Definition of centrally symmetric polytopes
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I'm doing this exercise and have trouble with the definition of centrally symmetric polytopes.
I understand what it means, but it just doesn't look like a workable definition in solving this problem. I thought of an alternative one, and I'd like to ask if it is equivalent to the one mentioned above.
Definition: A polytope P is said to be centrally symmetric if, for all points $x in P$, there exists a point $x' in P$ such that $x+x'=0.$
linear-algebra polytopes discrete-geometry
asked Dec 27 '18 at 17:08
ensbanaensbana
238214
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add a comment |
$begingroup$
I'm doing this exercise and have trouble with the definition of centrally symmetric polytopes.
I understand what it means, but it just doesn't look like a workable definition in solving this problem. I thought of an alternative one, and I'd like to ask if it is equivalent to the one mentioned above.
Definition: A polytope P is said to be centrally symmetric if, for all points $x in P$, there exists a point $x' in P$ such that $x+x'=0.$
linear-algebra polytopes discrete-geometry
asked Dec 27 '18 at 17:08
ensbanaensbana
238214
$endgroup$
add a comment |
$begingroup$
I'm doing this exercise and have trouble with the definition of centrally symmetric polytopes.
I understand what it means, but it just doesn't look like a workable definition in solving this problem. I thought of an alternative one, and I'd like to ask if it is equivalent to the one mentioned above.
Definition: A polytope P is said to be centrally symmetric if, for all points $x in P$, there exists a point $x' in P$ such that $x+x'=0.$
linear-algebra polytopes discrete-geometry
asked Dec 27 '18 at 17:08
ensbanaensbana
238214
$endgroup$
I'm doing this exercise and have trouble with the definition of centrally symmetric polytopes.
I understand what it means, but it just doesn't look like a workable definition in solving this problem. I thought of an alternative one, and I'd like to ask if it is equivalent to the one mentioned above.
Definition: A polytope P is said to be centrally symmetric if, for all points $x in P$, there exists a point $x' in P$ such that $x+x'=0.$
linear-algebra polytopes discrete-geometry
linear-algebra polytopes discrete-geometry
asked Dec 27 '18 at 17:08
ensbanaensbana
238214
asked Dec 27 '18 at 17:08
ensbanaensbana
238214
edited Dec 27 '18 at 18:51
ensbana
asked Dec 27 '18 at 17:08
ensbanaensbana
238214
asked Dec 27 '18 at 17:08
ensbanaensbana
238214
asked Dec 27 '18 at 17:08
ensbanaensbana
238214
238214
add a comment |
add a comment |
1 Answer
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When you have $x+x'=0$, then with the very same values you would have $x=-x'$, ain't it?
Therefore the whole set $V={x, x',...}$ thus satifies $V=-V$. And as $P=conv(V)$, and therefore $-P=conv(-V)$, you surely have $P=-P$.
--- rk
answered Dec 27 '18 at 17:56
Dr. Richard KlitzingDr. Richard Klitzing
1,78016
$endgroup$
$begingroup$
Thanks for verifying that definition. I got stuck because I didn't know how to think about the condition $P = -P$, but the equality of sets $V = -V$ that you wrote really cleared things up.
$endgroup$
– ensbana
Dec 27 '18 at 18:43
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
When you have $x+x'=0$, then with the very same values you would have $x=-x'$, ain't it?
Therefore the whole set $V={x, x',...}$ thus satifies $V=-V$. And as $P=conv(V)$, and therefore $-P=conv(-V)$, you surely have $P=-P$.
--- rk
answered Dec 27 '18 at 17:56
Dr. Richard KlitzingDr. Richard Klitzing
1,78016
$endgroup$
$begingroup$
Thanks for verifying that definition. I got stuck because I didn't know how to think about the condition $P = -P$, but the equality of sets $V = -V$ that you wrote really cleared things up.
$endgroup$
– ensbana
Dec 27 '18 at 18:43
add a comment |
$begingroup$
When you have $x+x'=0$, then with the very same values you would have $x=-x'$, ain't it?
Therefore the whole set $V={x, x',...}$ thus satifies $V=-V$. And as $P=conv(V)$, and therefore $-P=conv(-V)$, you surely have $P=-P$.
--- rk
answered Dec 27 '18 at 17:56
Dr. Richard KlitzingDr. Richard Klitzing
1,78016
$endgroup$
$begingroup$
Thanks for verifying that definition. I got stuck because I didn't know how to think about the condition $P = -P$, but the equality of sets $V = -V$ that you wrote really cleared things up.
$endgroup$
– ensbana
Dec 27 '18 at 18:43
add a comment |
$begingroup$
When you have $x+x'=0$, then with the very same values you would have $x=-x'$, ain't it?
Therefore the whole set $V={x, x',...}$ thus satifies $V=-V$. And as $P=conv(V)$, and therefore $-P=conv(-V)$, you surely have $P=-P$.
--- rk
answered Dec 27 '18 at 17:56
Dr. Richard KlitzingDr. Richard Klitzing
1,78016
$endgroup$
When you have $x+x'=0$, then with the very same values you would have $x=-x'$, ain't it?
Therefore the whole set $V={x, x',...}$ thus satifies $V=-V$. And as $P=conv(V)$, and therefore $-P=conv(-V)$, you surely have $P=-P$.
--- rk
answered Dec 27 '18 at 17:56
Dr. Richard KlitzingDr. Richard Klitzing
1,78016
answered Dec 27 '18 at 17:56
Dr. Richard KlitzingDr. Richard Klitzing
1,78016
answered Dec 27 '18 at 17:56
Dr. Richard KlitzingDr. Richard Klitzing
1,78016
answered Dec 27 '18 at 17:56
Dr. Richard KlitzingDr. Richard Klitzing
1,78016
1,78016
$begingroup$
Thanks for verifying that definition. I got stuck because I didn't know how to think about the condition $P = -P$, but the equality of sets $V = -V$ that you wrote really cleared things up.
$endgroup$
– ensbana
Dec 27 '18 at 18:43
add a comment |
$begingroup$
Thanks for verifying that definition. I got stuck because I didn't know how to think about the condition $P = -P$, but the equality of sets $V = -V$ that you wrote really cleared things up.
$endgroup$
– ensbana
Dec 27 '18 at 18:43
$begingroup$
Thanks for verifying that definition. I got stuck because I didn't know how to think about the condition $P = -P$, but the equality of sets $V = -V$ that you wrote really cleared things up.
$endgroup$
– ensbana
Dec 27 '18 at 18:43
$begingroup$
Thanks for verifying that definition. I got stuck because I didn't know how to think about the condition $P = -P$, but the equality of sets $V = -V$ that you wrote really cleared things up.
$endgroup$
– ensbana
Dec 27 '18 at 18:43
add a comment |
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