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For $n\geq 1$, endow $\mathbb{R}^n$ with the usual scalar product. Let $u=(1,1,\dots,1)\in\mathbb{R}^n$, $v\in {]0,+\infty[^n}$ and denote by $p_{u^\perp}$ and $p_{v^\perp}$ the orthographic projection onto the hyperplanes $u^\perp$ and $v^\perp$. For $x\in\mathbb{R}^n$ such that $x\notin [0,+\infty[^n$ and $x\notin {]-\infty,0]^n}$, do the following hold:

$$p_{u^\perp}(x)\cdot p_{v^\perp}(x)>0?$$

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  • $\begingroup$ I am never sure what symbols such as $\Bbb R_+$ and $\Bbb R_+^*$ mean. $\endgroup$ Apr 19 at 19:19
  • $\begingroup$ I edited, thank you. $\endgroup$
    – G. Panel
    Apr 19 at 19:29

3 Answers 3

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Imagine that $v$ is (almost) equal to $(1,0,\ldots,0)$, then for $x=(x_1,\ldots,x_n)$, $\sum x_i=:na$, we should check that $(x_1-a,x_2-a,\ldots,x_n-a)\cdot(0,x_2,\ldots,x_n)\geqslant 0$, in other words $x_2^2+\ldots+x_n^2-a(x_2+\ldots+x_n)\geqslant 0\,\,$ (A).

Assume that $n\geqslant 3$, fix $x_2,\ldots,x_n$ so that there are both positive and negative numbers between them and $x_2+\ldots+x_n>0$. Then for very large $x_1$ the guy $a$ is very large too, and (A) does not hold

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It doesn't hold, although it looks quite rare. For example $$u:=\begin{pmatrix}1\\1\\1\end{pmatrix}, \quad v:=\begin{pmatrix}\;1\\\;1\\27\end{pmatrix}, \quad x:=\begin{pmatrix}-1\\\;\;6\\\;25\end{pmatrix}$$

$$\begin{align} \\ \alpha &:= \frac{x\cdot u}{u\cdot u} &&= \frac{-1\cdot 1+6\cdot 1+25\cdot 1}{1^2+1^2+1^2} &&= \frac{30}{3} &&= 10\\ \\ \beta &:= \frac{x\cdot v}{v\cdot v} &&= \frac{-1\cdot 1+6\cdot 1+25\cdot 27}{1^2+1^2+27^2} &&= \frac{680}{731} &&= \frac{40}{43} \end{align}$$ $$ p_{u^\perp}(x) = x-\alpha u = \begin{pmatrix}-11\\-4\\\;15\end{pmatrix}, \qquad p_{v^\perp}(x) = x-\beta v = \begin{pmatrix}-\frac{83}{43}\\ \;\;\frac{218}{43}\\ -\frac{5}{43}\end{pmatrix}$$

$$p_{u^\perp}(x)\cdot p_{v^\perp}(x) \quad=11\cdot\frac{83}{43} - 4\cdot \frac{218}{43} - 15\cdot\frac{5}{43} \quad=-\frac{34}{43} \quad<0 $$

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There is nothing special about $\mathbf{u}=(1,\dots,1)$. I work with an arbitrary vector in $(0,\infty)^n$.

Claim) Let $n\geq 3$ and $\mathbf{u}=(u_1,\dots,u_n),\mathbf{v}=(v_1,\dots,v_n)\in (0,\infty)^n$ be two linearly independent vectors. Suppose $\mathbf{u},\mathbf{v}$ satisfy the followings:

  • $\frac{u_1}{u_2}<\frac{v_1}{v_2}$, $\frac{u_2}{u_3}>\frac{v_2}{v_3}$;
  • $\frac{A}{B}>\frac{4u_3v_3}{|u_1v_2-u_2v_1|^2}$ where $$ \begin{split} & A:=\langle\mathbf{u},\mathbf{v}\rangle \left(1-\frac{\langle\mathbf{u},\mathbf{v}\rangle^2}{|\mathbf{u}|^2|\mathbf{v}|^2}\right)>0,\\ &B:=|(u_1,u_2,u_3)\times(v_1,v_2,v_3)|^2>0. \end{split} \tag{$\star$}$$ Then there exists a vector $\mathbf{x}\in\Bbb{R}^n$ with both positive and negative components for which $\left\langle p_{\mathbf{u}^{\perp}}(\mathbf{x}),p_{\mathbf{v}^{\perp}}(\mathbf{x})\right\rangle<0$.

