Mixing
Difference between Ax-b and f(A)x-f(b)?
$begingroup$
I have an understanding how the solution of the following convex optimization behaves for a particular data: $min(norm(Ax-b),2)+norm(x,1))$. In this case $A$ is a known matrix and $b$ is a known vector. We want to find $x$. I want to analyse the case when both $A$ and $b$ are changed by a function. This means the new problem is $min(norm(f(A)x-f(b)),2)+norm(x,1))$.
This question may be vague as I have not specified the function $f$. Can there be a class of function for which the problem would not deviate much? Or is there any such $f$ which is actually used to get a better or useful optimization problem compare to the original one? Specifically, if the function $f$ is non-linear is there any application, or foe some non-linear $f$ the optimization problem become more attractive than the original one.
linear-algebra functional-analysis optimization soft-question
asked Jan 24 at 1:22
CreatorCreator
1,21811028
$endgroup$
|
show 1 more comment
$begingroup$
I have an understanding how the solution of the following convex optimization behaves for a particular data: $min(norm(Ax-b),2)+norm(x,1))$. In this case $A$ is a known matrix and $b$ is a known vector. We want to find $x$. I want to analyse the case when both $A$ and $b$ are changed by a function. This means the new problem is $min(norm(f(A)x-f(b)),2)+norm(x,1))$.
This question may be vague as I have not specified the function $f$. Can there be a class of function for which the problem would not deviate much? Or is there any such $f$ which is actually used to get a better or useful optimization problem compare to the original one? Specifically, if the function $f$ is non-linear is there any application, or foe some non-linear $f$ the optimization problem become more attractive than the original one.
linear-algebra functional-analysis optimization soft-question
asked Jan 24 at 1:22
CreatorCreator
1,21811028
$endgroup$
$begingroup$
If you specify a linear function $f$, it won't really change the problem significantly.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 1:41
$begingroup$
@nathan.j.mcdougall, yes the issue is my function is non-linear. I added a new sentence, thanks for pointing out.
$endgroup$
– Creator
Jan 24 at 1:46
$begingroup$
You use $f$ on both matrices and vectors. How should that extension be interpreted? Or is $f$ an element-wise function?
$endgroup$
– nathan.j.mcdougall
Jan 24 at 1:52
$begingroup$
@nathan.j.mcdougall element wise.
$endgroup$
– Creator
Jan 24 at 2:03
$begingroup$
If there is any solution to the original problem, the solution to the transformed problem will be identical, except it would have $f(A)$ instead of $A$, and $f(b)$ instead of $b$. The only way using $f$ would be useful, then, would be if there is a value of $A^*$ and $b^*$ which makes the problem significantly easier, which we can "force" $f$ to map to. The obvious example is $f=0$, but that's linear. I'm very skeptical that there's anything else.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 2:13
|
show 1 more comment
$begingroup$
I have an understanding how the solution of the following convex optimization behaves for a particular data: $min(norm(Ax-b),2)+norm(x,1))$. In this case $A$ is a known matrix and $b$ is a known vector. We want to find $x$. I want to analyse the case when both $A$ and $b$ are changed by a function. This means the new problem is $min(norm(f(A)x-f(b)),2)+norm(x,1))$.
This question may be vague as I have not specified the function $f$. Can there be a class of function for which the problem would not deviate much? Or is there any such $f$ which is actually used to get a better or useful optimization problem compare to the original one? Specifically, if the function $f$ is non-linear is there any application, or foe some non-linear $f$ the optimization problem become more attractive than the original one.
linear-algebra functional-analysis optimization soft-question
asked Jan 24 at 1:22
CreatorCreator
1,21811028
$endgroup$
I have an understanding how the solution of the following convex optimization behaves for a particular data: $min(norm(Ax-b),2)+norm(x,1))$. In this case $A$ is a known matrix and $b$ is a known vector. We want to find $x$. I want to analyse the case when both $A$ and $b$ are changed by a function. This means the new problem is $min(norm(f(A)x-f(b)),2)+norm(x,1))$.
This question may be vague as I have not specified the function $f$. Can there be a class of function for which the problem would not deviate much? Or is there any such $f$ which is actually used to get a better or useful optimization problem compare to the original one? Specifically, if the function $f$ is non-linear is there any application, or foe some non-linear $f$ the optimization problem become more attractive than the original one.
linear-algebra functional-analysis optimization soft-question
linear-algebra functional-analysis optimization soft-question
asked Jan 24 at 1:22
CreatorCreator
1,21811028
asked Jan 24 at 1:22
CreatorCreator
1,21811028
edited Jan 24 at 1:47
Creator
asked Jan 24 at 1:22
CreatorCreator
1,21811028
asked Jan 24 at 1:22
CreatorCreator
1,21811028
asked Jan 24 at 1:22
CreatorCreator
1,21811028
1,21811028
$begingroup$
If you specify a linear function $f$, it won't really change the problem significantly.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 1:41
$begingroup$
@nathan.j.mcdougall, yes the issue is my function is non-linear. I added a new sentence, thanks for pointing out.
