Subspace basis of $mathbb{R}^n$ only with positive values
$begingroup$
It seems obvious but I didn't find a proof yet:
Let $U$ be an arbitrary subspace of $mathbb{R}^n$. Set $m:=dim{U}$.
Can $U$ be written as $U=mathrm{span}{b_1,dotsc,b_m}$, $b_j=begin{pmatrix}b_{i,1}\vdots\b_{i,n}end{pmatrix}$ with $b_{i,j}ge0;forall iin{1,dotsc,m};forall jin{1,dotsc,n}$?
An equivalent formulation for this question:
$exists B=begin{pmatrix}b_{1,1}&cdots &b_{1,n}\vdots&&vdots\b_{m,1}&cdots &b_{m,n}end{pmatrix}inmathbb{R}^{mtimes n}: U=mathrm{img}(B)$ with $b_{i,j}ge0;forall iin{1,dotsc,m};forall jin{1,dotsc,n}$?
If yes, can you please give a construction algorithm for a given base ${u_1,dotsc,u_m}$ of $U$?
matrices vector-spaces
asked Dec 15 '18 at 15:20
quiliupquiliup
1799
$endgroup$
add a comment |
$begingroup$
It seems obvious but I didn't find a proof yet:
Let $U$ be an arbitrary subspace of $mathbb{R}^n$. Set $m:=dim{U}$.
Can $U$ be written as $U=mathrm{span}{b_1,dotsc,b_m}$, $b_j=begin{pmatrix}b_{i,1}\vdots\b_{i,n}end{pmatrix}$ with $b_{i,j}ge0;forall iin{1,dotsc,m};forall jin{1,dotsc,n}$?
An equivalent formulation for this question:
$exists B=begin{pmatrix}b_{1,1}&cdots &b_{1,n}\vdots&&vdots\b_{m,1}&cdots &b_{m,n}end{pmatrix}inmathbb{R}^{mtimes n}: U=mathrm{img}(B)$ with $b_{i,j}ge0;forall iin{1,dotsc,m};forall jin{1,dotsc,n}$?
If yes, can you please give a construction algorithm for a given base ${u_1,dotsc,u_m}$ of $U$?
matrices vector-spaces
asked Dec 15 '18 at 15:20
quiliupquiliup
1799
$endgroup$
add a comment |
$begingroup$
It seems obvious but I didn't find a proof yet:
Let $U$ be an arbitrary subspace of $mathbb{R}^n$. Set $m:=dim{U}$.
Can $U$ be written as $U=mathrm{span}{b_1,dotsc,b_m}$, $b_j=begin{pmatrix}b_{i,1}\vdots\b_{i,n}end{pmatrix}$ with $b_{i,j}ge0;forall iin{1,dotsc,m};forall jin{1,dotsc,n}$?
An equivalent formulation for this question:
$exists B=begin{pmatrix}b_{1,1}&cdots &b_{1,n}\vdots&&vdots\b_{m,1}&cdots &b_{m,n}end{pmatrix}inmathbb{R}^{mtimes n}: U=mathrm{img}(B)$ with $b_{i,j}ge0;forall iin{1,dotsc,m};forall jin{1,dotsc,n}$?
If yes, can you please give a construction algorithm for a given base ${u_1,dotsc,u_m}$ of $U$?
matrices vector-spaces
asked Dec 15 '18 at 15:20
quiliupquiliup
1799
$endgroup$
It seems obvious but I didn't find a proof yet:
Let $U$ be an arbitrary subspace of $mathbb{R}^n$. Set $m:=dim{U}$.
Can $U$ be written as $U=mathrm{span}{b_1,dotsc,b_m}$, $b_j=begin{pmatrix}b_{i,1}\vdots\b_{i,n}end{pmatrix}$ with $b_{i,j}ge0;forall iin{1,dotsc,m};forall jin{1,dotsc,n}$?
An equivalent formulation for this question:
$exists B=begin{pmatrix}b_{1,1}&cdots &b_{1,n}\vdots&&vdots\b_{m,1}&cdots &b_{m,n}end{pmatrix}inmathbb{R}^{mtimes n}: U=mathrm{img}(B)$ with $b_{i,j}ge0;forall iin{1,dotsc,m};forall jin{1,dotsc,n}$?
