Is the following claim true about the domain of solutions of a linear system with constraints?
Clash Royale CLAN TAG#URR8PPP
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I have the following problem:
$$beginarrayll textSolve A x = b\textsubject to & x_min le x_i le x_max, quadforall i in 1,2,3, M endarray$$
for $x in mathcalR^M times 1$, $b in mathcalR^3 times 1$,
$A in mathcalR^3 times M$, where A have row full rank.
Now I am looking for a method to check wether a solution to the above exist for all
$b = beginbmatrix b_1 \b_2\b_3endbmatrix $ in the following domain :
$ - beginbmatrix c_1 \c_2\c_3 endbmatrix le beginbmatrix b_1 \b_2\b_3endbmatrix le beginbmatrix c_1 \c_2\c_3 endbmatrix $
where the c's are positive constants
So my suggsted method is as follows:
1) Check $Ax=b_i$ has solution, for all boundaries of b:
$ b_1= beginbmatrix c_1 \c_2\c_3 endbmatrix ,b_2= beginbmatrix -c_1 \c_2\c_3 endbmatrix b_3= beginbmatrix c_1 \-c_2\c_3endbmatrix , cdots b_N= beginbmatrix -c_1 \-c_2\-c_3 endbmatrix $
Thus there are $2^3$ values of B to check.
2) I now want to claim that if I $Ax=b$ is solvable for all boundaries $b_i$(i.e.,the above 8 Equations), then a solution exists for all possible values of b in the domain
$ - beginbmatrix c_1 \c_2\c_3 endbmatrix le beginbmatrix b_1 \b_2\b_3endbmatrix le beginbmatrix c_1 \c_2\c_3 endbmatrix $
So my question is, can this be shown to be true?.
linear-algebra
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0
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I have the following problem:
$$beginarrayll textSolve A x = b\textsubject to & x_min le x_i le x_max, quadforall i in 1,2,3, M endarray$$
for $x in mathcalR^M times 1$, $b in mathcalR^3 times 1$,
$A in mathcalR^3 times M$, where A have row full rank.
Now I am looking for a method to check wether a solution to the above exist for all
$b = beginbmatrix b_1 \b_2\b_3endbmatrix $ in the following domain :
$ - beginbmatrix c_1 \c_2\c_3 endbmatrix le beginbmatrix b_1 \b_2\b_3endbmatrix le beginbmatrix c_1 \c_2\c_3 endbmatrix $
where the c's are positive constants
So my suggsted method is as follows:
1) Check $Ax=b_i$ has solution, for all boundaries of b:
$ b_1= beginbmatrix c_1 \c_2\c_3 endbmatrix ,b_2= beginbmatrix -c_1 \c_2\c_3 endbmatrix b_3= beginbmatrix c_1 \-c_2\c_3endbmatrix , cdots b_N= beginbmatrix -c_1 \-c_2\-c_3 endbmatrix $
Thus there are $2^3$ values of B to check.
2) I now want to claim that if I $Ax=b$ is solvable for all boundaries $b_i$(i.e.,the above 8 Equations), then a solution exists for all possible values of b in the domain
$ - beginbmatrix c_1 \c_2\c_3 endbmatrix le beginbmatrix b_1 \b_2\b_3endbmatrix le beginbmatrix c_1 \c_2\c_3 endbmatrix $
So my question is, can this be shown to be true?.
linear-algebra
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have the following problem:
$$beginarrayll textSolve A x = b\textsubject to & x_min le x_i le x_max, quadforall i in 1,2,3, M endarray$$
for $x in mathcalR^M times 1$, $b in mathcalR^3 times 1$,
$A in mathcalR^3 times M$, where A have row full rank.
