Proving that $Awin langle wrangle implies A$ is of the form $lambda I_n$












0












$begingroup$


If $w=begin{bmatrix} w_1\ vdots\w_n end{bmatrix}$ is a vector in $K^n$ for a field $K$ and
$A=
begin{bmatrix}
lambda & & \
& ddots & a_{ij}\
& & lambda
end{bmatrix}$ is a $ntimes n $ matrix with entries on $K$ and $lambdain K^{*}$ then $Aw=lambda w+e_i a_{ij} w_jnotinlangle wrangle$ if $ine j$. And so $langle A wranglene langle wrangle$ where $langle wrangle$ is the $1$ dimensional vector space generated by $w$.



But if $A$ is a general matrix in $GL_n(K)$ that is not of the form $lambda I_n$ then $Aw=sumlimits_{i=1}^nsumlimits_{j=1}^n a_{ij}w_je_i$.



How to show that $Awnotin langle wrangle$?



Can we decompose $A$ into $sum A_{ij}$ where each $A_{ij}$ has a simple form like above (so $A_{ij}wnotin langle wrangle$) and say that $Aw=underbrace{sum A_{ij}w}_{notin langle wrangle} notin langle wrangle$?



For context the question asks to prove that $AW=W~~forall Win{text{sub-vector-spaces of } K^n text{ with dimension 1}} iff A=lambda I_n$ for some $lambdain K^{*}$






linear-algebra group-theory linear-groups general-linear-group







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    If $w=begin{bmatrix} w_1\ vdots\w_n end{bmatrix}$ is a vector in $K^n$ for a field $K$ and
    $A=
    begin{bmatrix}
    lambda & & \
    & ddots & a_{ij}\
    & & lambda
    end{bmatrix}$     is a $ntimes n $ matrix with entries on $K$ and $lambdain K^{*}$ then $Aw=lambda w+e_i a_{ij} w_jnotinlangle wrangle$ if $ine j$. And so $langle A wranglene langle wrangle$ where $langle wrangle$ is the $1$ dimensional vector space generated by $w$.



    But if $A$ is a general matrix in $GL_n(K)$ that is not of the form $lambda I_n$ then $Aw=sumlimits_{i=1}^nsumlimits_{j=1}^n a_{ij}w_je_i$.



    How to show that $Awnotin langle wrangle$?



    Can we decompose $A$ into $sum A_{ij}$ where each $A_{ij}$ has a simple form like above (so $A_{ij}wnotin langle wrangle$) and say that $Aw=underbrace{sum A_{ij}w}_{notin langle wrangle} notin langle wrangle$?



    For context the question asks to prove that $AW=W~~forall Win{text{sub-vector-spaces of } K^n text{ with dimension 1}} iff A=lambda I_n$ for some $lambdain K^{*}$






    linear-algebra group-theory linear-groups general-linear-group







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      If $w=begin{bmatrix} w_1\ vdots\w_n end{bmatrix}$ is a vector in $K^n$ for a field $K$ and
      $A=
      begin{bmatrix}
      lambda & & \
      & ddots & a_{ij}\
      & & lambda
      end{bmatrix}$       is a $ntimes n $ matrix with entries on $K$ and $lambdain K^{*}$ then $Aw=lambda w+e_i a_{ij} w_jnotinlangle wrangle$ if $ine j$. And so $langle A wranglene langle wrangle$ where $langle wrangle$ is the $1$ dimensional vector space generated by $w$.



      But if $A$ is a general matrix in $GL_n(K)$ that is not of the form $lambda I_n$ then $Aw=sumlimits_{i=1}^nsumlimits_{j=1}^n a_{ij}w_je_i$.



      How to show that $Awnotin langle wrangle$?



      Can we decompose $A$ into $sum A_{ij}$ where each $A_{ij}$ has a simple form like above (so $A_{ij}wnotin langle wrangle$) and say that $Aw=underbrace{sum A_{ij}w}_{notin langle wrangle} notin langle wrangle$?