To see this, consider the quadratic form $$ \begin{split} &Q(\mathbf{x}):=\left\langle p_{\mathbf{u}^{\perp}}(\mathbf{x}),p_{\mathbf{v}^{\perp}}(\mathbf{x})\right\rangle= \left\langle\mathbf{x}-\frac{\langle\mathbf{x},\mathbf{u}\rangle}{|\mathbf{u}|^2}\mathbf{u}, \mathbf{x}-\frac{\langle\mathbf{x},\mathbf{v}\rangle}{|\mathbf{v}|^2}\mathbf{v}\right\rangle\\ &=|\mathbf{x}|^2-\frac{|\langle\mathbf{x},\mathbf{u}\rangle|^2}{|\mathbf{u}|^2}-\frac{|\langle\mathbf{x},\mathbf{v}\rangle|^2}{|\mathbf{v}|^2}+\frac{\langle\mathbf{x},\mathbf{u}\rangle\langle\mathbf{x},\mathbf{v}\rangle}{|\mathbf{u}|^2|\mathbf{v}|^2}\langle\mathbf{u},\mathbf{v}\rangle. \end{split} $$ on $\Bbb{R}^n$. If $\mathbf{x}$ is a linear combination of the form $\mathbf{u}+t\mathbf{v}$, then $$ Q(\mathbf{u}+t\mathbf{v})= \left\langle t\left(\mathbf{v}-\frac{\langle\mathbf{u},\mathbf{v}\rangle}{|\mathbf{u}|^2}\mathbf{u}\right), \mathbf{u}-\frac{\langle\mathbf{u},\mathbf{v}\rangle}{|\mathbf{v}|^2}\mathbf{v}\right\rangle =t\overbrace{\langle\mathbf{u},\mathbf{v}\rangle}^{>0} \overbrace{\left(-1+\frac{\langle\mathbf{u},\mathbf{v}\rangle^2}{|\mathbf{u}|^2|\mathbf{v}|^2}\right)}^{<0}.\tag{$\star\star$} $$ When $n=2$, the vectors $\mathbf{u},\mathbf{v}$ are in the first quadrant. For $t>0$, $Q$ is negative at $\pm(\mathbf{u}+t\mathbf{v})$ which belong to the first or the third quadrant. On the other hand, in order for $\pm(\mathbf{u}+t\mathbf{v})$ to belong to either the second or fourth quadrants, $t$ should be negative. But then $Q$ becomes positive by $(\star\star)$. This shows that when $n=2$ we have
$\left\langle p_{\mathbf{u}^{\perp}}(\mathbf{x}),p_{\mathbf{v}^{\perp}}(\mathbf{x})\right\rangle\geq 0$ if $\mathbf{x}\notin [0,\infty)^2\cup (-\infty,0]^2$.