Now I am looking for a method to check wether a solution to the above exist for all
$b = beginbmatrix b_1 \b_2\b_3endbmatrix $ in the following domain :
$ - beginbmatrix c_1 \c_2\c_3 endbmatrix le beginbmatrix b_1 \b_2\b_3endbmatrix le beginbmatrix c_1 \c_2\c_3 endbmatrix $
where the c's are positive constants
So my suggsted method is as follows:
1) Check $Ax=b_i$ has solution, for all boundaries of b:
$ b_1= beginbmatrix c_1 \c_2\c_3 endbmatrix ,b_2= beginbmatrix -c_1 \c_2\c_3 endbmatrix b_3= beginbmatrix c_1 \-c_2\c_3endbmatrix , cdots b_N= beginbmatrix -c_1 \-c_2\-c_3 endbmatrix $
Thus there are $2^3$ values of B to check.
2) I now want to claim that if I $Ax=b$ is solvable for all boundaries $b_i$(i.e.,the above 8 Equations), then a solution exists for all possible values of b in the domain
$ - beginbmatrix c_1 \c_2\c_3 endbmatrix le beginbmatrix b_1 \b_2\b_3endbmatrix le beginbmatrix c_1 \c_2\c_3 endbmatrix $
So my question is, can this be shown to be true?.
linear-algebra
I have the following problem:
$$beginarrayll textSolve A x = b\textsubject to & x_min le x_i le x_max, quadforall i in 1,2,3, M endarray$$
for $x in mathcalR^M times 1$, $b in mathcalR^3 times 1$,
$A in mathcalR^3 times M$, where A have row full rank.
Now I am looking for a method to check wether a solution to the above exist for all
$b = beginbmatrix b_1 \b_2\b_3endbmatrix $ in the following domain :
$ - beginbmatrix c_1 \c_2\c_3 endbmatrix le beginbmatrix b_1 \b_2\b_3endbmatrix le beginbmatrix c_1 \c_2\c_3 endbmatrix $
where the c's are positive constants
So my suggsted method is as follows:
1) Check $Ax=b_i$ has solution, for all boundaries of b:
$ b_1= beginbmatrix c_1 \c_2\c_3 endbmatrix ,b_2= beginbmatrix -c_1 \c_2\c_3 endbmatrix b_3= beginbmatrix c_1 \-c_2\c_3endbmatrix , cdots b_N= beginbmatrix -c_1 \-c_2\-c_3 endbmatrix $
Thus there are $2^3$ values of B to check.
2) I now want to claim that if I $Ax=b$ is solvable for all boundaries $b_i$(i.e.,the above 8 Equations), then a solution exists for all possible values of b in the domain
$ - beginbmatrix c_1 \c_2\c_3 endbmatrix le beginbmatrix b_1 \b_2\b_3endbmatrix le beginbmatrix c_1 \c_2\c_3 endbmatrix $
So my question is, can this be shown to be true?.
linear-algebra
asked Aug 29 at 8:27
Einar U
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1 Answer
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Yes. Suppose $Ax_i=b_i$ for all $b_i$ in the corners. Suppose you have any other point $b$. You represent it as a convex combination $b=sum t_ib_i$ with $t_igeq 0, sum t_i=1$. Then
$$A(sum t_ix_i) = sum t_i Ax_i = sum t_ib_i = b$$
and moreover the point $sum t_ix_i$ satisfies your box constrains because it is a convex combination of points $x_i$, all of which are in the box.
Seems reasonable!. This may just be the same (and is probably poorly formulated), but is it also possible to say that boundary of b solving the system, not subject to constraints c must form a convex polygon $C_max in mathcalR^3$. Therefore, if $-c<b<c$, also form a convex polygon $D in mathcalR^3$, and all verteces (corners) of the boundary of b are feasible (thus inside D), then due to convexity, we must have $D subset C$ to form a convex set
â Einar U
Aug 30 at 11:55
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Yes. Suppose $Ax_i=b_i$ for all $b_i$ in the corners. Suppose you have any other point $b$. You represent it as a convex combination $b=sum t_ib_i$ with $t_igeq 0, sum t_i=1$. Then
$$A(sum t_ix_i) = sum t_i Ax_i = sum t_ib_i = b$$
and moreover the point $sum t_ix_i$ satisfies your box constrains because it is a convex combination of points $x_i$, all of which are in the box.
