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I want to prove or contradict the following claim:

If we take two vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ in $\mathbb{R}^{d}$ ($d$ isn't neccesarily 2, so geometric proofs aren't available) and the angle between them, which is defined by $\cos(\alpha_{\mathbf{v}_1,\mathbf{v}_2}) = \frac{\mathbf{v}_1^T\mathbf{v}_2}{\Vert \mathbf{v}_1 \Vert \Vert \mathbf{v}_2 \Vert}$ the following holds:

  1. For any vector $\mathbf{u}$ s.t. $\text{sgn}(\mathbf{v}_1^T\mathbf{u}) = \text{sgn}(\mathbf{v}_2^T\mathbf{u}) = 1$ if we denote $\tilde{\mathbf{v}}_1 = \mathbf{v}_1+\mathbf{u}$ and $\tilde{\mathbf{v}}_2 = \mathbf{v}_2+\mathbf{u}$ we'll get $\alpha_{\tilde{\mathbf{v}}_1,\tilde{\mathbf{v}}_2}<\alpha_{\mathbf{v}_1,\mathbf{v}_2}$
  2. For any vector $\mathbf{u}$ s.t. $\text{sgn}(\mathbf{v}_1^T\mathbf{u}) = \text{sgn}(\mathbf{v}_2^T\mathbf{u}) = -1$ if we denote $\tilde{\mathbf{v}}_1 = \mathbf{v}_1-\mathbf{u}$ and $\tilde{\mathbf{v}}_2 = \mathbf{v}_2-\mathbf{u}$ we'll get $\alpha_{\tilde{\mathbf{v}}_1,\tilde{\mathbf{v}}_2}<\alpha_{\mathbf{v}_1,\mathbf{v}_2}$

I am pretty confident the above holds, since I ran a lot of numerical simulations and it does seem to hold, i.e. I believe the claim needs to be proved and not contradicted.

I attempted to use the algebraic definition of cosine with some algebraic tricks (triangle inequality etc) and it didn't work, same with the generalized cosine inequality (for vectors).

K.defaoite
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Roei Sarussi
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    The 2 statements are the same if we replace $\mathbf{u}$ with $-\mathbf{u}$ in statement 2. – Zerox Aug 19 '20 at 15:30
  • I suspect that something can be done here with the AM-GM inequality, i.e. $$

    \frac{v_1^Tv_2}{|v_1||v_2|} \leq \frac 12\left(\frac{v_1^Tv_2}{|v_1|^2} + \frac{v_1^Tv_2}{|v_2|^2} \right) $$

     – Ben Grossmann Aug 19 '20 at 15:37
  • Try $v_1=\left(-\frac{1}{16},-\frac{58}{207}\right)^T,v_2=\left(\frac{1}{16},\frac{1}{16}\right)^T,u=\left(\frac{62}{71},-\frac{43}{210}\right)^T$. – Alexey Burdin Aug 19 '20 at 22:29

1 Answers1

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Both claims are false. Since we can obtain one claim from the other by replacing $u$ by $-u$, it suffices to disprove that the first claim.

Pick two linearly independent vectors $u$ and $v_1$ such that $v_1^Tu>0$. Let $v_2=2v_1$. Then $v_2^Tu>0$ but $$ \alpha_{v_1,v_2}=0<\alpha_{\tilde{v}_1,\tilde{v}_2}. $$ For a concrete counterexample, let \begin{aligned} u&=(1,1)^T,\\ v_1&=(1,0)^T,\\ v_2&=(2,0)^T,\\ \tilde{v_1}=u+v_1&=(2,1)^T,\\ \tilde{v_2}=u+v_2&=(3,1)^T. \end{aligned} Then $$ \frac{v_1^Tv_2}{\|v_1\|\|v_2\|}=1 >\frac{7}{\sqrt{50}}=\frac{\tilde{v}_1^T\tilde{v}_2}{\|\tilde{v}_1\|\|\tilde{v}_2\|} $$ and hence $$ \alpha_{v_1,v_2} =\arccos\frac{v_1^Tv_2}{\|v_1\|\|v_2\|} <\arccos\frac{\tilde{v}_1^T\tilde{v}_2}{\|\tilde{v}_1\|\|\tilde{v}_2\|} =\alpha_{\tilde{v}_1,\tilde{v}_2}. $$ By perturbing $v_2$ slightly along a direction normal to itself, one can also obtain a counterexample in which $v_1$ and $v_2$ are not linearly dependent.

user1551
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