Can three vectors $ v_1,v_2,v_3 in mathbb R^2$ may be chosen from ${{ u_1,u_2,u_3,u_4}}$ such that $ u=...
Let $u_1,u_2,u_3,u_4$ be vectors in $mathbb{ R^2}$ and
$$ u= sum_{j=1}^{4} t_ju_j;text{ }t_j>0 text{ and } sum_{j=1}^{4} t_j=1$$
Then three vectors $ v_1,v_2,v_3 in mathbb{R}^2$ may be chosen from ${{ u_1,u_2,u_3,u_4}}$ such that
$$ u= sum_{j=1}^{3}s_jv_j ;text{ } s_jgeq 0 text{ and } sum_{j=1}^{3} s_j=1$$
Since $u_1,u_2,u_3,u_4$ are linearly dependent, i can replace $u_4$ (assuming $u_4$ is dependent one) by linear combination of $u_1,u_2,u_3$ so that $ u= sum_{j=1}^{3}s_jv_j$ but I can't claim $sum_{j=1}^{3} s_j=1$ .i'm stuck
Please give me a hint! (Using linear algebra )
I didn't studied topology yet, so it's hard for me to understand topological proof!
Thanks.
linear-algebra vector-spaces linear-transformations
asked Nov 30 at 12:44
Cloud JR
820417
add a comment |
Let $u_1,u_2,u_3,u_4$ be vectors in $mathbb{ R^2}$ and
$$ u= sum_{j=1}^{4} t_ju_j;text{ }t_j>0 text{ and } sum_{j=1}^{4} t_j=1$$
Then three vectors $ v_1,v_2,v_3 in mathbb{R}^2$ may be chosen from ${{ u_1,u_2,u_3,u_4}}$ such that
$$ u= sum_{j=1}^{3}s_jv_j ;text{ } s_jgeq 0 text{ and } sum_{j=1}^{3} s_j=1$$
Since $u_1,u_2,u_3,u_4$ are linearly dependent, i can replace $u_4$ (assuming $u_4$ is dependent one) by linear combination of $u_1,u_2,u_3$ so that $ u= sum_{j=1}^{3}s_jv_j$ but I can't claim $sum_{j=1}^{3} s_j=1$ .i'm stuck
Please give me a hint! (Using linear algebra )
I didn't studied topology yet, so it's hard for me to understand topological proof!
Thanks.
linear-algebra vector-spaces linear-transformations
asked Nov 30 at 12:44
Cloud JR
820417
Do you mean chosen from or constructed from? Maybe "chosen from the span of.."?
– Paul
Nov 30 at 12:47
@Paul chosen from given vectors. And not from span
– Cloud JR
Nov 30 at 12:52
Source : part A 6th question. univ.tifr.res.in/gs2019/Files/GS2012_QP_MTH.pdf
– Cloud JR
Nov 30 at 12:53
@CloudJR you can't do what you say you can. For example, if $;u_1=u_2=0;,;;u_3=(1,0);,;;u_4=(0,1);$ , there in $;u=u_4;$ you can't dispose of $;u_4;$...unless some other conditions are given.
– DonAntonio
Nov 30 at 13:04
@DonAntonio, well i actually assume u4 is dependent, let me edit it thanks
– Cloud JR
Nov 30 at 13:45
add a comment |
Let $u_1,u_2,u_3,u_4$ be vectors in $mathbb{ R^2}$ and
$$ u= sum_{j=1}^{4} t_ju_j;text{ }t_j>0 text{ and } sum_{j=1}^{4} t_j=1$$
Then three vectors $ v_1,v_2,v_3 in mathbb{R}^2$ may be chosen from ${{ u_1,u_2,u_3,u_4}}$ such that
$$ u= sum_{j=1}^{3}s_jv_j ;text{ } s_jgeq 0 text{ and } sum_{j=1}^{3} s_j=1$$
Since $u_1,u_2,u_3,u_4$ are linearly dependent, i can replace $u_4$ (assuming $u_4$ is dependent one) by linear combination of $u_1,u_2,u_3$ so that $ u= sum_{j=1}^{3}s_jv_j$ but I can't claim $sum_{j=1}^{3} s_j=1$ .i'm stuck
Please give me a hint! (Using linear algebra )
I didn't studied topology yet, so it's hard for me to understand topological proof!
