Mixing
Vectors with pairwise bounded angular distance
$begingroup$
Take a subset $V$ of $mathbb{R}^n$ for which no two vectors can have an angle greater than $pi/2$ between them. Does it follow that $V$ is contained in a hypercone? To be specific, is there a vector $uin mathbb{R}^n$ such that for all $vin V$ the angle between $u$ and $v$ is at most $pi/4$?
It seems fairly intuitive that this should be so, but I couldn't think of an argument. If not, a counter example would be nice.
geometry euclidean-geometry
asked Feb 1 at 15:23
Joshua TilleyJoshua Tilley
553313
$endgroup$
add a comment |
$begingroup$
Take a subset $V$ of $mathbb{R}^n$ for which no two vectors can have an angle greater than $pi/2$ between them. Does it follow that $V$ is contained in a hypercone? To be specific, is there a vector $uin mathbb{R}^n$ such that for all $vin V$ the angle between $u$ and $v$ is at most $pi/4$?
It seems fairly intuitive that this should be so, but I couldn't think of an argument. If not, a counter example would be nice.
geometry euclidean-geometry
asked Feb 1 at 15:23
Joshua TilleyJoshua Tilley
553313
$endgroup$
add a comment |
$begingroup$
Take a subset $V$ of $mathbb{R}^n$ for which no two vectors can have an angle greater than $pi/2$ between them. Does it follow that $V$ is contained in a hypercone? To be specific, is there a vector $uin mathbb{R}^n$ such that for all $vin V$ the angle between $u$ and $v$ is at most $pi/4$?
It seems fairly intuitive that this should be so, but I couldn't think of an argument. If not, a counter example would be nice.
geometry euclidean-geometry
asked Feb 1 at 15:23
Joshua TilleyJoshua Tilley
553313
$endgroup$
Take a subset $V$ of $mathbb{R}^n$ for which no two vectors can have an angle greater than $pi/2$ between them. Does it follow that $V$ is contained in a hypercone? To be specific, is there a vector $uin mathbb{R}^n$ such that for all $vin V$ the angle between $u$ and $v$ is at most $pi/4$?
It seems fairly intuitive that this should be so, but I couldn't think of an argument. If not, a counter example would be nice.
geometry euclidean-geometry
geometry euclidean-geometry
asked Feb 1 at 15:23
Joshua TilleyJoshua Tilley
553313
asked Feb 1 at 15:23
Joshua TilleyJoshua Tilley
553313
asked Feb 1 at 15:23
Joshua TilleyJoshua Tilley
553313
asked Feb 1 at 15:23
Joshua TilleyJoshua Tilley
553313
asked Feb 1 at 15:23
Joshua TilleyJoshua Tilley
553313
553313
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The answer is negative in all its generality because. Here is a counter-example : take as subset the three vectors of the canonical basis $e_1,e_2,e_3$ (see figure),
Fig. 1 : The intersection of the three cones should have contained a non zero vector $V$.
We are looking for a vector $V neq 0$ with coordinates $(x,y,z)$ belonging to the intersection of the three cones ("solid cones" = their surface and their interior) with resp. equations (see explanation below) :
$$begin{cases}x^2+y^2&leq&z^2\x^2+z^2&leq&y^2\y^2+z^2&leq&x^2\end{cases}tag{1}$$
But no such vector exists : the intersection is reduced to the null vector ; this is convincing on Fig. 1 but an algebraic proof is at our fingertips : indeed, if the coordinates $(x,y,z)$ of $V$ satisfy these three inequations, they satisfy their sum, i.e., are such that $2(x^2+y^2+z^2) leq (x^2+y^2+z^2)$ which is possible only for the null vector.
