Suppose $A$ is positive definite, such that for all $x \in \mathbb{R}^n$, we have:
$$x^TAx > 0$$
Do we say $A$ is positive definite on $\mathbb{R}^n$ or $\mathbb{R}^n \times \mathbb{R}^n$
Do we say a matrix $A$ is positive definite on $\mathbb{R}^n$ or $\mathbb{R}^n \times \mathbb{R}^n$
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1On $\Bbb R^n$ because the operator is defined *on* $\Bbb R^n$. – 2017-01-15
2 Answers
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Remember that $x^T$ is the transpose of the vector $x$, so ultimately, the input is just one vector in $\mathbb{R}^n$. Think about the quadratic form as $q(x) = \langle Ax,x\rangle$.
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On $\mathbb{R}^n$ makes more sense to me since that's what $A$ is acting on.
Looking at the Google search results for "positive definite on", whenever the context is linear algebra it seem to always mention the vector space next.