Thanks.
linear-algebra vector-spaces linear-transformations
asked Nov 30 at 12:44
Cloud JR
820417
Let $u_1,u_2,u_3,u_4$ be vectors in $mathbb{ R^2}$ and
$$ u= sum_{j=1}^{4} t_ju_j;text{ }t_j>0 text{ and } sum_{j=1}^{4} t_j=1$$
Then three vectors $ v_1,v_2,v_3 in mathbb{R}^2$ may be chosen from ${{ u_1,u_2,u_3,u_4}}$ such that
$$ u= sum_{j=1}^{3}s_jv_j ;text{ } s_jgeq 0 text{ and } sum_{j=1}^{3} s_j=1$$
Since $u_1,u_2,u_3,u_4$ are linearly dependent, i can replace $u_4$ (assuming $u_4$ is dependent one) by linear combination of $u_1,u_2,u_3$ so that $ u= sum_{j=1}^{3}s_jv_j$ but I can't claim $sum_{j=1}^{3} s_j=1$ .i'm stuck
Please give me a hint! (Using linear algebra )
I didn't studied topology yet, so it's hard for me to understand topological proof!
Thanks.
linear-algebra vector-spaces linear-transformations
linear-algebra vector-spaces linear-transformations
asked Nov 30 at 12:44
Cloud JR
820417
asked Nov 30 at 12:44
Cloud JR
820417
edited Nov 30 at 13:52
asked Nov 30 at 12:44
Cloud JR
820417
asked Nov 30 at 12:44
Cloud JR
820417
asked Nov 30 at 12:44
Cloud JR
820417
820417
Do you mean chosen from or constructed from? Maybe "chosen from the span of.."?
– Paul
Nov 30 at 12:47
@Paul chosen from given vectors. And not from span
– Cloud JR
Nov 30 at 12:52
Source : part A 6th question. univ.tifr.res.in/gs2019/Files/GS2012_QP_MTH.pdf
– Cloud JR
Nov 30 at 12:53
@CloudJR you can't do what you say you can. For example, if $;u_1=u_2=0;,;;u_3=(1,0);,;;u_4=(0,1);$ , there in $;u=u_4;$ you can't dispose of $;u_4;$...unless some other conditions are given.
– DonAntonio
Nov 30 at 13:04
@DonAntonio, well i actually assume u4 is dependent, let me edit it thanks
– Cloud JR
Nov 30 at 13:45
add a comment |
Do you mean chosen from or constructed from? Maybe "chosen from the span of.."?
– Paul
Nov 30 at 12:47
@Paul chosen from given vectors. And not from span
– Cloud JR
Nov 30 at 12:52
Source : part A 6th question. univ.tifr.res.in/gs2019/Files/GS2012_QP_MTH.pdf
– Cloud JR
Nov 30 at 12:53
@CloudJR you can't do what you say you can. For example, if $;u_1=u_2=0;,;;u_3=(1,0);,;;u_4=(0,1);$ , there in $;u=u_4;$ you can't dispose of $;u_4;$...unless some other conditions are given.
– DonAntonio
Nov 30 at 13:04
@DonAntonio, well i actually assume u4 is dependent, let me edit it thanks
– Cloud JR
Nov 30 at 13:45
Do you mean chosen from or constructed from? Maybe "chosen from the span of.."?
– Paul
Nov 30 at 12:47
Do you mean chosen from or constructed from? Maybe "chosen from the span of.."?