      For context the question asks to prove that $AW=W~~forall Win{text{sub-vector-spaces of } K^n text{ with dimension 1}} iff A=lambda I_n$ for some $lambdain K^{*}$






      linear-algebra group-theory linear-groups general-linear-group







      share|cite|improve this question











      $endgroup$




      If $w=begin{bmatrix} w_1\ vdots\w_n end{bmatrix}$ is a vector in $K^n$ for a field $K$ and
      $A=
      begin{bmatrix}
      lambda & & \
      & ddots & a_{ij}\
      & & lambda
      end{bmatrix}$       is a $ntimes n $ matrix with entries on $K$ and $lambdain K^{*}$ then $Aw=lambda w+e_i a_{ij} w_jnotinlangle wrangle$ if $ine j$. And so $langle A wranglene langle wrangle$ where $langle wrangle$ is the $1$ dimensional vector space generated by $w$.



      But if $A$ is a general matrix in $GL_n(K)$ that is not of the form $lambda I_n$ then $Aw=sumlimits_{i=1}^nsumlimits_{j=1}^n a_{ij}w_je_i$.



      How to show that $Awnotin langle wrangle$?



      Can we decompose $A$ into $sum A_{ij}$ where each $A_{ij}$ has a simple form like above (so $A_{ij}wnotin langle wrangle$) and say that $Aw=underbrace{sum A_{ij}w}_{notin langle wrangle} notin langle wrangle$?



      For context the question asks to prove that $AW=W~~forall Win{text{sub-vector-spaces of } K^n text{ with dimension 1}} iff A=lambda I_n$ for some $lambdain K^{*}$






      linear-algebra group-theory linear-groups general-linear-group




      linear-algebra group-theory linear-groups general-linear-group






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 8 at 7:37









      Martin Sleziak

      44.8k10119273




      44.8k10119273










      asked Jan 5 at 10:39









      John CataldoJohn Cataldo

      1,1881316




      1,1881316






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          The condition $Awinlangle wrangle$, applied to elements of the standard basis of $K^n$, implies that $A$ is a diagonal matrix. Now, for elements $w_1neq w_2$ of the basis, being linearly independent, $Aw_1=lambda_1 w_1$ and $Aw_2=lambda_2 w_2$ together with $A(w_1+w_2)=lambda(w_1+w_2)$ imply $lambda_1=lambda_2(=lambda)$.






          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%2f3062599%2fproving-that-aw-in-langle-w-rangle-implies-a-is-of-the-form-lambda-i-n%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









            2












            $begingroup$

            The condition $Awinlangle wrangle$, applied to elements of the standard basis of $K^n$, implies that $A$ is a diagonal matrix. Now, for elements $w_1neq w_2$ of the basis, being linearly independent, $Aw_1=lambda_1 w_1$ and $Aw_2=lambda_2 w_2$ together with $A(w_1+w_2)=lambda(w_1+w_2)$ imply $lambda_1=lambda_2(=lambda)$.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              The condition $Awinlangle wrangle$, applied to elements of the standard basis of $K^n$, implies that $A$ is a diagonal matrix. Now, for elements $w_1neq w_2$ of the basis, being linearly independent, $Aw_1=lambda_1 w_1$ and $Aw_2=lambda_2 w_2$ together with $A(w_1+w_2)=lambda(w_1+w_2)$ imply $lambda_1=lambda_2(=lambda)$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                The condition $Awinlangle wrangle$, applied to elements of the standard basis of $K^n$, implies that $A$ is a diagonal matrix. Now, for elements $w_1neq w_2$ of the basis, being linearly independent, $Aw_1=lambda_1 w_1$ and $Aw_2=lambda_2 w_2$ together with $A(w_1+w_2)=lambda(w_1+w_2)$ imply $lambda_1=lambda_2(=lambda)$.






                share|cite|improve this answer









                $endgroup$



                The condition $Awinlangle wrangle$, applied to elements of the standard basis of $K^n$, implies that $A$ is a diagonal matrix. Now, for elements $w_1neq w_2$ of the basis, being linearly independent, $Aw_1=lambda_1 w_1$ and $Aw_2=lambda_2 w_2$ together with $A(w_1+w_2)=lambda(w_1+w_2)$ imply $lambda_1=lambda_2(=lambda)$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 5 at 11:14









                metamorphymetamorphy

                3,7021621




                3,7021621






























                    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%2f3062599%2fproving-that-aw-in-langle-w-rangle-implies-a-is-of-the-form-lambda-i-n%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