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Showing that the solution set of a system of linear equations is an affine set
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Convexity of k pointsPartial solution to linear algebra underdetermined systemConvex Hull definition and counterexample?Find a linear map $f_theta: mathbb R^2 to mathbb R^2$ which describes rotation by $theta$ in counterclockwise directionIs an expectation of a concave function still concaveHow to show that the convex hull of a set describes a polyhedron?Affine combination also in affine set: do one of the coefficients have to not be $1$?Every affine set can be expressed as the solution set of a system of linear equationsQuestion about proof of Affine Hull being the smallest Affine SetConvex Combination Generalized to Infinite Sums
$begingroup$
I'm currently studying convex optimization using the textbook written by Boyd. I came across an example where the author demonstrates how an equivalent way to express affine sets is that they are a solution set for a system of linear equations.
The example given in the book is in two variables (i.e. $x_1$ and $x_2$). I was attempting to prove this using the same method used by the author, but this time in $k$ variables. I'm not sure if my approach is correct, however as it seems a bit naive and was hoping to get some feedback on it.
My Approach:
Suppose that $C = x $, $x_1, cdots , x_k in C$, and $theta_1, cdots , theta_k in BbbR$ with $sum_i = 1^k theta_i = 1$. Since each $x_i$ is in $C$, we can say that $Ax_k = b$. We want to show that $A(theta_1 x_1 + theta_2 x_2 + cdots + theta_k x_k) = b$.
We could rewrite the last equation as
$$A(theta_1 x_1 + (1 - theta_1)(fractheta_21 - theta_1x_2 + cdots + fractheta_k1 - theta_1 x_k))$$
which is
$$theta_1 Ax_1 + (1 - theta_1)A(fractheta_21 - theta_1x_2 + cdots + fractheta_k1 - theta_1x_k)$$
We can conclude that $A(fractheta_21 - theta_1x_2 + cdots + fractheta_k1 - theta_1x_k) in C$ because each $x_i$ is in $C$ and the coefficients are greater than $0$ and sum up to $1$. Thus, $A(fractheta_21 - theta_1x_2 + cdots + fractheta_k1 - theta_1x_k) = b$.
If we put these values into the previous equation we get
$$theta_1 Ax_1 + (1 - theta_1)A(fractheta_21 - theta_1x_2 + cdots + fractheta_k1 - theta_1x_k) = theta_1 b + (1 - theta_1)b = b$$
Thus we can conclude that this set is affine.
Is my approach correct? I'm not sure if the conclusion I drew that $A(fractheta_21 - theta_1x_2 + cdots + fractheta_k1 - theta_1x_k) = b$ is acceptable or not.
Thank you.
linear-algebra convex-optimization
asked Apr 2 at 1:06
SeankalaSeankala
26511
$endgroup$
add a comment |
$begingroup$
I'm currently studying convex optimization using the textbook written by Boyd. I came across an example where the author demonstrates how an equivalent way to express affine sets is that they are a solution set for a system of linear equations.
The example given in the book is in two variables (i.e. $x_1$ and $x_2$). I was attempting to prove this using the same method used by the author, but this time in $k$ variables. I'm not sure if my approach is correct, however as it seems a bit naive and was hoping to get some feedback on it.
My Approach:
Suppose that $C = x $, $x_1, cdots , x_k in C$, and $theta_1, cdots , theta_k in BbbR$ with $sum_i = 1^k theta_i = 1$. Since each $x_i$ is in $C$, we can say that $Ax_k = b$. We want to show that $A(theta_1 x_1 + theta_2 x_2 + cdots + theta_k x_k) = b$.
We could rewrite the last equation as
$$A(theta_1 x_1 + (1 - theta_1)(fractheta_21 - theta_1x_2 + cdots + fractheta_k1 - theta_1 x_k))$$
which is
$$theta_1 Ax_1 + (1 - theta_1)A(fractheta_21 - theta_1x_2 + cdots + fractheta_k1 - theta_1x_k)$$
We can conclude that $A(fractheta_21 - theta_1x_2 + cdots + fractheta_k1 - theta_1x_k) in C$ because each $x_i$ is in $C$ and the coefficients are greater than $0$ and sum up to $1$. Thus, $A(fractheta_21 - theta_1x_2 + cdots + fractheta_k1 - theta_1x_k) = b$.