Explanation of inequalities (1) :
Let us take the example of inequality
$$x^2+y^2leq z^2.tag{2}$$
We start from the constraint :
$$0 leq alpha leq pi/4 iff cos(alpha) geq cos(pi/4),$$
itself equivalent to the following condition involving a dot product :
$$cos(alpha) = dfrac{V cdot e_3}{|V||e_3|}=dfrac{0x+0y+1z}{sqrt{x^2+y^2+z^2} . 1} geq dfrac{1}{sqrt{2}}$$
squaring, we get : $dfrac{z^2}{x^2+y^2+z^2} geq dfrac12$
which itself is equivalent to inequality (2), which was our objective.
Remark 1 : more generaly, this brings a counter-example even for a set of $n$ vectors ; as long as this set contains these 3 vectors, there is no hope...
Remark 2 : this solution has much in common with that of @Floris Claassens, but thinking in terms of cones brings a global view.
answered Feb 2 at 13:39
Jean MarieJean Marie
31.5k42355
$endgroup$
1
$begingroup$
Thanks! As a follow up: the set is clearly contained in a (degenerate) cone, namely a cone with angle $pi$. The set need not be contained in a cone with angle $pi/2$ as you showed, is $pi$ the smallest cone that necessarily contains the set, I wonder?
$endgroup$
– Joshua Tilley
Feb 2 at 13:57
$begingroup$
I must think about that because it is so easy in these matters to utter false things... besides, I have modified my initial figure : in fact the 3 cones are mutually tangent.
$endgroup$
– Jean Marie
Feb 2 at 14:01
add a comment |
$begingroup$
This is actually not the case. Consider ${e_{1}=(1,0,0),e_{2}=(0,1,0),e_{3}=(0,0,1)}subsetmathbb{R}^{3}$. Clearly
$$frac{langle e_{i},e_{j}rangle}{|e_{i}||e_{j}|}=cos(theta)geq0$$
for all $i,jin{1,2,3}$ where $theta$ is the angle between $e_{i}$ and $e_{j}$. Now suppose there is some $u=(u_{1},u_{2},u_{3})$ such that the angle between $u$ and $e_{i}$ is at most $pi/4$, i.e.
$$frac{langle e_{i},urangle}{|e_{i}||u|}geqfrac{1}{2}sqrt{2}$$
Clearly $x,y,z>0$. We may assume wlog that $u_{1}^{2}+u_{2}^{2}+u_{3}^{2}=3$. Then $u_{i}leq 1$ for some $i$. We find
$$frac{1}{2}sqrt{2}leqfrac{langle e_{i},urangle}{|e_{i}||u|}=frac{u_{i}}{sqrt{3}}leqfrac{1}{3}sqrt{3}$$
which is a contradiction.
answered Feb 1 at 21:49
Floris ClaassensFloris Claassens
1,47429
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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$begingroup$
The answer is negative in all its generality because. Here is a counter-example : take as subset the three vectors of the canonical basis $e_1,e_2,e_3$ (see figure),
Fig. 1 : The intersection of the three cones should have contained a non zero vector $V$.
We are looking for a vector $V neq 0$ with coordinates $(x,y,z)$ belonging to the intersection of the three cones ("solid cones" = their surface and their interior) with resp. equations (see explanation below) :
$$begin{cases}x^2+y^2&leq&z^2\x^2+z^2&leq&y^2\y^2+z^2&leq&x^2\end{cases}tag{1}$$
But no such vector exists : the intersection is reduced to the null vector ; this is convincing on Fig. 1 but an algebraic proof is at our fingertips : indeed, if the coordinates $(x,y,z)$ of $V$ satisfy these three inequations, they satisfy their sum, i.e., are such that $2(x^2+y^2+z^2) leq (x^2+y^2+z^2)$ which is possible only for the null vector.
Explanation of inequalities (1) :
Let us take the example of inequality
$$x^2+y^2leq z^2.tag{2}$$
We start from the constraint :
$$0 leq alpha leq pi/4 iff cos(alpha) geq cos(pi/4),$$
itself equivalent to the following condition involving a dot product :
$$cos(alpha) = dfrac{V cdot e_3}{|V||e_3|}=dfrac{0x+0y+1z}{sqrt{x^2+y^2+z^2} . 1} geq dfrac{1}{sqrt{2}}$$
squaring, we get : $dfrac{z^2}{x^2+y^2+z^2} geq dfrac12$
which itself is equivalent to inequality (2), which was our objective.
