An equality with vectors












0












$begingroup$


Here is a problem with its solution.



Let $mathbf{s}_{ntimes1}$ is a vector with elements $s_{j}inleft[-1,1right]$
for $j=1,2,ldots,n$. I have the equality
$$
lambdaalphamathbf{s}=mathbf{a}
$$
where $lambda$ is an unknown parameter, $alphainleft(0,1right)$
is a fixed constant and $mathbf{a}$ is an $ntimes1$ vector. We
can show that the minimum value of $lambda$ that satisfies this
equality for some $s_{j}inleft[-1,1right]$ is
$$
lambda_{min}=frac{1}{alpha}leftVert mathbf{a}rightVert _{infty}
$$
where $leftVert mathbf{a}rightVert _{infty}=max_{j}left|a_{j}right|$.
I would like to extend this equality with a fixed vector $mathbf{b}$ like this:
$$
lambda(alphamathbf{s}+left(1-alpharight)mathbf{b})=mathbf{a}
$$
The problem is the same: What is the minimum value of $lambda$ (if
exists) that satisfies this equality for some $s_{j}inleft[-1,1right]$?



I tried to use inequalities with absolute values but could not get a solution.






optimization vectors norm







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    is $b$ fixed? do you mean 'for some s_j' (instead of 'for all')?
    $endgroup$
    – LinAlg
    Dec 27 '18 at 21:02










  • $begingroup$
    Yes, thanks. I edited the question.
    $endgroup$
    – mert
    Dec 28 '18 at 20:47










  • $begingroup$
    did you appreciate my answer?
    $endgroup$
    – LinAlg
    Jan 8 at 15:03
















0












$begingroup$


Here is a problem with its solution.



Let $mathbf{s}_{ntimes1}$ is a vector with elements $s_{j}inleft[-1,1right]$
for $j=1,2,ldots,n$. I have the equality
$$
lambdaalphamathbf{s}=mathbf{a}
$$
where $lambda$ is an unknown parameter, $alphainleft(0,1right)$
is a fixed constant and $mathbf{a}$ is an $ntimes1$ vector. We
can show that the minimum value of $lambda$ that satisfies this
equality for some $s_{j}inleft[-1,1right]$ is
$$
lambda_{min}=frac{1}{alpha}leftVert mathbf{a}rightVert _{infty}
$$
where $leftVert mathbf{a}rightVert _{infty}=max_{j}left|a_{j}right|$.
I would like to extend this equality with a fixed vector $mathbf{b}$ like this:
$$
lambda(alphamathbf{s}+left(1-alpharight)mathbf{b})=mathbf{a}
$$
The problem is the same: What is the minimum value of $lambda$ (if
exists) that satisfies this equality for some $s_{j}inleft[-1,1right]$?



I tried to use inequalities with absolute values but could not get a solution.






optimization vectors norm







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    is $b$ fixed? do you mean 'for some s_j' (instead of 'for all')?
    $endgroup$
    – LinAlg
    Dec 27 '18 at 21:02










  • $begingroup$
    Yes, thanks. I edited the question.
    $endgroup$
    – mert
    Dec 28 '18 at 20:47










  • $begingroup$
    did you appreciate my answer?
    $endgroup$
    – LinAlg
    Jan 8 at 15:03














0












0








0





$begingroup$


Here is a problem with its solution.



Let $mathbf{s}_{ntimes1}$ is a vector with elements $s_{j}inleft[-1,1right]$
for $j=1,2,ldots,n$. I have the equality
$$
lambdaalphamathbf{s}=mathbf{a}
$$
where $lambda$ is an unknown parameter, $alphainleft(0,1right)$
is a fixed constant and $mathbf{a}$ is an $ntimes1$ vector. We
can show that the minimum value of $lambda$ that satisfies this
equality for some $s_{j}inleft[-1,1right]$ is
$$
lambda_{min}=frac{1}{alpha}leftVert mathbf{a}rightVert _{infty}
$$
where $leftVert mathbf{a}rightVert _{infty}=max_{j}left|a_{j}right|$.
I would like to extend this equality with a fixed vector $mathbf{b}$ like this:
$$
lambda(alphamathbf{s}+left(1-alpharight)mathbf{b})=mathbf{a}
$$
The problem is the same: What is the minimum value of $lambda$ (if
exists) that satisfies this equality for some $s_{j}inleft[-1,1right]$?



I tried to use inequalities with absolute values but could not get a solution.






optimization vectors norm







share|cite|improve this question











$endgroup$




Here is a problem with its solution.



Let $mathbf{s}_{ntimes1}$ is a vector with elements $s_{j}inleft[-1,1right]$
for $j=1,2,ldots,n$. I have the equality
$$
lambdaalphamathbf{s}=mathbf{a}
$$
where $lambda$ is an unknown parameter, $alphainleft(0,1right)$
is a fixed constant and $mathbf{a}$ is an $ntimes1$ vector. We
can show that the minimum value of $lambda$ that satisfies this
equality for some $s_{j}inleft[-1,1right]$ is
$$
lambda_{min}=frac{1}{alpha}leftVert mathbf{a}rightVert _{infty}
$$
where $leftVert mathbf{a}rightVert _{infty}=max_{j}left|a_{j}right|$.
I would like to extend this equality with a fixed vector $mathbf{b}$ like this:
$$
lambda(alphamathbf{s}+left(1-alpharight)mathbf{b})=mathbf{a}
$$
The problem is the same: What is the minimum value of $lambda$ (if
exists) that satisfies this equality for some $s_{j}inleft[-1,1right]$?



