How to equally balance $|X|_1$ over the columns of $X$?
up vote
0
down vote
favorite
I have the following optimization
$$min_X f(X)+lambda |X|_1,$$
in which $Xin mathbb{R}^{n times k}$, and $|X|_1=sum_{i,j} |x_{ij}|$, and $f(X)$ is differentiable in $X$.
It is ideal for me to have a small value for $|X|_1$ which is equally balanced over the columns of $X$.
For example, I like to have $|X_{:,i}|_1 approx |X_{:,j}|_1~forall i,j$, where $X_{:,i}$ denotes colimn $i$ of $X$.
Specifically, I want to prevent situations in which $X$ has a few columns with large entries and a lot of columns with very small entries, assuming $f(X)$ allows those situations.
linear-algebra optimization convex-optimization
add a comment |
up vote
0
down vote
favorite
I have the following optimization
$$min_X f(X)+lambda |X|_1,$$
in which $Xin mathbb{R}^{n times k}$, and $|X|_1=sum_{i,j} |x_{ij}|$, and $f(X)$ is differentiable in $X$.
It is ideal for me to have a small value for $|X|_1$ which is equally balanced over the columns of $X$.
For example, I like to have $|X_{:,i}|_1 approx |X_{:,j}|_1~forall i,j$, where $X_{:,i}$ denotes colimn $i$ of $X$.
Specifically, I want to prevent situations in which $X$ has a few columns with large entries and a lot of columns with very small entries, assuming $f(X)$ allows those situations.
linear-algebra optimization convex-optimization
you could have a different $lambda$ for each column, and keep adjusting $lambda$ until you are satisfied
– LinAlg
14 hours ago
@LinAlg yes, but it would not be easy especially when $k$ is large (e.g 100 or more)
– Babak
13 hours ago
1
that the problem is not easy is to be expected, since setting a lower bound on a norm is difficult from an optimization perspective
– LinAlg
13 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have the following optimization
$$min_X f(X)+lambda |X|_1,$$
in which $Xin mathbb{R}^{n times k}$, and $|X|_1=sum_{i,j} |x_{ij}|$, and $f(X)$ is differentiable in $X$.
It is ideal for me to have a small value for $|X|_1$ which is equally balanced over the columns of $X$.
For example, I like to have $|X_{:,i}|_1 approx |X_{:,j}|_1~forall i,j$, where $X_{:,i}$ denotes colimn $i$ of $X$.
Specifically, I want to prevent situations in which $X$ has a few columns with large entries and a lot of columns with very small entries, assuming $f(X)$ allows those situations.
linear-algebra optimization convex-optimization
I have the following optimization
$$min_X f(X)+lambda |X|_1,$$
in which $Xin mathbb{R}^{n times k}$, and $|X|_1=sum_{i,j} |x_{ij}|$, and $f(X)$ is differentiable in $X$.
It is ideal for me to have a small value for $|X|_1$ which is equally balanced over the columns of $X$.
For example, I like to have $|X_{:,i}|_1 approx |X_{:,j}|_1~forall i,j$, where $X_{:,i}$ denotes colimn $i$ of $X$.
Specifically, I want to prevent situations in which $X$ has a few columns with large entries and a lot of columns with very small entries, assuming $f(X)$ allows those situations.
linear-algebra optimization convex-optimization
linear-algebra optimization convex-optimization
edited 13 hours ago
asked 15 hours ago
Babak
297110
297110
you could have a different $lambda$ for each column, and keep adjusting $lambda$ until you are satisfied
– LinAlg
14 hours ago
@LinAlg yes, but it would not be easy especially when $k$ is large (e.g 100 or more)
– Babak
13 hours ago
1
that the problem is not easy is to be expected, since setting a lower bound on a norm is difficult from an optimization perspective
– LinAlg
13 hours ago
add a comment |
you could have a different $lambda$ for each column, and keep adjusting $lambda$ until you are satisfied
– LinAlg
14 hours ago
@LinAlg yes, but it would not be easy especially when $k$ is large (e.g 100 or more)
– Babak
13 hours ago
1
that the problem is not easy is to be expected, since setting a lower bound on a norm is difficult from an optimization perspective
– LinAlg
13 hours ago
you could have a different $lambda$ for each column, and keep adjusting $lambda$ until you are satisfied
– LinAlg
14 hours ago
you could have a different $lambda$ for each column, and keep adjusting $lambda$ until you are satisfied
– LinAlg
14 hours ago
@LinAlg yes, but it would not be easy especially when $k$ is large (e.g 100 or more)
– Babak
13 hours ago
@LinAlg yes, but it would not be easy especially when $k$ is large (e.g 100 or more)
– Babak
13 hours ago
1
1
that the problem is not easy is to be expected, since setting a lower bound on a norm is difficult from an optimization perspective
– LinAlg
13 hours ago
that the problem is not easy is to be expected, since setting a lower bound on a norm is difficult from an optimization perspective
– LinAlg
13 hours ago
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2995312%2fhow-to-equally-balance-x-1-over-the-columns-of-x%23new-answer', 'question_page');
}
);
Post as a guest
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
you could have a different $lambda$ for each column, and keep adjusting $lambda$ until you are satisfied
– LinAlg
14 hours ago
@LinAlg yes, but it would not be easy especially when $k$ is large (e.g 100 or more)
– Babak
13 hours ago
1
that the problem is not easy is to be expected, since setting a lower bound on a norm is difficult from an optimization perspective
– LinAlg
13 hours ago