Remark 1 : more generaly, this brings a counter-example even for a set of $n$ vectors ; as long as this set contains these 3 vectors, there is no hope...
Remark 2 : this solution has much in common with that of @Floris Claassens, but thinking in terms of cones brings a global view.
answered Feb 2 at 13:39
Jean MarieJean Marie
31.5k42355
$endgroup$
1
$begingroup$
Thanks! As a follow up: the set is clearly contained in a (degenerate) cone, namely a cone with angle $pi$. The set need not be contained in a cone with angle $pi/2$ as you showed, is $pi$ the smallest cone that necessarily contains the set, I wonder?
$endgroup$
– Joshua Tilley
Feb 2 at 13:57
$begingroup$
I must think about that because it is so easy in these matters to utter false things... besides, I have modified my initial figure : in fact the 3 cones are mutually tangent.
$endgroup$
– Jean Marie
Feb 2 at 14:01
add a comment |
$begingroup$
The answer is negative in all its generality because. Here is a counter-example : take as subset the three vectors of the canonical basis $e_1,e_2,e_3$ (see figure),
Fig. 1 : The intersection of the three cones should have contained a non zero vector $V$.
We are looking for a vector $V neq 0$ with coordinates $(x,y,z)$ belonging to the intersection of the three cones ("solid cones" = their surface and their interior) with resp. equations (see explanation below) :
$$begin{cases}x^2+y^2&leq&z^2\x^2+z^2&leq&y^2\y^2+z^2&leq&x^2\end{cases}tag{1}$$
But no such vector exists : the intersection is reduced to the null vector ; this is convincing on Fig. 1 but an algebraic proof is at our fingertips : indeed, if the coordinates $(x,y,z)$ of $V$ satisfy these three inequations, they satisfy their sum, i.e., are such that $2(x^2+y^2+z^2) leq (x^2+y^2+z^2)$ which is possible only for the null vector.
Explanation of inequalities (1) :
Let us take the example of inequality
$$x^2+y^2leq z^2.tag{2}$$
We start from the constraint :
$$0 leq alpha leq pi/4 iff cos(alpha) geq cos(pi/4),$$
itself equivalent to the following condition involving a dot product :
$$cos(alpha) = dfrac{V cdot e_3}{|V||e_3|}=dfrac{0x+0y+1z}{sqrt{x^2+y^2+z^2} . 1} geq dfrac{1}{sqrt{2}}$$
squaring, we get : $dfrac{z^2}{x^2+y^2+z^2} geq dfrac12$
which itself is equivalent to inequality (2), which was our objective.
Remark 1 : more generaly, this brings a counter-example even for a set of $n$ vectors ; as long as this set contains these 3 vectors, there is no hope...
Remark 2 : this solution has much in common with that of @Floris Claassens, but thinking in terms of cones brings a global view.
answered Feb 2 at 13:39
Jean MarieJean Marie
31.5k42355
$endgroup$
1
$begingroup$
Thanks! As a follow up: the set is clearly contained in a (degenerate) cone, namely a cone with angle $pi$. The set need not be contained in a cone with angle $pi/2$ as you showed, is $pi$ the smallest cone that necessarily contains the set, I wonder?
$endgroup$
– Joshua Tilley
Feb 2 at 13:57
$begingroup$
I must think about that because it is so easy in these matters to utter false things... besides, I have modified my initial figure : in fact the 3 cones are mutually tangent.
$endgroup$
– Jean Marie
Feb 2 at 14:01
add a comment |
$begingroup$
The answer is negative in all its generality because. Here is a counter-example : take as subset the three vectors of the canonical basis $e_1,e_2,e_3$ (see figure),
Fig. 1 : The intersection of the three cones should have contained a non zero vector $V$.