– Paul
Nov 30 at 12:47
@Paul chosen from given vectors. And not from span
– Cloud JR
Nov 30 at 12:52
@Paul chosen from given vectors. And not from span
– Cloud JR
Nov 30 at 12:52
Source : part A 6th question. univ.tifr.res.in/gs2019/Files/GS2012_QP_MTH.pdf
– Cloud JR
Nov 30 at 12:53
Source : part A 6th question. univ.tifr.res.in/gs2019/Files/GS2012_QP_MTH.pdf
– Cloud JR
Nov 30 at 12:53
@CloudJR you can't do what you say you can. For example, if $;u_1=u_2=0;,;;u_3=(1,0);,;;u_4=(0,1);$ , there in $;u=u_4;$ you can't dispose of $;u_4;$...unless some other conditions are given.
– DonAntonio
Nov 30 at 13:04
@CloudJR you can't do what you say you can. For example, if $;u_1=u_2=0;,;;u_3=(1,0);,;;u_4=(0,1);$ , there in $;u=u_4;$ you can't dispose of $;u_4;$...unless some other conditions are given.
– DonAntonio
Nov 30 at 13:04
@DonAntonio, well i actually assume u4 is dependent, let me edit it thanks
– Cloud JR
Nov 30 at 13:45
@DonAntonio, well i actually assume u4 is dependent, let me edit it thanks
– Cloud JR
Nov 30 at 13:45
add a comment |
1 Answer
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By its definition, $u$ lies in the convex hull of the $u_i$. It suffices to show that for any $u$ there are three $u_i$ whose convex hull contains $u$.
- If there are four points on the convex hull, without loss of generality take them to be $u_1u_2u_3u_4$ in that order. Then $u_1u_2u_3$ and $u_3u_4u_1$ partition the hull, so $u$ lies in at least one of them; the points of the enclosing triangle may be taken as the $v_i$.
- If there are only three points on the convex hull, a correct choice of $v_i$ is simply those three points.
- The two- and one-point cases are trivial.
answered Nov 30 at 13:19
Parcly Taxel
41.2k137199
I didn't studied convex hull before. Your proof looks geometrically intuitive, but it's hard for me to understand
– Cloud JR
Nov 30 at 13:50
@CloudJR Perhaps you studied "barycentric coordinates" and stuff, which is just next to convex hulls and etc. ?
– DonAntonio
Nov 30 at 13:52
But i will try to understand, and i will comment if i understood or having any problem. Please help ! Thanks
– Cloud JR
Nov 30 at 13:54
@DonAntonio nope. I'm not that good at geometry , i am just a second year undergraduate. I think it was a linear algebra problem, but when i tried no linear algebraic tools works... So i ask help here.I'm gonna check it out , in wiki or some books and let you know if i understood. Thanks
– Cloud JR
Nov 30 at 13:59
As far as I remember this is standard 2nd. undergraduate year stuff, and many times it is taught within linear algebra and not until topology...
– DonAntonio
Nov 30 at 14:08
|
show 3 more comments
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By its definition, $u$ lies in the convex hull of the $u_i$. It suffices to show that for any $u$ there are three $u_i$ whose convex hull contains $u$.
- If there are four points on the convex hull, without loss of generality take them to be $u_1u_2u_3u_4$ in that order. Then $u_1u_2u_3$ and $u_3u_4u_1$ partition the hull, so $u$ lies in at least one of them; the points of the enclosing triangle may be taken as the $v_i$.
- If there are only three points on the convex hull, a correct choice of $v_i$ is simply those three points.
- The two- and one-point cases are trivial.
answered Nov 30 at 13:19
Parcly Taxel
41.2k137199
I didn't studied convex hull before. Your proof looks geometrically intuitive, but it's hard for me to understand
– Cloud JR
Nov 30 at 13:50
@CloudJR Perhaps you studied "barycentric coordinates" and stuff, which is just next to convex hulls and etc. ?
– DonAntonio
Nov 30 at 13:52
But i will try to understand, and i will comment if i understood or having any problem. Please help ! Thanks
– Cloud JR
Nov 30 at 13:54
@DonAntonio nope. I'm not that good at geometry , i am just a second year undergraduate. I think it was a linear algebra problem, but when i tried no linear algebraic tools works... So i ask help here.I'm gonna check it out , in wiki or some books and let you know if i understood. Thanks
– Cloud JR
Nov 30 at 13:59
As far as I remember this is standard 2nd. undergraduate year stuff, and many times it is taught within linear algebra and not until topology...
