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Let matrix $A \in M_n(\mathbb{R})$ be positive definite. And let $f : \mathbb{R}^n \to \mathbb{R}$ be

$$f(x) := \frac12 \langle A x, x \rangle - \langle b, x \rangle$$

where $b \in \mathbb{R}^n$. So absolutely $H f(x)=A$. Why is then $f$ elliptic?

I think $f$ should only be strictly convex , but as it's written in my course elliptic! Need some help please and thanks.


Definition: $f$ being elliptic means that for all $x \in \mathbb{R^n}$, there is $a > 0$ such that $$\langle Hf(x) h, h \rangle \ge a \|h\|^2$$ where $h \in \mathbb{R^n}$.


My proof

If $A$ positive definite, then $\langle Au, u \rangle > 0$, for all $u \ne 0$. Let $u =\frac y{\|y\|}$, then:

$$\left\langle A\frac y{\|y\|},\frac y{\|y\|} \right\rangle > 0 \Rightarrow \langle Ay, y \rangle > a \|y\|^2$$

where $a > 0$. But $Hf(x)=A$, so $f$ is elliptic. Is that true?

1 Answers1

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Yes your proof seems okay. However, the first expression should be $\langle A\frac{y}{\|y\|}, \frac{y}{\|y\|} \rangle$. I see you have used two $A$. Also, $a$ is the smallest eigenvalue.

Shiv Tavker
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  • Where did I use two A ? And how did you know that a is the smallest eigen value? Can you explain pls? – user806015 Jul 05 '20 at 08:52
  • Sorry for the mess. Do you want a proof for $a$ being the smallest eigenvalue? https://math.stackexchange.com/questions/433169/minimize-xtax-subject-to-x-1-show-that-xtax-is-the-smallest – Shiv Tavker Jul 05 '20 at 13:48