{2} x^{*T} Ax^* - x^{*T} b) = -(\frac{1}{2} x^{*T} Ax^* - b^{T} x^*) = -f(x^*)$