Mixing
How to minimize $L_1$ norm in a subspace when $L_2$ norm is fixed?
I want to solve the following optimization problem
$$text{minimize}~~lVert vec{x} rVert_1$$
$$text{subject to}~~lVert vec{x} rVert_2 = 1$$
$$text{and}~~A vec{x} = vec{0}$$
where $vec{x} in mathbb R^n$, $lVert vec{x} rVert_1 = sum_{i=1}^n |x_i|$, $lVert vec{x} rVert_2 = (sum_{i=1}^n |x_i|^2)^{1/2}$, $A in mathbb R^{mtimes n}$ and $m<n$.
This is not a convex optimization problem because the domain of $vec{x}$ is not convex. I wonder if there is any method to solve this problem, or even some method in special cases would be good.
linear-algebra optimization convex-optimization nonlinear-optimization
asked Nov 20 '18 at 21:04
Doris
404212
add a comment |
I want to solve the following optimization problem
$$text{minimize}~~lVert vec{x} rVert_1$$
$$text{subject to}~~lVert vec{x} rVert_2 = 1$$
$$text{and}~~A vec{x} = vec{0}$$
where $vec{x} in mathbb R^n$, $lVert vec{x} rVert_1 = sum_{i=1}^n |x_i|$, $lVert vec{x} rVert_2 = (sum_{i=1}^n |x_i|^2)^{1/2}$, $A in mathbb R^{mtimes n}$ and $m<n$.
This is not a convex optimization problem because the domain of $vec{x}$ is not convex. I wonder if there is any method to solve this problem, or even some method in special cases would be good.
linear-algebra optimization convex-optimization nonlinear-optimization
asked Nov 20 '18 at 21:04
Doris
404212
would you mind changing the constraint into $||x||_2 leq 1$?
– LinAlg
Nov 20 '18 at 21:21
1
@LinAlg Then the solution would be zero.
– Doris
Nov 20 '18 at 21:40
Related: math.stackexchange.com/questions/900564/…
– LinAlg
Nov 20 '18 at 22:05
@WillM.: This does not work. If $A = (1 1)$, the constraint is $x_1 + x_2 = 0$. Obviously, you cannot restrict to $x_1, x_2 ge 0$.
– gerw
Nov 22 '18 at 7:59
Ok, right. Apply Lagrange multipliers then.
– Will M.
Nov 22 '18 at 15:54
add a comment |
I want to solve the following optimization problem
$$text{minimize}~~lVert vec{x} rVert_1$$
$$text{subject to}~~lVert vec{x} rVert_2 = 1$$
$$text{and}~~A vec{x} = vec{0}$$
where $vec{x} in mathbb R^n$, $lVert vec{x} rVert_1 = sum_{i=1}^n |x_i|$, $lVert vec{x} rVert_2 = (sum_{i=1}^n |x_i|^2)^{1/2}$, $A in mathbb R^{mtimes n}$ and $m<n$.
This is not a convex optimization problem because the domain of $vec{x}$ is not convex. I wonder if there is any method to solve this problem, or even some method in special cases would be good.
linear-algebra optimization convex-optimization nonlinear-optimization
asked Nov 20 '18 at 21:04
Doris
404212
I want to solve the following optimization problem
$$text{minimize}~~lVert vec{x} rVert_1$$
$$text{subject to}~~lVert vec{x} rVert_2 = 1$$
$$text{and}~~A vec{x} = vec{0}$$
where $vec{x} in mathbb R^n$, $lVert vec{x} rVert_1 = sum_{i=1}^n |x_i|$, $lVert vec{x} rVert_2 = (sum_{i=1}^n |x_i|^2)^{1/2}$, $A in mathbb R^{mtimes n}$ and $m<n$.
This is not a convex optimization problem because the domain of $vec{x}$ is not convex. I wonder if there is any method to solve this problem, or even some method in special cases would be good.
linear-algebra optimization convex-optimization nonlinear-optimization
linear-algebra optimization convex-optimization nonlinear-optimization
asked Nov 20 '18 at 21:04
Doris
404212
asked Nov 20 '18 at 21:04
Doris
404212
asked Nov 20 '18 at 21:04
Doris
404212
asked Nov 20 '18 at 21:04
Doris
404212
asked Nov 20 '18 at 21:04
Doris
404212
404212
would you mind changing the constraint into $||x||_2 leq 1$?
– LinAlg
Nov 20 '18 at 21:21
1
@LinAlg Then the solution would be zero.
– Doris
Nov 20 '18 at 21:40
Related: math.stackexchange.com/questions/900564/…
– LinAlg
Nov 20 '18 at 22:05
@WillM.: This does not work. If $A = (1 1)$, the constraint is $x_1 + x_2 = 0$. Obviously, you cannot restrict to $x_1, x_2 ge 0$.
– gerw
Nov 22 '18 at 7:59
Ok, right. Apply Lagrange multipliers then.
– Will M.
Nov 22 '18 at 15:54
add a comment |
would you mind changing the constraint into $||x||_2 leq 1$?
– LinAlg
Nov 20 '18 at 21:21
1
@LinAlg Then the solution would be zero.
– Doris
Nov 20 '18 at 21:40
Related: math.stackexchange.com/questions/900564/…
– LinAlg
Nov 20 '18 at 22:05
@WillM.: This does not work. If $A = (1 1)$, the constraint is $x_1 + x_2 = 0$. Obviously, you cannot restrict to $x_1, x_2 ge 0$.
– gerw
Nov 22 '18 at 7:59
Ok, right. Apply Lagrange multipliers then.
– Will M.
Nov 22 '18 at 15:54
would you mind changing the constraint into $||x||_2 leq 1$?
– LinAlg
Nov 20 '18 at 21:21
would you mind changing the constraint into $||x||_2 leq 1$?
– LinAlg
Nov 20 '18 at 21:21
1
1
@LinAlg Then the solution would be zero.
– Doris
Nov 20 '18 at 21:40
@LinAlg Then the solution would be zero.
– Doris
Nov 20 '18 at 21:40
Related: math.stackexchange.com/questions/900564/…
– LinAlg
Nov 20 '18 at 22:05
Related: math.stackexchange.com/questions/900564/…
– LinAlg
Nov 20 '18 at 22:05
@WillM.: This does not work. If $A = (1 1)$, the constraint is $x_1 + x_2 = 0$. Obviously, you cannot restrict to $x_1, x_2 ge 0$.
– gerw
Nov 22 '18 at 7:59
@WillM.: This does not work. If $A = (1 1)$, the constraint is $x_1 + x_2 = 0$. Obviously, you cannot restrict to $x_1, x_2 ge 0$.
– gerw
Nov 22 '18 at 7:59
Ok, right. Apply Lagrange multipliers then.
– Will M.
Nov 22 '18 at 15:54
Ok, right. Apply Lagrange multipliers then.
– Will M.
Nov 22 '18 at 15:54
add a comment |
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would you mind changing the constraint into $||x||_2 leq 1$?
– LinAlg
Nov 20 '18 at 21:21
1
@LinAlg Then the solution would be zero.
– Doris
Nov 20 '18 at 21:40
Related: math.stackexchange.com/questions/900564/…
– LinAlg
Nov 20 '18 at 22:05
@WillM.: This does not work. If $A = (1 1)$, the constraint is $x_1 + x_2 = 0$. Obviously, you cannot restrict to $x_1, x_2 ge 0$.
– gerw
Nov 22 '18 at 7:59
Ok, right. Apply Lagrange multipliers then.
– Will M.
Nov 22 '18 at 15:54