We are looking for a vector $V neq 0$ with coordinates $(x,y,z)$ belonging to the intersection of the three cones ("solid cones" = their surface and their interior) with resp. equations (see explanation below) :
$$begin{cases}x^2+y^2&leq&z^2\x^2+z^2&leq&y^2\y^2+z^2&leq&x^2\end{cases}tag{1}$$
But no such vector exists : the intersection is reduced to the null vector ; this is convincing on Fig. 1 but an algebraic proof is at our fingertips : indeed, if the coordinates $(x,y,z)$ of $V$ satisfy these three inequations, they satisfy their sum, i.e., are such that $2(x^2+y^2+z^2) leq (x^2+y^2+z^2)$ which is possible only for the null vector.
Explanation of inequalities (1) :
Let us take the example of inequality
$$x^2+y^2leq z^2.tag{2}$$
We start from the constraint :
$$0 leq alpha leq pi/4 iff cos(alpha) geq cos(pi/4),$$
itself equivalent to the following condition involving a dot product :
$$cos(alpha) = dfrac{V cdot e_3}{|V||e_3|}=dfrac{0x+0y+1z}{sqrt{x^2+y^2+z^2} . 1} geq dfrac{1}{sqrt{2}}$$
squaring, we get : $dfrac{z^2}{x^2+y^2+z^2} geq dfrac12$
which itself is equivalent to inequality (2), which was our objective.
Remark 1 : more generaly, this brings a counter-example even for a set of $n$ vectors ; as long as this set contains these 3 vectors, there is no hope...
Remark 2 : this solution has much in common with that of @Floris Claassens, but thinking in terms of cones brings a global view.
answered Feb 2 at 13:39
Jean MarieJean Marie
31.5k42355
$endgroup$
The answer is negative in all its generality because. Here is a counter-example : take as subset the three vectors of the canonical basis $e_1,e_2,e_3$ (see figure),
Fig. 1 : The intersection of the three cones should have contained a non zero vector $V$.
We are looking for a vector $V neq 0$ with coordinates $(x,y,z)$ belonging to the intersection of the three cones ("solid cones" = their surface and their interior) with resp. equations (see explanation below) :
$$begin{cases}x^2+y^2&leq&z^2\x^2+z^2&leq&y^2\y^2+z^2&leq&x^2\end{cases}tag{1}$$
But no such vector exists : the intersection is reduced to the null vector ; this is convincing on Fig. 1 but an algebraic proof is at our fingertips : indeed, if the coordinates $(x,y,z)$ of $V$ satisfy these three inequations, they satisfy their sum, i.e., are such that $2(x^2+y^2+z^2) leq (x^2+y^2+z^2)$ which is possible only for the null vector.
Explanation of inequalities (1) :
Let us take the example of inequality
$$x^2+y^2leq z^2.tag{2}$$
We start from the constraint :
$$0 leq alpha leq pi/4 iff cos(alpha) geq cos(pi/4),$$
itself equivalent to the following condition involving a dot product :
$$cos(alpha) = dfrac{V cdot e_3}{|V||e_3|}=dfrac{0x+0y+1z}{sqrt{x^2+y^2+z^2} . 1} geq dfrac{1}{sqrt{2}}$$
squaring, we get : $dfrac{z^2}{x^2+y^2+z^2} geq dfrac12$
which itself is equivalent to inequality (2), which was our objective.
Remark 1 : more generaly, this brings a counter-example even for a set of $n$ vectors ; as long as this set contains these 3 vectors, there is no hope...
