Generalization of "symmetric positive definite" to higher dimensions?

Asked Jan 13 '22 at 22:28

Active Jan 14 '22 at 10:48

Viewed 214 times

Suppose I have a positive-definite function $f:\mathbb{R}^n\to \mathbb{R}^1$.

Then the Hessian $\nabla^2 f$ is symmetric positive definite and I can write $\nabla^2 f=QQ^T$ for some real-valued matrix $Q$. It allows me to write $q'Hq$ as $\|u\|^2$ with $u=Q^Tq$

Is there an analogue to "symmetric positive definite" for $\nabla^4 f$ and a corresponding decomposition? IE, something that would allow me to evaluate the quartic form efficiently, for instance, as $\|u\|^4$

edited Jan 14 '22 at 10:48

Rodrigo de Azevedo

23,223

asked Jan 13 '22 at 22:28

Yaroslav Bulatov

5,579

tangentially related: here (And in general $\nabla^2 f$ is not positive definite) – Arctic Char Jan 13 '22 at 22:38
@ArcticChar good point, I've been looking at loss functions like cross-entropy loss for so long I forgot other functions exist. Edited to specify "positive definite" function – Yaroslav Bulatov Jan 13 '22 at 22:50
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This reminds me of a previous question of mine, which received an interesting comment of Rodrigo de Azevedo. P.S. Interestingly I wrote this comment simultaneously with Rodrigo. – Giuseppe Negro Jan 13 '22 at 23:39
A $p$-linear symmetric form is called positive definite if $\varphi(x^{(p)}) > 0$ for all $x \neq 0$ and $(x_1, \ldots, x_p) \mapsto \varphi(x_1, \ldots, x_p)$ is linear in each entry. – William M. Jan 13 '22 at 23:56
@WilliamM. What's $x^{(p)}$? – Arctic Char Jan 14 '22 at 11:15
@ArcticChar $x^{(p)}$ is short-hand for $(x, \ldots, x) \in \mathrm{V}^p,$ and $\mathrm{V}^p$ is the vector space where $x$ lies. – William M. Jan 14 '22 at 16:23

Generalization of "symmetric positive definite" to higher dimensions?

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