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I read:

The bilinear form $(Dg)^{-1}[a](Dg)^{-T}[a]$ is positive definite on the tangent space $T_a S$ uniformly in $a \in S$

where $g:S\subset \mathbb{R}^n \to T \subset \mathbb{R}^n$.

I don't understand what the bilinear form is.. and how can it be defined on the tangent space? does the map $(Dg)^{-1}[a](\cdot)$ go from $S \to T$ or am I wrong?

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The $Dg$ at here is just a matrix. You can prove $AA^{T}$ is positive definite for any non-singular matrix $A$. – Bombyx mori Nov 11 '12 at 17:52
@user32240 Thanks, but what is the bilinear form? What arguments does it take? – soup Nov 11 '12 at 17:54
up vote 2 down vote accepted

Your typesetting cannot be quite right. What I see is the Jacobian $Dg$ evaluated at $a.$ They then take the inverse, call it $M = Dg^{-1}.$ The object of interest is then $M M^T,$ which is a square matrix of real numbers and is symmetric. If we take some nonzero row vector $v,$ with column vector $v^T,$ we may look at the quadratic form $$ v M M^T v^T = vM \cdot vM = |vM|^2 > 0. $$ The result is nonzero when $v\neq 0$ because $M$ is nonsingular.

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This clarified my confusion as well; thanks. – Bombyx mori Nov 11 '12 at 18:11
Thanks. I think what they mean is $aM(x)M(x)^Ta^T > 0$ for vectors $a$. So the matrix is evaluated at $x$. With regards to the "uniformly in $a$" condition, under this interpretation, do you think this means that $aM(x)M(x)^Ta^T > Caa^T$ holds for all $a$ and some fixed constant $C$? – soup Nov 19 '12 at 10:26

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