Mixing
How to correctly use KKT conditions?
$begingroup$
Let $Ain mathbb{R}^{ntimes n}$ be a positive definite matrix. Then find
begin{equation}
max_{|x_i|leq1}x^TAx
end{equation}
Here I want to use KKT conditions to show that $|x_i|=1$ is an optimal solution.
Derivative of objective function $nabla f(x^*)=2Ax$ and derivative of inequality constraint function is $sum_{i=1}^{n}mu_inabla g_i(x^*)=[pmmu_1 quad pmmu_2 quad cdots quad pmmu_n]'$, so KKT conditions are
begin{equation}
2Ax-[pmmu_1 quad pmmu_2 quad cdots quad pmmu_n]'=0;
end{equation}
begin{equation}
mu_igeq0;
end{equation}
begin{equation}
mu_ig_i(x^*)=0;
end{equation}
So if $2Ax$ is non-zero element-wise, then $mu_i neq0$, thus $g_i(x^*)=0quad$ i.e. $|x_i|=1$ is optimal solution.
But how to show that $2Ax$ is indeed non-zero element-wise?
linear-algebra optimization karush-kuhn-tucker
edited Jan 8 at 3:06
LinAlg
9,0111521
asked Jan 7 at 6:14
LeeLee
328111
$endgroup$
add a comment |
$begingroup$
Let $Ain mathbb{R}^{ntimes n}$ be a positive definite matrix. Then find
begin{equation}
max_{|x_i|leq1}x^TAx
end{equation}
Here I want to use KKT conditions to show that $|x_i|=1$ is an optimal solution.
Derivative of objective function $nabla f(x^*)=2Ax$ and derivative of inequality constraint function is $sum_{i=1}^{n}mu_inabla g_i(x^*)=[pmmu_1 quad pmmu_2 quad cdots quad pmmu_n]'$, so KKT conditions are
begin{equation}
2Ax-[pmmu_1 quad pmmu_2 quad cdots quad pmmu_n]'=0;
end{equation}
begin{equation}
mu_igeq0;
end{equation}
begin{equation}
mu_ig_i(x^*)=0;
end{equation}
So if $2Ax$ is non-zero element-wise, then $mu_i neq0$, thus $g_i(x^*)=0quad$ i.e. $|x_i|=1$ is optimal solution.
But how to show that $2Ax$ is indeed non-zero element-wise?
linear-algebra optimization karush-kuhn-tucker
edited Jan 8 at 3:06
LinAlg
9,0111521
asked Jan 7 at 6:14
LeeLee
328111
$endgroup$
$begingroup$
If $(Ax)_i = 0$, you do not need $|x_i|=1$. What makes you think $|x_i|=1$ for all $i$?
$endgroup$
– LinAlg
Jan 7 at 23:11
$begingroup$
I did a simulation that shows optimal solution is always at $|x_i|=1$. Even though, $(Ax)_i=0$ for some $x$, you can change signs of some $x_i$ to get $(Ax)_ineq 0$ for all $i$.
$endgroup$
– Lee
Jan 8 at 1:22
$begingroup$
If $A=begin{bmatrix}2 & -1 & -1\-1 & 2 &1 \-1 & 1 &2end{bmatrix}>0$, then $Ax=[0 quad 2 quad 2]'$ for $x=[1 quad 1quad 1]'$, but $Ax=[4quad-4quad-4]'$ for $x=[1 quad -1quad -1]'$.
$endgroup$
– Lee
Jan 8 at 1:31
$begingroup$
are you saying the KKT conditions are not sufficient for optimality?
$endgroup$
– LinAlg
Jan 8 at 1:41
$begingroup$
no, I am saying that to use KKT conditions, I have to show that there exist $xin{pm1}^n$ such that $(Ax)_ineq0$ for all $i$. But I don't know how. I want to show that $mu_i>0$ for all $i$
$endgroup$
– Lee
Jan 8 at 1:46
add a comment |
$begingroup$
Let $Ain mathbb{R}^{ntimes n}$ be a positive definite matrix. Then find
begin{equation}
max_{|x_i|leq1}x^TAx
end{equation}
Here I want to use KKT conditions to show that $|x_i|=1$ is an optimal solution.
Derivative of objective function $nabla f(x^*)=2Ax$ and derivative of inequality constraint function is $sum_{i=1}^{n}mu_inabla g_i(x^*)=[pmmu_1 quad pmmu_2 quad cdots quad pmmu_n]'$, so KKT conditions are
begin{equation}
2Ax-[pmmu_1 quad pmmu_2 quad cdots quad pmmu_n]'=0;
end{equation}
begin{equation}
mu_igeq0;
end{equation}
begin{equation}
mu_ig_i(x^*)=0;
end{equation}
So if $2Ax$ is non-zero element-wise, then $mu_i neq0$, thus $g_i(x^*)=0quad$ i.e. $|x_i|=1$ is optimal solution.