If we put these values into the previous equation we get
$$theta_1 Ax_1 + (1 - theta_1)A(fractheta_21 - theta_1x_2 + cdots + fractheta_k1 - theta_1x_k) = theta_1 b + (1 - theta_1)b = b$$
Thus we can conclude that this set is affine.
Is my approach correct? I'm not sure if the conclusion I drew that $A(fractheta_21 - theta_1x_2 + cdots + fractheta_k1 - theta_1x_k) = b$ is acceptable or not.
Thank you.
linear-algebra convex-optimization
asked Apr 2 at 1:06
SeankalaSeankala
26511
$endgroup$
1
$begingroup$
You are using mathematical induction, so I suggest that you mention that explicitly at an appropriate place in the proof. But let me give you a hint for what could be a simpler proof: In $A(theta_1 x_1 + cdots theta_k x_k)$, All the $theta_i$ are scalars, and each $x_i in C$. So how can you rewrite this expression using the former, and what can you conclude from the latter (look at the definition of $C$ again if it helps)?
$endgroup$
– M. Vinay
Apr 2 at 2:14
add a comment |
$begingroup$
I'm currently studying convex optimization using the textbook written by Boyd. I came across an example where the author demonstrates how an equivalent way to express affine sets is that they are a solution set for a system of linear equations.
The example given in the book is in two variables (i.e. $x_1$ and $x_2$). I was attempting to prove this using the same method used by the author, but this time in $k$ variables. I'm not sure if my approach is correct, however as it seems a bit naive and was hoping to get some feedback on it.
My Approach:
Suppose that $C = x $, $x_1, cdots , x_k in C$, and $theta_1, cdots , theta_k in BbbR$ with $sum_i = 1^k theta_i = 1$. Since each $x_i$ is in $C$, we can say that $Ax_k = b$. We want to show that $A(theta_1 x_1 + theta_2 x_2 + cdots + theta_k x_k) = b$.
We could rewrite the last equation as
$$A(theta_1 x_1 + (1 - theta_1)(fractheta_21 - theta_1x_2 + cdots + fractheta_k1 - theta_1 x_k))$$
which is
$$theta_1 Ax_1 + (1 - theta_1)A(fractheta_21 - theta_1x_2 + cdots + fractheta_k1 - theta_1x_k)$$
We can conclude that $A(fractheta_21 - theta_1x_2 + cdots + fractheta_k1 - theta_1x_k) in C$ because each $x_i$ is in $C$ and the coefficients are greater than $0$ and sum up to $1$. Thus, $A(fractheta_21 - theta_1x_2 + cdots + fractheta_k1 - theta_1x_k) = b$.
If we put these values into the previous equation we get
$$theta_1 Ax_1 + (1 - theta_1)A(fractheta_21 - theta_1x_2 + cdots + fractheta_k1 - theta_1x_k) = theta_1 b + (1 - theta_1)b = b$$
Thus we can conclude that this set is affine.
Is my approach correct? I'm not sure if the conclusion I drew that $A(fractheta_21 - theta_1x_2 + cdots + fractheta_k1 - theta_1x_k) = b$ is acceptable or not.
Thank you.
linear-algebra convex-optimization
asked Apr 2 at 1:06
SeankalaSeankala
26511
$endgroup$
I'm currently studying convex optimization using the textbook written by Boyd. I came across an example where the author demonstrates how an equivalent way to express affine sets is that they are a solution set for a system of linear equations.
The example given in the book is in two variables (i.e. $x_1$ and $x_2$). I was attempting to prove this using the same method used by the author, but this time in $k$ variables. I'm not sure if my approach is correct, however as it seems a bit naive and was hoping to get some feedback on it.
My Approach:
Suppose that $C = x $, $x_1, cdots , x_k in C$, and $theta_1, cdots , theta_k in BbbR$ with $sum_i = 1^k theta_i = 1$. Since each $x_i$ is in $C$, we can say that $Ax_k = b$. We want to show that $A(theta_1 x_1 + theta_2 x_2 + cdots + theta_k x_k) = b$.
