How to equally balance $|X|_1$ over the columns of $X$?

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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
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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 |
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0
down vote
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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 |
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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