But how to show that $2Ax$ is indeed non-zero element-wise?
linear-algebra optimization karush-kuhn-tucker
edited Jan 8 at 3:06
LinAlg
9,0111521
asked Jan 7 at 6:14
LeeLee
328111
$endgroup$
Let $Ain mathbb{R}^{ntimes n}$ be a positive definite matrix. Then find
begin{equation}
max_{|x_i|leq1}x^TAx
end{equation}
Here I want to use KKT conditions to show that $|x_i|=1$ is an optimal solution.
Derivative of objective function $nabla f(x^*)=2Ax$ and derivative of inequality constraint function is $sum_{i=1}^{n}mu_inabla g_i(x^*)=[pmmu_1 quad pmmu_2 quad cdots quad pmmu_n]'$, so KKT conditions are
begin{equation}
2Ax-[pmmu_1 quad pmmu_2 quad cdots quad pmmu_n]'=0;
end{equation}
begin{equation}
mu_igeq0;
end{equation}
begin{equation}
mu_ig_i(x^*)=0;
end{equation}
So if $2Ax$ is non-zero element-wise, then $mu_i neq0$, thus $g_i(x^*)=0quad$ i.e. $|x_i|=1$ is optimal solution.
But how to show that $2Ax$ is indeed non-zero element-wise?
linear-algebra optimization karush-kuhn-tucker
linear-algebra optimization karush-kuhn-tucker
edited Jan 8 at 3:06
LinAlg
9,0111521
asked Jan 7 at 6:14
LeeLee
328111
edited Jan 8 at 3:06
LinAlg
9,0111521
asked Jan 7 at 6:14
LeeLee
328111
edited Jan 8 at 3:06
LinAlg
9,0111521
edited Jan 8 at 3:06
LinAlg
9,0111521
edited Jan 8 at 3:06
LinAlg
9,0111521
9,0111521
asked Jan 7 at 6:14
LeeLee
328111
asked Jan 7 at 6:14
LeeLee
328111
asked Jan 7 at 6:14
LeeLee
328111
328111
$begingroup$
If $(Ax)_i = 0$, you do not need $|x_i|=1$. What makes you think $|x_i|=1$ for all $i$?
$endgroup$
– LinAlg
Jan 7 at 23:11
$begingroup$
I did a simulation that shows optimal solution is always at $|x_i|=1$. Even though, $(Ax)_i=0$ for some $x$, you can change signs of some $x_i$ to get $(Ax)_ineq 0$ for all $i$.
$endgroup$
– Lee
Jan 8 at 1:22
$begingroup$
If $A=begin{bmatrix}2 & -1 & -1\-1 & 2 &1 \-1 & 1 &2end{bmatrix}>0$, then $Ax=[0 quad 2 quad 2]'$ for $x=[1 quad 1quad 1]'$, but $Ax=[4quad-4quad-4]'$ for $x=[1 quad -1quad -1]'$.
$endgroup$
– Lee
Jan 8 at 1:31
$begingroup$
are you saying the KKT conditions are not sufficient for optimality?
$endgroup$
– LinAlg
Jan 8 at 1:41
$begingroup$
no, I am saying that to use KKT conditions, I have to show that there exist $xin{pm1}^n$ such that $(Ax)_ineq0$ for all $i$. But I don't know how. I want to show that $mu_i>0$ for all $i$
$endgroup$
– Lee
Jan 8 at 1:46
add a comment |
$begingroup$
If $(Ax)_i = 0$, you do not need $|x_i|=1$. What makes you think $|x_i|=1$ for all $i$?
$endgroup$
– LinAlg
Jan 7 at 23:11
$begingroup$
I did a simulation that shows optimal solution is always at $|x_i|=1$. Even though, $(Ax)_i=0$ for some $x$, you can change signs of some $x_i$ to get $(Ax)_ineq 0$ for all $i$.
$endgroup$
– Lee
Jan 8 at 1:22
$begingroup$
If $A=begin{bmatrix}2 & -1 & -1\-1 & 2 &1 \-1 & 1 &2end{bmatrix}>0$, then $Ax=[0 quad 2 quad 2]'$ for $x=[1 quad 1quad 1]'$, but $Ax=[4quad-4quad-4]'$ for $x=[1 quad -1quad -1]'$.
