A very basic question about unconstrained quadratic program

Question

In an unconstrained quadratic program given below $$\min.~~ \frac{1}{2}x^TAx+b^Tx$$ why the optimal value is $-\infty$ if $b$ is not in $R(A)$?

Answer 1

Assume $b\notin R(A)$. It is well-known that we can decompose ${\mathbf R}^n$ as $$ {\mathbf R}^n = N(A^T)\oplus R(A). $$ (It is easy to see that $N(A^T)\subseteq R(A)^\perp$ and both spaces have the same dimension.)

In particular, we can decompose $$ b = z + Ac, $$ with $A^Tz=0$. Since $b\notin R(A)$, it must be $z\ne0$.

Now take $\lambda\in\mathbf R$. For $x=\lambda z$ we have \begin{align*} \frac{1}{2}x^TAx + b^Tx &= \frac{\lambda^2}{2}z^TAz + \lambda b^Tz\\ &= \lambda b^Tz &&;z^TA=0\\ &= \lambda z^Tz + \lambda c^TA^Tz &&;b = z + Ac\\ &= \lambda z^Tz &&;A^Tz=0, \end{align*} which shows that the quadratic form has no upper or lower bound.