Conditions for inner product sign with a linear transformation

Question

I need to find properties for the following inner products situation.

Let $u,v\in\mathbb{R}^{n}$ with inner product usual in $\mathbb{R}$ and let $A\in\mathbb{M}_{n\times n}(\mathbb{R})$ a real matrix with real coeficients. I need to find conditions (whatever they are) such as $$\langle u,v \rangle >0 \implies \langle Au,Av \rangle >0$$

For example if $A$ is a orthogonal matriz the sign is preserve and even more the value is the same but I would like to know if there is a "minimum" condition for this situation to happen.

On the other hand I want to emphasize that $A$ is the linearization of any diffeomorphism, that is, A is the derivative of someone, for example I understand that if my diffeomorphism is convex then its linearization will be positive definite (I could be wrong)

Answer 1

We can equivalently write your requirement as $$ \langle u,v \rangle > 0 \implies \langle u, A^TAv\rangle > 0. $$ So, it suffices to find conditions on $M = A^TA$ such that $\langle u,v \rangle > 0 \implies \langle u, M v\rangle > 0$. The answer to this question is that $M$ must be a multiple of the identity.

If $M$ is not a multiple of the identity matrix, then there exists a (non-zero) vector $v$ such that $Mv$ is not a multiple of $v$ (cf. this post for instance). Let $u$ be given by $$ u = \frac 12 \left(\frac{v}{\|v\|} - \frac{Mv}{\|Mv\|}\right). $$ Using the Cauchy-Schwarz inequality, deduce that $\langle u,v\rangle > 0$, but $\langle u,Mv \rangle < 0$.

Thus, we conclude that $A^TA$ must be a multiple of the identity matrix. This in turn implies that $A$ must be a multiple of an orthogonal matrix.