We could rewrite the last equation as
$$A(theta_1 x_1 + (1 - theta_1)(fractheta_21 - theta_1x_2 + cdots + fractheta_k1 - theta_1 x_k))$$
which is
$$theta_1 Ax_1 + (1 - theta_1)A(fractheta_21 - theta_1x_2 + cdots + fractheta_k1 - theta_1x_k)$$
We can conclude that $A(fractheta_21 - theta_1x_2 + cdots + fractheta_k1 - theta_1x_k) in C$ because each $x_i$ is in $C$ and the coefficients are greater than $0$ and sum up to $1$. Thus, $A(fractheta_21 - theta_1x_2 + cdots + fractheta_k1 - theta_1x_k) = b$.
If we put these values into the previous equation we get
$$theta_1 Ax_1 + (1 - theta_1)A(fractheta_21 - theta_1x_2 + cdots + fractheta_k1 - theta_1x_k) = theta_1 b + (1 - theta_1)b = b$$
Thus we can conclude that this set is affine.
Is my approach correct? I'm not sure if the conclusion I drew that $A(fractheta_21 - theta_1x_2 + cdots + fractheta_k1 - theta_1x_k) = b$ is acceptable or not.
Thank you.
linear-algebra convex-optimization
linear-algebra convex-optimization
asked Apr 2 at 1:06
SeankalaSeankala
26511
asked Apr 2 at 1:06
SeankalaSeankala
26511
asked Apr 2 at 1:06
SeankalaSeankala
26511
asked Apr 2 at 1:06
SeankalaSeankala
26511
asked Apr 2 at 1:06
SeankalaSeankala
26511
26511
1
$begingroup$
You are using mathematical induction, so I suggest that you mention that explicitly at an appropriate place in the proof. But let me give you a hint for what could be a simpler proof: In $A(theta_1 x_1 + cdots theta_k x_k)$, All the $theta_i$ are scalars, and each $x_i in C$. So how can you rewrite this expression using the former, and what can you conclude from the latter (look at the definition of $C$ again if it helps)?
$endgroup$
– M. Vinay
Apr 2 at 2:14
add a comment |
1
$begingroup$
You are using mathematical induction, so I suggest that you mention that explicitly at an appropriate place in the proof. But let me give you a hint for what could be a simpler proof: In $A(theta_1 x_1 + cdots theta_k x_k)$, All the $theta_i$ are scalars, and each $x_i in C$. So how can you rewrite this expression using the former, and what can you conclude from the latter (look at the definition of $C$ again if it helps)?
$endgroup$
– M. Vinay
Apr 2 at 2:14
1
1
$begingroup$
You are using mathematical induction, so I suggest that you mention that explicitly at an appropriate place in the proof. But let me give you a hint for what could be a simpler proof: In $A(theta_1 x_1 + cdots theta_k x_k)$, All the $theta_i$ are scalars, and each $x_i in C$. So how can you rewrite this expression using the former, and what can you conclude from the latter (look at the definition of $C$ again if it helps)?
$endgroup$
– M. Vinay
Apr 2 at 2:14
$begingroup$
You are using mathematical induction, so I suggest that you mention that explicitly at an appropriate place in the proof. But let me give you a hint for what could be a simpler proof: In $A(theta_1 x_1 + cdots theta_k x_k)$, All the $theta_i$ are scalars, and each $x_i in C$. So how can you rewrite this expression using the former, and what can you conclude from the latter (look at the definition of $C$ again if it helps)?
$endgroup$
– M. Vinay
Apr 2 at 2:14
add a comment |
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$begingroup$
You are using mathematical induction, so I suggest that you mention that explicitly at an appropriate place in the proof. But let me give you a hint for what could be a simpler proof: In $A(theta_1 x_1 + cdots theta_k x_k)$, All the $theta_i$ are scalars, and each $x_i in C$. So how can you rewrite this expression using the former, and what can you conclude from the latter (look at the definition of $C$ again if it helps)?
$endgroup$
– M. Vinay
Apr 2 at 2:14