If yes, can you please give a construction algorithm for a given base ${u_1,dotsc,u_m}$ of $U$?
matrices vector-spaces
matrices vector-spaces
asked Dec 15 '18 at 15:20
quiliupquiliup
1799
asked Dec 15 '18 at 15:20
quiliupquiliup
1799
asked Dec 15 '18 at 15:20
quiliupquiliup
1799
asked Dec 15 '18 at 15:20
quiliupquiliup
1799
asked Dec 15 '18 at 15:20
quiliupquiliup
1799
1799
add a comment |
add a comment |
2 Answers
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oldest
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$begingroup$
Consider the line $y=-x$ as a subspace of $Bbb R^2$. Any basis consists of one vector, which must be of the form $(alpha,-alpha)$ for some non-zero $alphainBbb R$.
answered Dec 15 '18 at 15:53
TonyKTonyK
42.2k355134
$endgroup$
add a comment |
$begingroup$
It's not true. For a counter example, consider $Bbb R^3$ and the 2D subspace $U$ whose normal vector is $(1,1,1)$. Any two non-negative non-zero vectors in $Bbb R^3_{ge 0}$ will span a subspace part of which is out of $U$ because $UcapBbb R^3_{ge 0} = {0}$, hence they cannot span $U$.
answered Dec 15 '18 at 15:37
VimVim
8,05631348
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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$begingroup$
Consider the line $y=-x$ as a subspace of $Bbb R^2$. Any basis consists of one vector, which must be of the form $(alpha,-alpha)$ for some non-zero $alphainBbb R$.
answered Dec 15 '18 at 15:53
TonyKTonyK
42.2k355134
$endgroup$
add a comment |
$begingroup$
Consider the line $y=-x$ as a subspace of $Bbb R^2$. Any basis consists of one vector, which must be of the form $(alpha,-alpha)$ for some non-zero $alphainBbb R$.
answered Dec 15 '18 at 15:53
TonyKTonyK
42.2k355134
$endgroup$
add a comment |
$begingroup$
Consider the line $y=-x$ as a subspace of $Bbb R^2$. Any basis consists of one vector, which must be of the form $(alpha,-alpha)$ for some non-zero $alphainBbb R$.
answered Dec 15 '18 at 15:53
TonyKTonyK
42.2k355134
$endgroup$
Consider the line $y=-x$ as a subspace of $Bbb R^2$. Any basis consists of one vector, which must be of the form $(alpha,-alpha)$ for some non-zero $alphainBbb R$.
answered Dec 15 '18 at 15:53
TonyKTonyK
42.2k355134
answered Dec 15 '18 at 15:53
TonyKTonyK
42.2k355134
answered Dec 15 '18 at 15:53
TonyKTonyK
42.2k355134
answered Dec 15 '18 at 15:53
TonyKTonyK
42.2k355134
42.2k355134
add a comment |
add a comment |
$begingroup$
It's not true. For a counter example, consider $Bbb R^3$ and the 2D subspace $U$ whose normal vector is $(1,1,1)$. Any two non-negative non-zero vectors in $Bbb R^3_{ge 0}$ will span a subspace part of which is out of $U$ because $UcapBbb R^3_{ge 0} = {0}$, hence they cannot span $U$.
answered Dec 15 '18 at 15:37
VimVim
8,05631348
$endgroup$
add a comment |
$begingroup$
It's not true. For a counter example, consider $Bbb R^3$ and the 2D subspace $U$ whose normal vector is $(1,1,1)$. Any two non-negative non-zero vectors in $Bbb R^3_{ge 0}$ will span a subspace part of which is out of $U$ because $UcapBbb R^3_{ge 0} = {0}$, hence they cannot span $U$.
answered Dec 15 '18 at 15:37
VimVim
8,05631348
$endgroup$
add a comment |
$begingroup$
It's not true. For a counter example, consider $Bbb R^3$ and the 2D subspace $U$ whose normal vector is $(1,1,1)$. Any two non-negative non-zero vectors in $Bbb R^3_{ge 0}$ will span a subspace part of which is out of $U$ because $UcapBbb R^3_{ge 0} = {0}$, hence they cannot span $U$.
answered Dec 15 '18 at 15:37
VimVim
8,05631348
$endgroup$
It's not true. For a counter example, consider $Bbb R^3$ and the 2D subspace $U$ whose normal vector is $(1,1,1)$. Any two non-negative non-zero vectors in $Bbb R^3_{ge 0}$ will span a subspace part of which is out of $U$ because $UcapBbb R^3_{ge 0} = {0}$, hence they cannot span $U$.
answered Dec 15 '18 at 15:37
VimVim
8,05631348
answered Dec 15 '18 at 15:37
VimVim
8,05631348
answered Dec 15 '18 at 15:37
VimVim
8,05631348
answered Dec 15 '18 at 15:37
VimVim
8,05631348
8,05631348
add a comment |
add a comment |
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