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my question is about how the inner product (usual in $\mathbb{R}^{n}$) changes when applying some transformation to the vectors of said inner product. I explain

Suppose we have $v_{1}$ and $v_{2}$ vectors in $\mathbb{R}^{n}$ and we will assume that, for example

$$\langle v_{1},v_{2}\rangle > 0$$

And we will also assume that there is some transformation $h:\mathbb{R}^{n} \to \mathbb{R}^{n}$ (not necessarily linear, but the information is useful if it is necessary to add additional properties).

So my question is, what would the sign (positive or negative) of this product look like?

$$\langle h(v_{1}), h(v_{2})\rangle$$

I would particularly like to know what would be the properties that the transformation $h$ needs to maintain the sign or when it changes

A particular case that I am analyzing is what happens when I multiply the vectors by any matrix, this would also help me if someone can help me

EDIT: Thanks to the clarifications in the comments, I can now more confidently ask the following:

We will suppose that $h$ corresponds to a transformation such as a change of coordinates for example, and we will say that $A$ is its Jacobian matrix, so under the same previous assumptions I want to know if

$$\langle v_{1},v_{2}\rangle > 0$$

so what about the sign of $\langle Av_{1}, Av_{2}\rangle$

StrinoX3
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  • Multiplying $v_1$ and $v_2$ by a matrix, say $A$, is the same as applying a linear map $h$ whose associated matrix (relative to a fixed basis $\beta$) is $A$, that is $[h]_{\beta}^{\beta} = A$. So you are in fact investigating the sign of $\langle h(v_1), h(v_2)\rangle$ when $h:\mathbb{R}^n\to\mathbb{R}^n$ is a linear map. This is consistent with your use of the word 'transformation' for $h$ (transformation tends to mean a linear map). – Bailey May 31 '22 at 01:31
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    Unfortunately this question does not have a definitive answer. For example, say $n=2$, $v_1=(1,0)$, $v_2=(3,0)$, and we use the regular scalar product. Then $<v_1,v_2>=3$. Now let's choose $h(v)=v-(2,0)$. You can immediately see that $<h(v_1),h(v_2)><0$. But this depends on the values of $v_1$ and $v_2$. – Andrei May 31 '22 at 01:33
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    I'm not sure about necessary conditions on $h$ for $\langle v_1, v_2\rangle$ and $\langle h(v_1),h(v_2)\rangle$ to share the same sign, but there are a few sufficient conditions. One is that $h$ is an isometry. We say that $h$ is an isometry if $|v|=|h(v)|$ for all $v\in \mathbb{R}^n$. Then $h$ is an isometry if and only if $\langle h(v_1),h(v_2)\rangle =\langle v_1,v_2\rangle$ for all $v_1,v_2\in\mathbb{R}^n$. See here. – Bailey May 31 '22 at 01:37
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    Thank you very much for your answers! it helps me a lot. Imagine that I would find that it is a bit ambiguous to think that independent of the transformation there is a general condition to keep or change the sign. In fact, if it is possible that you can help me, my current problem (which is when h is a topological equivalence in dynamic systems) is when I have the Jacobian matrix of $h$, seeing it as a change of coordinates. Any idea to show that the sign changes or stays in that case would help me a lot – StrinoX3 May 31 '22 at 01:45

1 Answers1

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This answers the reverse problem, what are the requirements on $A$ if you have $$\langle v_{1},v_{2}\rangle > 0 \implies\langle Av_{1}, Av_{2}\rangle>0$$ i.e. your linear transform is sign preserving. In short, sign preserving $\Leftrightarrow$ angle preserving $\Leftrightarrow$ $A^TA=\lambda I,\lambda>0$, $A$ is scaled rotation.

Basis Setup

As you said, we assume $A$ is invertible (basis change), then $A^TA$ is a positive definite matrix, which have a full set of orthogonal eigenvectors. Let's use these eigenvectors as our basis. $$ \begin{align} A^TA&=U\Lambda U^T\\ U&=[u_1,...u_n],UU^T=I\\ \Lambda&=diag(\lambda_1,...\lambda_n),\lambda_i>0 \end{align} $$ Then decompose $v_1,v_2$ on this basis, the condition becomes $$ v_1=\sum_ia_i u_i,\;\;v_2=\sum_ib_i u_i,\\ \sum_ia_ib_i>0\implies\sum_i \lambda_ia_ib_i>0 $$ Next I'll show by contradiction that if $\lambda_1=\lambda_2=...=\lambda_n$ is not true, then there exist $v_1,v_2$ such that the sign preserving condition is violated.

Proof

Assume existing two eigenvalues with different values, let them be $\lambda_1\neq\lambda_2$. Without loss of generality, let $\lambda_1>\lambda_2$. Then we can construct such example in their eigenspace $Span(v_1,v_2)$.

Set $$ a_1b_1=-\frac{\lambda_1+\lambda_2}{2}<-\lambda_2\\ a_2b_2=\lambda_1\\ $$ Then $$ \langle v_{1},v_{2}\rangle=a_1b_1+ a_2b_2=\frac{\lambda_1-\lambda_2}{2}>0 $$ but $$ \langle Av_{1},Av_{2}\rangle=\lambda_1a_1b_1+ \lambda_2a_2b_2=-\lambda_1\frac{\lambda_1+\lambda_2}{2}+\lambda_1\lambda_2\\ =\lambda_1\frac{-\lambda_1+\lambda_2}{2}<0 $$ which contradicts the sign preserving condition.

Implication

Thus if $A$ preserves sign of any pair of $v_1,v_2\in \mathbb R^n$ then $A^TA=\lambda I,\lambda>0 $, i.e. $A$ has to be a scaled version of an orthogonal matrix / rotation matrix. Obviously, this is also sufficient for sign preserving and angle preserving.

So we proved one equivalent condition for the inner product sign preserving linear transform: $A$ has to be a scaled version of rotation. Otherwise, we cannot predict the sign of $\langle Av_{1}, Av_{2}\rangle$ from $\langle v_{1},v_{2}\rangle$.

This is related to another question, which says the orthogonality preserving linear mapping has to be a scaled rotation.

ALL Orthogonality preserving linear maps from $\mathbb R^n$ to $\mathbb R^n$?

Prove that there is a positive real number $\lambda$, so that $A =\lambda B$, for two positive definite square matrices

Binxu Wang 王彬旭
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