$endgroup$
– Lee
Jan 8 at 1:31
$begingroup$
are you saying the KKT conditions are not sufficient for optimality?
$endgroup$
– LinAlg
Jan 8 at 1:41
$begingroup$
no, I am saying that to use KKT conditions, I have to show that there exist $xin{pm1}^n$ such that $(Ax)_ineq0$ for all $i$. But I don't know how. I want to show that $mu_i>0$ for all $i$
$endgroup$
– Lee
Jan 8 at 1:46
$begingroup$
If $(Ax)_i = 0$, you do not need $|x_i|=1$. What makes you think $|x_i|=1$ for all $i$?
$endgroup$
– LinAlg
Jan 7 at 23:11
$begingroup$
If $(Ax)_i = 0$, you do not need $|x_i|=1$. What makes you think $|x_i|=1$ for all $i$?
$endgroup$
– LinAlg
Jan 7 at 23:11
$begingroup$
I did a simulation that shows optimal solution is always at $|x_i|=1$. Even though, $(Ax)_i=0$ for some $x$, you can change signs of some $x_i$ to get $(Ax)_ineq 0$ for all $i$.
$endgroup$
– Lee
Jan 8 at 1:22
$begingroup$
I did a simulation that shows optimal solution is always at $|x_i|=1$. Even though, $(Ax)_i=0$ for some $x$, you can change signs of some $x_i$ to get $(Ax)_ineq 0$ for all $i$.
$endgroup$
– Lee
Jan 8 at 1:22
$begingroup$
If $A=begin{bmatrix}2 & -1 & -1\-1 & 2 &1 \-1 & 1 &2end{bmatrix}>0$, then $Ax=[0 quad 2 quad 2]'$ for $x=[1 quad 1quad 1]'$, but $Ax=[4quad-4quad-4]'$ for $x=[1 quad -1quad -1]'$.
$endgroup$
– Lee
Jan 8 at 1:31
$begingroup$
If $A=begin{bmatrix}2 & -1 & -1\-1 & 2 &1 \-1 & 1 &2end{bmatrix}>0$, then $Ax=[0 quad 2 quad 2]'$ for $x=[1 quad 1quad 1]'$, but $Ax=[4quad-4quad-4]'$ for $x=[1 quad -1quad -1]'$.
$endgroup$
– Lee
Jan 8 at 1:31
$begingroup$
are you saying the KKT conditions are not sufficient for optimality?
$endgroup$
– LinAlg
Jan 8 at 1:41
$begingroup$
are you saying the KKT conditions are not sufficient for optimality?
$endgroup$
– LinAlg
Jan 8 at 1:41
$begingroup$
no, I am saying that to use KKT conditions, I have to show that there exist $xin{pm1}^n$ such that $(Ax)_ineq0$ for all $i$. But I don't know how. I want to show that $mu_i>0$ for all $i$
$endgroup$
– Lee
Jan 8 at 1:46
$begingroup$
no, I am saying that to use KKT conditions, I have to show that there exist $xin{pm1}^n$ such that $(Ax)_ineq0$ for all $i$. But I don't know how. I want to show that $mu_i>0$ for all $i$
$endgroup$
– Lee
Jan 8 at 1:46
add a comment |
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$begingroup$
If $(Ax)_i = 0$, you do not need $|x_i|=1$. What makes you think $|x_i|=1$ for all $i$?
$endgroup$
– LinAlg
Jan 7 at 23:11
$begingroup$
I did a simulation that shows optimal solution is always at $|x_i|=1$. Even though, $(Ax)_i=0$ for some $x$, you can change signs of some $x_i$ to get $(Ax)_ineq 0$ for all $i$.
$endgroup$
– Lee
Jan 8 at 1:22
$begingroup$
If $A=begin{bmatrix}2 & -1 & -1\-1 & 2 &1 \-1 & 1 &2end{bmatrix}>0$, then $Ax=[0 quad 2 quad 2]'$ for $x=[1 quad 1quad 1]'$, but $Ax=[4quad-4quad-4]'$ for $x=[1 quad -1quad -1]'$.
$endgroup$
– Lee
Jan 8 at 1:31
$begingroup$
are you saying the KKT conditions are not sufficient for optimality?
$endgroup$
– LinAlg
Jan 8 at 1:41
$begingroup$
no, I am saying that to use KKT conditions, I have to show that there exist $xin{pm1}^n$ such that $(Ax)_ineq0$ for all $i$. But I don't know how. I want to show that $mu_i>0$ for all $i$
$endgroup$
– Lee
Jan 8 at 1:46