$endgroup$
– Creator
Jan 24 at 1:46
$begingroup$
You use $f$ on both matrices and vectors. How should that extension be interpreted? Or is $f$ an element-wise function?
$endgroup$
– nathan.j.mcdougall
Jan 24 at 1:52
$begingroup$
@nathan.j.mcdougall element wise.
$endgroup$
– Creator
Jan 24 at 2:03
$begingroup$
If there is any solution to the original problem, the solution to the transformed problem will be identical, except it would have $f(A)$ instead of $A$, and $f(b)$ instead of $b$. The only way using $f$ would be useful, then, would be if there is a value of $A^*$ and $b^*$ which makes the problem significantly easier, which we can "force" $f$ to map to. The obvious example is $f=0$, but that's linear. I'm very skeptical that there's anything else.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 2:13
|
show 1 more comment
$begingroup$
If you specify a linear function $f$, it won't really change the problem significantly.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 1:41
$begingroup$
@nathan.j.mcdougall, yes the issue is my function is non-linear. I added a new sentence, thanks for pointing out.
$endgroup$
– Creator
Jan 24 at 1:46
$begingroup$
You use $f$ on both matrices and vectors. How should that extension be interpreted? Or is $f$ an element-wise function?
$endgroup$
– nathan.j.mcdougall
Jan 24 at 1:52
$begingroup$
@nathan.j.mcdougall element wise.
$endgroup$
– Creator
Jan 24 at 2:03
$begingroup$
If there is any solution to the original problem, the solution to the transformed problem will be identical, except it would have $f(A)$ instead of $A$, and $f(b)$ instead of $b$. The only way using $f$ would be useful, then, would be if there is a value of $A^*$ and $b^*$ which makes the problem significantly easier, which we can "force" $f$ to map to. The obvious example is $f=0$, but that's linear. I'm very skeptical that there's anything else.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 2:13
$begingroup$
If you specify a linear function $f$, it won't really change the problem significantly.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 1:41
$begingroup$
If you specify a linear function $f$, it won't really change the problem significantly.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 1:41
$begingroup$
@nathan.j.mcdougall, yes the issue is my function is non-linear. I added a new sentence, thanks for pointing out.
$endgroup$
– Creator
Jan 24 at 1:46
$begingroup$
@nathan.j.mcdougall, yes the issue is my function is non-linear. I added a new sentence, thanks for pointing out.
$endgroup$
– Creator
Jan 24 at 1:46
$begingroup$
You use $f$ on both matrices and vectors. How should that extension be interpreted? Or is $f$ an element-wise function?
$endgroup$
– nathan.j.mcdougall
Jan 24 at 1:52
$begingroup$
You use $f$ on both matrices and vectors. How should that extension be interpreted? Or is $f$ an element-wise function?
$endgroup$
– nathan.j.mcdougall
Jan 24 at 1:52
$begingroup$
@nathan.j.mcdougall element wise.
$endgroup$
– Creator
Jan 24 at 2:03
$begingroup$
@nathan.j.mcdougall element wise.
$endgroup$
– Creator
Jan 24 at 2:03
$begingroup$
If there is any solution to the original problem, the solution to the transformed problem will be identical, except it would have $f(A)$ instead of $A$, and $f(b)$ instead of $b$. The only way using $f$ would be useful, then, would be if there is a value of $A^*$ and $b^*$ which makes the problem significantly easier, which we can "force" $f$ to map to. The obvious example is $f=0$, but that's linear. I'm very skeptical that there's anything else.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 2:13
$begingroup$
If there is any solution to the original problem, the solution to the transformed problem will be identical, except it would have $f(A)$ instead of $A$, and $f(b)$ instead of $b$. The only way using $f$ would be useful, then, would be if there is a value of $A^*$ and $b^*$ which makes the problem significantly easier, which we can "force" $f$ to map to. The obvious example is $f=0$, but that's linear. I'm very skeptical that there's anything else.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 2:13
|
show 1 more comment
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$begingroup$
If you specify a linear function $f$, it won't really change the problem significantly.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 1:41
$begingroup$
@nathan.j.mcdougall, yes the issue is my function is non-linear. I added a new sentence, thanks for pointing out.
$endgroup$
– Creator
Jan 24 at 1:46
$begingroup$
You use $f$ on both matrices and vectors. How should that extension be interpreted? Or is $f$ an element-wise function?
$endgroup$
– nathan.j.mcdougall
Jan 24 at 1:52
$begingroup$
@nathan.j.mcdougall element wise.
$endgroup$
– Creator
Jan 24 at 2:03
$begingroup$
If there is any solution to the original problem, the solution to the transformed problem will be identical, except it would have $f(A)$ instead of $A$, and $f(b)$ instead of $b$. The only way using $f$ would be useful, then, would be if there is a value of $A^*$ and $b^*$ which makes the problem significantly easier, which we can "force" $f$ to map to. The obvious example is $f=0$, but that's linear. I'm very skeptical that there's anything else.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 2:13