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The last question I asked regarding this was not understood, so I will try to be clearer, based on an answer they gave me in another post (About sign positive or negative of inner product)

Suppose we have $v_1$ and $v_2$ vectors in $\mathbb{R}^{n}$ and we will assume that, for example

$$\langle v_1, v_2 \rangle >0$$

In addition, we will assume that there is a linear transformation $A:\mathbb{R}^{n} \to \mathbb{R}^{n}$ which is scaled rotation. In the previous answer the following assertion is made.

If $\langle v_1, v_2 \rangle >0$, then $\langle Av_1, Av_2 \rangle >0$. In short sign preserving $\iff$ angle preserving $\iff$ $A^{T}A = \lambda I, \lambda >0$, but the last equivalence is not clear to me and an implication is missing, the truth is that since I do not understand the last property it is difficult for me to follow the proof. The my question is

Why sign preserving $\iff$ angle preserving $\iff$ $A^{T}A = \lambda I, \lambda >0$?, and how is related to $$\langle v_1, v_2 \rangle >0, \implies \langle Av_1, Av_2 \rangle >0$$

I hope my question can be understood

StrinoX3
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1 Answers1

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The property that $A^T A = \lambda I$ stems from the fact that $A$ is a scaled rotation. That is, $A = c R$, where $R$ is a rotation matrix and $c$ is a positive scalar (if it were negative, then $A$ would no longer just involve rotation, but also reflection). $R$ is also an orthonormal matrix, which means that $R^T R = I$. Substituting in $R = \frac{1}{c}A$ into this implies that $\frac{1}{c^2} A^T A = I$, or $A^T A = c^2 I$. Let $\lambda = c^2$, such that $A^T A = \lambda I$.

mhdadk
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