I tried to use inequalities with absolute values but could not get a solution.






optimization vectors norm




optimization vectors norm






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '18 at 20:48







mert

















asked Dec 27 '18 at 8:03









mertmert

53529




53529








  • 1




    $begingroup$
    is $b$ fixed? do you mean 'for some s_j' (instead of 'for all')?
    $endgroup$
    – LinAlg
    Dec 27 '18 at 21:02










  • $begingroup$
    Yes, thanks. I edited the question.
    $endgroup$
    – mert
    Dec 28 '18 at 20:47










  • $begingroup$
    did you appreciate my answer?
    $endgroup$
    – LinAlg
    Jan 8 at 15:03














  • 1




    $begingroup$
    is $b$ fixed? do you mean 'for some s_j' (instead of 'for all')?
    $endgroup$
    – LinAlg
    Dec 27 '18 at 21:02










  • $begingroup$
    Yes, thanks. I edited the question.
    $endgroup$
    – mert
    Dec 28 '18 at 20:47










  • $begingroup$
    did you appreciate my answer?
    $endgroup$
    – LinAlg
    Jan 8 at 15:03








1




1




$begingroup$
is $b$ fixed? do you mean 'for some s_j' (instead of 'for all')?
$endgroup$
– LinAlg
Dec 27 '18 at 21:02




$begingroup$
is $b$ fixed? do you mean 'for some s_j' (instead of 'for all')?
$endgroup$
– LinAlg
Dec 27 '18 at 21:02












$begingroup$
Yes, thanks. I edited the question.
$endgroup$
– mert
Dec 28 '18 at 20:47




$begingroup$
Yes, thanks. I edited the question.
$endgroup$
– mert
Dec 28 '18 at 20:47












$begingroup$
did you appreciate my answer?
$endgroup$
– LinAlg
Jan 8 at 15:03




$begingroup$
did you appreciate my answer?
$endgroup$
– LinAlg
Jan 8 at 15:03










1 Answer
1






active

oldest

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0












$begingroup$

Your problem can be summarized as:
$$min_{x in mathbb{R},y in [-alpha,alpha]^n} { x : xy + x(1-alpha)b = a}$$
Substitute $xy=z$:
$$min_{x in mathbb{R},z in [-alpha x,alpha x ]^n} { x : z + x(1-alpha)b = a}$$
The dual problem is:
$$ max_{v in mathbb{R}^n,w_1 in mathbb{R}_+^n,w_2 in mathbb{R}_+^n} { a^Tv : (1-alpha)b^Tv + alpha e^T(w_1+w_2) = 1, ; e^T(v + w_1-w_2) = 0}$$
I do not see an immediate solution to any of these problems, but the last two problems you can just feed to a linear optimization solver.






share|cite|improve this answer









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    $begingroup$

    Your problem can be summarized as:
    $$min_{x in mathbb{R},y in [-alpha,alpha]^n} { x : xy + x(1-alpha)b = a}$$
    Substitute $xy=z$:
    $$min_{x in mathbb{R},z in [-alpha x,alpha x ]^n} { x : z + x(1-alpha)b = a}$$
    The dual problem is:
    $$ max_{v in mathbb{R}^n,w_1 in mathbb{R}_+^n,w_2 in mathbb{R}_+^n} { a^Tv : (1-alpha)b^Tv + alpha e^T(w_1+w_2) = 1, ; e^T(v + w_1-w_2) = 0}$$
    I do not see an immediate solution to any of these problems, but the last two problems you can just feed to a linear optimization solver.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Your problem can be summarized as:
      $$min_{x in mathbb{R},y in [-alpha,alpha]^n} { x : xy + x(1-alpha)b = a}$$
      Substitute $xy=z$:
      $$min_{x in mathbb{R},z in [-alpha x,alpha x ]^n} { x : z + x(1-alpha)b = a}$$
      The dual problem is:
      $$ max_{v in mathbb{R}^n,w_1 in mathbb{R}_+^n,w_2 in mathbb{R}_+^n} { a^Tv : (1-alpha)b^Tv + alpha e^T(w_1+w_2) = 1, ; e^T(v + w_1-w_2) = 0}$$
      I do not see an immediate solution to any of these problems, but the last two problems you can just feed to a linear optimization solver.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Your problem can be summarized as:
        $$min_{x in mathbb{R},y in [-alpha,alpha]^n} { x : xy + x(1-alpha)b = a}$$
        Substitute $xy=z$:
        $$min_{x in mathbb{R},z in [-alpha x,alpha x ]^n} { x : z + x(1-alpha)b = a}$$
        The dual problem is:
        $$ max_{v in mathbb{R}^n,w_1 in mathbb{R}_+^n,w_2 in mathbb{R}_+^n} { a^Tv : (1-alpha)b^Tv + alpha e^T(w_1+w_2) = 1, ; e^T(v + w_1-w_2) = 0}$$
        I do not see an immediate solution to any of these problems, but the last two problems you can just feed to a linear optimization solver.






        share|cite|improve this answer









        $endgroup$



        Your problem can be summarized as:
        $$min_{x in mathbb{R},y in [-alpha,alpha]^n} { x : xy + x(1-alpha)b = a}$$
        Substitute $xy=z$:
        $$min_{x in mathbb{R},z in [-alpha x,alpha x ]^n} { x : z + x(1-alpha)b = a}$$
        The dual problem is:
        $$ max_{v in mathbb{R}^n,w_1 in mathbb{R}_+^n,w_2 in mathbb{R}_+^n} { a^Tv : (1-alpha)b^Tv + alpha e^T(w_1+w_2) = 1, ; e^T(v + w_1-w_2) = 0}$$
        I do not see an immediate solution to any of these problems, but the last two problems you can just feed to a linear optimization solver.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 28 '18 at 21:50









        LinAlgLinAlg

        9,7541521




        9,7541521






























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