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Let $s^*=\mathrm{inf}\Bigl\lbrace{\frac{1}{2}x^TMx+v^Tx+w:x \in \mathbb{R}^n\Bigr\rbrace}$ be a minimization problem with $M \in \mathbb{R}^{n \times n}$ symmetric, $v \in \mathbb{R}^n$ and $w \in \mathbb{R}$.

How to show:

If $M$ is positive semidefinite and $Mx^*=-v$ has no solution, it's $s^*=-\infty$.

My idea was:

Let $f(x)=\frac{1}{2}x^TMx+v^Tx+w$.

It's $\nabla f(x)=Mx+v$, so there is no maximum or minimum point which means it has to be unbounded. But I don't see why it's $s^*=-\infty$.

Is this way corrrect or how can it be shown properly?

Tartulop
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    Write $v=v_1+v_2$ where $v_1 \in \ker M$ and $v_2 \in ({\cal R}M)^\bot$. Then consider $f(-tv_1)= -t |v_1|^2 +w$. You need to understand and justify these steps, of course. – copper.hat Oct 13 '22 at 18:04
  • What happens with $$M = \left( \begin{array}{cc} 1&0 \ 0&0 \ \end{array} \right) $$ and $v = \left( \begin{array}{c} 0 \ 1 \ \end{array} \right) $ and $w = 17 ; ; ; ? $ – Will Jagy Oct 13 '22 at 19:21
  • @copper.hat What do you mean with $\mathcal{R}$? – Tartulop Oct 16 '22 at 14:22
  • It is the range space of the following matrix. – copper.hat Oct 16 '22 at 14:32

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