– DonAntonio
Nov 30 at 14:08
|
show 3 more comments
By its definition, $u$ lies in the convex hull of the $u_i$. It suffices to show that for any $u$ there are three $u_i$ whose convex hull contains $u$.
- If there are four points on the convex hull, without loss of generality take them to be $u_1u_2u_3u_4$ in that order. Then $u_1u_2u_3$ and $u_3u_4u_1$ partition the hull, so $u$ lies in at least one of them; the points of the enclosing triangle may be taken as the $v_i$.
- If there are only three points on the convex hull, a correct choice of $v_i$ is simply those three points.
- The two- and one-point cases are trivial.
answered Nov 30 at 13:19
Parcly Taxel
41.2k137199
I didn't studied convex hull before. Your proof looks geometrically intuitive, but it's hard for me to understand
– Cloud JR
Nov 30 at 13:50
@CloudJR Perhaps you studied "barycentric coordinates" and stuff, which is just next to convex hulls and etc. ?
– DonAntonio
Nov 30 at 13:52
But i will try to understand, and i will comment if i understood or having any problem. Please help ! Thanks
– Cloud JR
Nov 30 at 13:54
@DonAntonio nope. I'm not that good at geometry , i am just a second year undergraduate. I think it was a linear algebra problem, but when i tried no linear algebraic tools works... So i ask help here.I'm gonna check it out , in wiki or some books and let you know if i understood. Thanks
– Cloud JR
Nov 30 at 13:59
As far as I remember this is standard 2nd. undergraduate year stuff, and many times it is taught within linear algebra and not until topology...
– DonAntonio
Nov 30 at 14:08
|
show 3 more comments
By its definition, $u$ lies in the convex hull of the $u_i$. It suffices to show that for any $u$ there are three $u_i$ whose convex hull contains $u$.
- If there are four points on the convex hull, without loss of generality take them to be $u_1u_2u_3u_4$ in that order. Then $u_1u_2u_3$ and $u_3u_4u_1$ partition the hull, so $u$ lies in at least one of them; the points of the enclosing triangle may be taken as the $v_i$.
- If there are only three points on the convex hull, a correct choice of $v_i$ is simply those three points.
- The two- and one-point cases are trivial.
answered Nov 30 at 13:19
Parcly Taxel
41.2k137199
By its definition, $u$ lies in the convex hull of the $u_i$. It suffices to show that for any $u$ there are three $u_i$ whose convex hull contains $u$.
- If there are four points on the convex hull, without loss of generality take them to be $u_1u_2u_3u_4$ in that order. Then $u_1u_2u_3$ and $u_3u_4u_1$ partition the hull, so $u$ lies in at least one of them; the points of the enclosing triangle may be taken as the $v_i$.
- If there are only three points on the convex hull, a correct choice of $v_i$ is simply those three points.
- The two- and one-point cases are trivial.
answered Nov 30 at 13:19
Parcly Taxel
41.2k137199
answered Nov 30 at 13:19
Parcly Taxel
41.2k137199
answered Nov 30 at 13:19
Parcly Taxel
41.2k137199
answered Nov 30 at 13:19
Parcly Taxel
41.2k137199
41.2k137199
I didn't studied convex hull before. Your proof looks geometrically intuitive, but it's hard for me to understand
– Cloud JR
Nov 30 at 13:50
@CloudJR Perhaps you studied "barycentric coordinates" and stuff, which is just next to convex hulls and etc. ?