Finally, suppose $n>2$. Then there exist non-zero vectors perpendicular to both $\mathbf{u}$ and $\mathbf{w}$. For any such vector $\mathbf{w}$, the formula for $Q$ implies $Q(\mathbf{x}+\mathbf{w})=Q(\mathbf{x})+|\mathbf{w}|^2$. The idea is to find $t>0$ and $\mathbf{w}\perp\mathbf{u},\mathbf{v}$ such that $\mathbf{u}+t\mathbf{v}+\mathbf{w}$ has both positive and negative components (which should be possible since components of $\mathbf{w}$ cannot be all positive or all negative) and
$$ Q(\mathbf{u}+t\mathbf{v}+\mathbf{w}) =\overbrace{t\langle\mathbf{u},\mathbf{v}\rangle}^{>0} \overbrace{\left(-1+\frac{\langle\mathbf{u},\mathbf{v}\rangle^2}{|\mathbf{u}|^2|\mathbf{v}|^2}\right)}^{<0}+|\mathbf{w}|^2<0. \tag{$\star\star\star$} $$ We take $\mathbf{w}$ to be $$ \mathbf{w}:=s(u_2v_3-u_3v_2,u_3v_1-u_1v_3,u_1v_2-u_2v_1,\overbrace{0,\dots,0}^{n-3})\quad s>0. $$ Clearly, $\mathbf{w}\perp\mathbf{u},\mathbf{v}$, and by our hypotheses, its third component is negative while the first one is positive. For any $s,t>0$ the first component of $\mathbf{u}+t\mathbf{v}+\mathbf{w}$ is positive, and it suffices to arrange $s,t$ so that the third component is negative, and moreover $(\star\star\star)$ holds. The former amounts to $$ u_3+tv_3<s|u_1v_2-u_2v_1| $$ while, with the notation from $(\star)$, the latter means $$ -tA+s^2B<0. $$ So the problem boils down to $\frac{u_3+tv_3}{|u_1v_2-u_2v_1|}<\sqrt{\frac{tA}{B}}$, because then any $s$ between them qualifies. The last inequality means that
$$ r\mapsto{\frac{v_3}{|u_1v_2-u_2v_1|}}r^2-\sqrt{\frac{A}{B}}r+\frac{u_3}{|u_1v_2-u_2v_1|} $$ attains negative values over $(0,\infty)$. Since all the coefficients of this quadratic are positive except the linear one, this is equivalent to having positive discriminant: $$ \frac{A}{B}>\frac{4u_3v_3}{|u_1v_2-u_2v_1|^2}; $$ an inequality which was assumed. $\blacksquare$

Corollary) Suppose $n\geq 3$ and $\mathbf{u}\in (0,\infty)^n$. Then there exist vectors $\mathbf{v}\in (0,\infty)^n$ and $\mathbf{x}\in\Bbb{R}^n\setminus([0,\infty)^n\cup(-\infty,0]^n)$ such that $\left\langle p_{\mathbf{u}^{\perp}}(\mathbf{x}),p_{\mathbf{v}^{\perp}}(\mathbf{x})\right\rangle<0$.

To show this, one only needs to pick $\mathbf{v}=(v_1,\dots,v_n)$ with positive entries such that $\frac{u_1}{u_2}<\frac{v_1}{v_2}$, $\frac{u_2}{u_3}>\frac{v_2}{v_3}$ and $\frac{A}{B}>\frac{4u_3v_3}{|u_1v_2-u_2v_1|^2}$. If $v_1>0$ is constant and $v_2,v_3$ get small so that $\frac{v_2}{v_3}$ becomes small too, then $\frac{u_1}{u_2}<\frac{v_1}{v_2}$, $\frac{u_2}{u_3}>\frac{v_2}{v_3}$ hold; and $\frac{4u_3v_3}{|u_1v_2-u_2v_1|^2}\to 0$. But quantities $A$ and $B$ remain bounded from below. Consequently, when $v_2,v_3\to 0^{+}$ such that $\frac{v_2}{v_3}<\frac{u_2}{u_3}$ as $v_1,v_4,\dots,v_n$ are constant, we obtain vectors $\mathbf{v}$ satisfying the desired properties. $\blacksquare$

This argument also provides a recipe for constructing vectors $\mathbf{x}$ and $\mathbf{v}$ for which the dot product under consideration is negative.

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  • $\begingroup$ This doesn't seem to work (the projections of $x$ to $u^\perp$ and $v^\perp$ are both in the second quadrant). In general, it seems the statement is true for $n=2$ $\endgroup$
    – Saúl RM
    Apr 19 at 19:23
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    $\begingroup$ Sorry, I had the projection along $u$ and $v$ in mind. $\endgroup$
    – KhashF
    Apr 19 at 19:25
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    $\begingroup$ @SaúlRM I rewrote my answer. $\endgroup$
    – KhashF
    Apr 20 at 1:43

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