Seems reasonable!. This may just be the same (and is probably poorly formulated), but is it also possible to say that boundary of b solving the system, not subject to constraints c must form a convex polygon $C_max in mathcalR^3$. Therefore, if $-c<b<c$, also form a convex polygon $D in mathcalR^3$, and all verteces (corners) of the boundary of b are feasible (thus inside D), then due to convexity, we must have $D subset C$ to form a convex set
â Einar U
Aug 30 at 11:55
add a comment |Â
up vote
0
down vote
accepted
Yes. Suppose $Ax_i=b_i$ for all $b_i$ in the corners. Suppose you have any other point $b$. You represent it as a convex combination $b=sum t_ib_i$ with $t_igeq 0, sum t_i=1$. Then
$$A(sum t_ix_i) = sum t_i Ax_i = sum t_ib_i = b$$
and moreover the point $sum t_ix_i$ satisfies your box constrains because it is a convex combination of points $x_i$, all of which are in the box.
Seems reasonable!. This may just be the same (and is probably poorly formulated), but is it also possible to say that boundary of b solving the system, not subject to constraints c must form a convex polygon $C_max in mathcalR^3$. Therefore, if $-c<b<c$, also form a convex polygon $D in mathcalR^3$, and all verteces (corners) of the boundary of b are feasible (thus inside D), then due to convexity, we must have $D subset C$ to form a convex set
â Einar U
Aug 30 at 11:55
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Yes. Suppose $Ax_i=b_i$ for all $b_i$ in the corners. Suppose you have any other point $b$. You represent it as a convex combination $b=sum t_ib_i$ with $t_igeq 0, sum t_i=1$. Then
$$A(sum t_ix_i) = sum t_i Ax_i = sum t_ib_i = b$$
and moreover the point $sum t_ix_i$ satisfies your box constrains because it is a convex combination of points $x_i$, all of which are in the box.
Yes. Suppose $Ax_i=b_i$ for all $b_i$ in the corners. Suppose you have any other point $b$. You represent it as a convex combination $b=sum t_ib_i$ with $t_igeq 0, sum t_i=1$. Then
$$A(sum t_ix_i) = sum t_i Ax_i = sum t_ib_i = b$$
and moreover the point $sum t_ix_i$ satisfies your box constrains because it is a convex combination of points $x_i$, all of which are in the box.
answered Aug 29 at 9:15
Michal Adamaszek
1,71948
1,71948
Seems reasonable!. This may just be the same (and is probably poorly formulated), but is it also possible to say that boundary of b solving the system, not subject to constraints c must form a convex polygon $C_max in mathcalR^3$. Therefore, if $-c<b<c$, also form a convex polygon $D in mathcalR^3$, and all verteces (corners) of the boundary of b are feasible (thus inside D), then due to convexity, we must have $D subset C$ to form a convex set
â Einar U
Aug 30 at 11:55
add a comment |Â
Seems reasonable!. This may just be the same (and is probably poorly formulated), but is it also possible to say that boundary of b solving the system, not subject to constraints c must form a convex polygon $C_max in mathcalR^3$. Therefore, if $-c<b<c$, also form a convex polygon $D in mathcalR^3$, and all verteces (corners) of the boundary of b are feasible (thus inside D), then due to convexity, we must have $D subset C$ to form a convex set
â Einar U
Aug 30 at 11:55
Seems reasonable!. This may just be the same (and is probably poorly formulated), but is it also possible to say that boundary of b solving the system, not subject to constraints c must form a convex polygon $C_max in mathcalR^3$. Therefore, if $-c<b<c$, also form a convex polygon $D in mathcalR^3$, and all verteces (corners) of the boundary of b are feasible (thus inside D), then due to convexity, we must have $D subset C$ to form a convex set
â Einar U
Aug 30 at 11:55
Seems reasonable!. This may just be the same (and is probably poorly formulated), but is it also possible to say that boundary of b solving the system, not subject to constraints c must form a convex polygon $C_max in mathcalR^3$. Therefore, if $-c<b<c$, also form a convex polygon $D in mathcalR^3$, and all verteces (corners) of the boundary of b are feasible (thus inside D), then due to convexity, we must have $D subset C$ to form a convex set
â Einar U
Aug 30 at 11:55
add a comment |Â
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