– DonAntonio
Nov 30 at 13:52
But i will try to understand, and i will comment if i understood or having any problem. Please help ! Thanks
– Cloud JR
Nov 30 at 13:54
@DonAntonio nope. I'm not that good at geometry , i am just a second year undergraduate. I think it was a linear algebra problem, but when i tried no linear algebraic tools works... So i ask help here.I'm gonna check it out , in wiki or some books and let you know if i understood. Thanks
– Cloud JR
Nov 30 at 13:59
As far as I remember this is standard 2nd. undergraduate year stuff, and many times it is taught within linear algebra and not until topology...
– DonAntonio
Nov 30 at 14:08
|
show 3 more comments
I didn't studied convex hull before. Your proof looks geometrically intuitive, but it's hard for me to understand
– Cloud JR
Nov 30 at 13:50
@CloudJR Perhaps you studied "barycentric coordinates" and stuff, which is just next to convex hulls and etc. ?
– DonAntonio
Nov 30 at 13:52
But i will try to understand, and i will comment if i understood or having any problem. Please help ! Thanks
– Cloud JR
Nov 30 at 13:54
@DonAntonio nope. I'm not that good at geometry , i am just a second year undergraduate. I think it was a linear algebra problem, but when i tried no linear algebraic tools works... So i ask help here.I'm gonna check it out , in wiki or some books and let you know if i understood. Thanks
– Cloud JR
Nov 30 at 13:59
As far as I remember this is standard 2nd. undergraduate year stuff, and many times it is taught within linear algebra and not until topology...
– DonAntonio
Nov 30 at 14:08
I didn't studied convex hull before. Your proof looks geometrically intuitive, but it's hard for me to understand
– Cloud JR
Nov 30 at 13:50
I didn't studied convex hull before. Your proof looks geometrically intuitive, but it's hard for me to understand
– Cloud JR
Nov 30 at 13:50
@CloudJR Perhaps you studied "barycentric coordinates" and stuff, which is just next to convex hulls and etc. ?
– DonAntonio
Nov 30 at 13:52
@CloudJR Perhaps you studied "barycentric coordinates" and stuff, which is just next to convex hulls and etc. ?
– DonAntonio
Nov 30 at 13:52
But i will try to understand, and i will comment if i understood or having any problem. Please help ! Thanks
– Cloud JR
Nov 30 at 13:54
But i will try to understand, and i will comment if i understood or having any problem. Please help ! Thanks
– Cloud JR
Nov 30 at 13:54
@DonAntonio nope. I'm not that good at geometry , i am just a second year undergraduate. I think it was a linear algebra problem, but when i tried no linear algebraic tools works... So i ask help here.I'm gonna check it out , in wiki or some books and let you know if i understood. Thanks
– Cloud JR
Nov 30 at 13:59
@DonAntonio nope. I'm not that good at geometry , i am just a second year undergraduate. I think it was a linear algebra problem, but when i tried no linear algebraic tools works... So i ask help here.I'm gonna check it out , in wiki or some books and let you know if i understood. Thanks
– Cloud JR
Nov 30 at 13:59
As far as I remember this is standard 2nd. undergraduate year stuff, and many times it is taught within linear algebra and not until topology...
– DonAntonio
Nov 30 at 14:08
As far as I remember this is standard 2nd. undergraduate year stuff, and many times it is taught within linear algebra and not until topology...
– DonAntonio
Nov 30 at 14:08
|
show 3 more comments
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Do you mean chosen from or constructed from? Maybe "chosen from the span of.."?
– Paul
Nov 30 at 12:47
@Paul chosen from given vectors. And not from span
– Cloud JR
Nov 30 at 12:52
Source : part A 6th question. univ.tifr.res.in/gs2019/Files/GS2012_QP_MTH.pdf
– Cloud JR
Nov 30 at 12:53
@CloudJR you can't do what you say you can. For example, if $;u_1=u_2=0;,;;u_3=(1,0);,;;u_4=(0,1);$ , there in $;u=u_4;$ you can't dispose of $;u_4;$...unless some other conditions are given.
– DonAntonio
Nov 30 at 13:04
@DonAntonio, well i actually assume u4 is dependent, let me edit it thanks
– Cloud JR
Nov 30 at 13:45