Remark 2 : this solution has much in common with that of @Floris Claassens, but thinking in terms of cones brings a global view.
answered Feb 2 at 13:39
Jean MarieJean Marie
31.5k42355
edited Feb 2 at 15:24
answered Feb 2 at 13:39
Jean MarieJean Marie
31.5k42355
answered Feb 2 at 13:39
Jean MarieJean Marie
31.5k42355
answered Feb 2 at 13:39
Jean MarieJean Marie
31.5k42355
31.5k42355
1
$begingroup$
Thanks! As a follow up: the set is clearly contained in a (degenerate) cone, namely a cone with angle $pi$. The set need not be contained in a cone with angle $pi/2$ as you showed, is $pi$ the smallest cone that necessarily contains the set, I wonder?
$endgroup$
– Joshua Tilley
Feb 2 at 13:57
$begingroup$
I must think about that because it is so easy in these matters to utter false things... besides, I have modified my initial figure : in fact the 3 cones are mutually tangent.
$endgroup$
– Jean Marie
Feb 2 at 14:01
add a comment |
1
$begingroup$
Thanks! As a follow up: the set is clearly contained in a (degenerate) cone, namely a cone with angle $pi$. The set need not be contained in a cone with angle $pi/2$ as you showed, is $pi$ the smallest cone that necessarily contains the set, I wonder?
$endgroup$
– Joshua Tilley
Feb 2 at 13:57
$begingroup$
I must think about that because it is so easy in these matters to utter false things... besides, I have modified my initial figure : in fact the 3 cones are mutually tangent.
$endgroup$
– Jean Marie
Feb 2 at 14:01
1
1
$begingroup$
Thanks! As a follow up: the set is clearly contained in a (degenerate) cone, namely a cone with angle $pi$. The set need not be contained in a cone with angle $pi/2$ as you showed, is $pi$ the smallest cone that necessarily contains the set, I wonder?
$endgroup$
– Joshua Tilley
Feb 2 at 13:57
$begingroup$
Thanks! As a follow up: the set is clearly contained in a (degenerate) cone, namely a cone with angle $pi$. The set need not be contained in a cone with angle $pi/2$ as you showed, is $pi$ the smallest cone that necessarily contains the set, I wonder?
$endgroup$
– Joshua Tilley
Feb 2 at 13:57
$begingroup$
I must think about that because it is so easy in these matters to utter false things... besides, I have modified my initial figure : in fact the 3 cones are mutually tangent.
$endgroup$
– Jean Marie
Feb 2 at 14:01
$begingroup$
I must think about that because it is so easy in these matters to utter false things... besides, I have modified my initial figure : in fact the 3 cones are mutually tangent.
$endgroup$
– Jean Marie
Feb 2 at 14:01
add a comment |
$begingroup$
This is actually not the case. Consider ${e_{1}=(1,0,0),e_{2}=(0,1,0),e_{3}=(0,0,1)}subsetmathbb{R}^{3}$. Clearly
$$frac{langle e_{i},e_{j}rangle}{|e_{i}||e_{j}|}=cos(theta)geq0$$
for all $i,jin{1,2,3}$ where $theta$ is the angle between $e_{i}$ and $e_{j}$. Now suppose there is some $u=(u_{1},u_{2},u_{3})$ such that the angle between $u$ and $e_{i}$ is at most $pi/4$, i.e.
$$frac{langle e_{i},urangle}{|e_{i}||u|}geqfrac{1}{2}sqrt{2}$$
Clearly $x,y,z>0$. We may assume wlog that $u_{1}^{2}+u_{2}^{2}+u_{3}^{2}=3$. Then $u_{i}leq 1$ for some $i$. We find
$$frac{1}{2}sqrt{2}leqfrac{langle e_{i},urangle}{|e_{i}||u|}=frac{u_{i}}{sqrt{3}}leqfrac{1}{3}sqrt{3}$$
which is a contradiction.
