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

votes


















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 Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3053694%2fan-equality-with-vectors%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    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












      $begingroup$

      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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3053694%2fan-equality-with-vectors%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Ellipse (mathématiques)

            Image
            Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
            Read more

            Quarter-circle Tiles

            Image
            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
            Read more

            Mont Emei

            Image
            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
            Read more