answered Feb 1 at 21:49
Floris ClaassensFloris Claassens
1,47429
$endgroup$
add a comment |
$begingroup$
This is actually not the case. Consider ${e_{1}=(1,0,0),e_{2}=(0,1,0),e_{3}=(0,0,1)}subsetmathbb{R}^{3}$. Clearly
$$frac{langle e_{i},e_{j}rangle}{|e_{i}||e_{j}|}=cos(theta)geq0$$
for all $i,jin{1,2,3}$ where $theta$ is the angle between $e_{i}$ and $e_{j}$. Now suppose there is some $u=(u_{1},u_{2},u_{3})$ such that the angle between $u$ and $e_{i}$ is at most $pi/4$, i.e.
$$frac{langle e_{i},urangle}{|e_{i}||u|}geqfrac{1}{2}sqrt{2}$$
Clearly $x,y,z>0$. We may assume wlog that $u_{1}^{2}+u_{2}^{2}+u_{3}^{2}=3$. Then $u_{i}leq 1$ for some $i$. We find
$$frac{1}{2}sqrt{2}leqfrac{langle e_{i},urangle}{|e_{i}||u|}=frac{u_{i}}{sqrt{3}}leqfrac{1}{3}sqrt{3}$$
which is a contradiction.
answered Feb 1 at 21:49
Floris ClaassensFloris Claassens
1,47429
$endgroup$
add a comment |
$begingroup$
This is actually not the case. Consider ${e_{1}=(1,0,0),e_{2}=(0,1,0),e_{3}=(0,0,1)}subsetmathbb{R}^{3}$. Clearly
$$frac{langle e_{i},e_{j}rangle}{|e_{i}||e_{j}|}=cos(theta)geq0$$
for all $i,jin{1,2,3}$ where $theta$ is the angle between $e_{i}$ and $e_{j}$. Now suppose there is some $u=(u_{1},u_{2},u_{3})$ such that the angle between $u$ and $e_{i}$ is at most $pi/4$, i.e.
$$frac{langle e_{i},urangle}{|e_{i}||u|}geqfrac{1}{2}sqrt{2}$$
Clearly $x,y,z>0$. We may assume wlog that $u_{1}^{2}+u_{2}^{2}+u_{3}^{2}=3$. Then $u_{i}leq 1$ for some $i$. We find
$$frac{1}{2}sqrt{2}leqfrac{langle e_{i},urangle}{|e_{i}||u|}=frac{u_{i}}{sqrt{3}}leqfrac{1}{3}sqrt{3}$$
which is a contradiction.
answered Feb 1 at 21:49
Floris ClaassensFloris Claassens
1,47429
$endgroup$
This is actually not the case. Consider ${e_{1}=(1,0,0),e_{2}=(0,1,0),e_{3}=(0,0,1)}subsetmathbb{R}^{3}$. Clearly
$$frac{langle e_{i},e_{j}rangle}{|e_{i}||e_{j}|}=cos(theta)geq0$$
for all $i,jin{1,2,3}$ where $theta$ is the angle between $e_{i}$ and $e_{j}$. Now suppose there is some $u=(u_{1},u_{2},u_{3})$ such that the angle between $u$ and $e_{i}$ is at most $pi/4$, i.e.
$$frac{langle e_{i},urangle}{|e_{i}||u|}geqfrac{1}{2}sqrt{2}$$
Clearly $x,y,z>0$. We may assume wlog that $u_{1}^{2}+u_{2}^{2}+u_{3}^{2}=3$. Then $u_{i}leq 1$ for some $i$. We find
$$frac{1}{2}sqrt{2}leqfrac{langle e_{i},urangle}{|e_{i}||u|}=frac{u_{i}}{sqrt{3}}leqfrac{1}{3}sqrt{3}$$
which is a contradiction.
answered Feb 1 at 21:49
Floris ClaassensFloris Claassens
1,47429
answered Feb 1 at 21:49
Floris ClaassensFloris Claassens
1,47429
answered Feb 1 at 21:49
Floris ClaassensFloris Claassens
1,47429
answered Feb 1 at 21:49
Floris ClaassensFloris Claassens
1,47429
1,47429
add a comment |
add a comment |
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Required, but never shown
Required, but never shown
