Finding the norm of a matrix given an orthonormal basis












1












$begingroup$


So I am working on practice problems out of a textbook I am working through over break. One of the problems asks to show that given a complex $ntimes n$ matrix $X$ and any orthonormal basis ${u_1,u_2,...,u_n}$ of $mathbb{C}^n$ that the Hilbert-Schmidt norm of the matrix given as
$$| X |=bigg(sum_{j,k=1}^n|X_{jk}|^2bigg)^{1/2}$$
can be expressed as
$$| X |^2=sum_{j,k=1}^n|langle u_j,Xu_krangle|^2$$



I figured that, if we consider the coordinate vectors w.r.t. our basis, we essentially have
$$langle u_j,Xu_krangle rightarrowlangle e_j,Xe_krangle$$
where it follows that $$langle e_j,Xe_krangle=X_{jk}$$
However, I am rusty on my linear algebra and I know this is not fully correct. I was thinking that, in order to use the coordinate vectors, I need to change $X$ to be w.r.t. our orthonormal basis but I keep confusing myself in the process.






linear-algebra norm inner-product-space







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    So I am working on practice problems out of a textbook I am working through over break. One of the problems asks to show that given a complex $ntimes n$ matrix $X$ and any orthonormal basis ${u_1,u_2,...,u_n}$ of $mathbb{C}^n$ that the Hilbert-Schmidt norm of the matrix given as
    $$| X |=bigg(sum_{j,k=1}^n|X_{jk}|^2bigg)^{1/2}$$
    can be expressed as
    $$| X |^2=sum_{j,k=1}^n|langle u_j,Xu_krangle|^2$$



    I figured that, if we consider the coordinate vectors w.r.t. our basis, we essentially have
    $$langle u_j,Xu_krangle rightarrowlangle e_j,Xe_krangle$$
    where it follows that $$langle e_j,Xe_krangle=X_{jk}$$
    However, I am rusty on my linear algebra and I know this is not fully correct. I was thinking that, in order to use the coordinate vectors, I need to change $X$ to be w.r.t. our orthonormal basis but I keep confusing myself in the process.






    linear-algebra norm inner-product-space







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      So I am working on practice problems out of a textbook I am working through over break. One of the problems asks to show that given a complex $ntimes n$ matrix $X$ and any orthonormal basis ${u_1,u_2,...,u_n}$ of $mathbb{C}^n$ that the Hilbert-Schmidt norm of the matrix given as
      $$| X |=bigg(sum_{j,k=1}^n|X_{jk}|^2bigg)^{1/2}$$
      can be expressed as
      $$| X |^2=sum_{j,k=1}^n|langle u_j,Xu_krangle|^2$$



      I figured that, if we consider the coordinate vectors w.r.t. our basis, we essentially have
      $$langle u_j,Xu_krangle rightarrowlangle e_j,Xe_krangle$$
      where it follows that $$langle e_j,Xe_krangle=X_{jk}$$
      However, I am rusty on my linear algebra and I know this is not fully correct. I was thinking that, in order to use the coordinate vectors, I need to change $X$ to be w.r.t. our orthonormal basis but I keep confusing myself in the process.






      linear-algebra norm inner-product-space







      share|cite|improve this question









      $endgroup$




      So I am working on practice problems out of a textbook I am working through over break. One of the problems asks to show that given a complex $ntimes n$ matrix $X$ and any orthonormal basis ${u_1,u_2,...,u_n}$ of $mathbb{C}^n$ that the Hilbert-Schmidt norm of the matrix given as
      $$| X |=bigg(sum_{j,k=1}^n|X_{jk}|^2bigg)^{1/2}$$
      can be expressed as
      $$| X |^2=sum_{j,k=1}^n|langle u_j,Xu_krangle|^2$$



      I figured that, if we consider the coordinate vectors w.r.t. our basis, we essentially have
      $$langle u_j,Xu_krangle rightarrowlangle e_j,Xe_krangle$$
      where it follows that $$langle e_j,Xe_krangle=X_{jk}$$
      However, I am rusty on my linear algebra and I know this is not fully correct. I was thinking that, in order to use the coordinate vectors, I need to change $X$ to be w.r.t. our orthonormal basis but I keep confusing myself in the process.






      linear-algebra norm inner-product-space




      linear-algebra norm inner-product-space






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 2 at 22:25









      Ian B.Ian B.

      898




      898






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          To complete you may use two things:




          1. To figure out that
            $$
            |X|^2=bigg(sum_{j,k=1}^n|X_{jk}|^2bigg)=operatorname{trace}X^*X
            $$
            ($X^*$ is the Hermitian conjugate of $X$)

          2. To know the cyclic trace property
            $$
            operatorname{trace}(AB)=operatorname{trace}(BA).
            $$
            Then everything comes nicely: for any unitary matrix $U$ (the columns are an orthonormal basis $u_1ldots,u_n$) we have $UU^*=I$, so
            begin{align}
            |X|^2&=operatorname{trace}X^*X=operatorname{trace}X^*XUU^*=\
            &=operatorname{trace}U^*X^*XU=operatorname{trace}U^*X^*UU^*XU=|U^*XU|^2.
            end{align}
            Here any element of $U^*XU$ is
            $$
            u_j^*Xu_k=langle u_j,Xu_krangle.
            $$






          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%2f3060044%2ffinding-the-norm-of-a-matrix-given-an-orthonormal-basis%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









            1












            $begingroup$

            To complete you may use two things:




            1. To figure out that
              $$
              |X|^2=bigg(sum_{j,k=1}^n|X_{jk}|^2bigg)=operatorname{trace}X^*X
              $$
              ($X^*$ is the Hermitian conjugate of $X$)

            2. To know the cyclic trace property
              $$
              operatorname{trace}(AB)=operatorname{trace}(BA).
              $$
              Then everything comes nicely: for any unitary matrix $U$ (the columns are an orthonormal basis $u_1ldots,u_n$) we have $UU^*=I$, so
              begin{align}
              |X|^2&=operatorname{trace}X^*X=operatorname{trace}X^*XUU^*=\
              &=operatorname{trace}U^*X^*XU=operatorname{trace}U^*X^*UU^*XU=|U^*XU|^2.
              end{align}
              Here any element of $U^*XU$ is
              $$
              u_j^*Xu_k=langle u_j,Xu_krangle.
              $$






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              To complete you may use two things:




              1. To figure out that
                $$
                |X|^2=bigg(sum_{j,k=1}^n|X_{jk}|^2bigg)=operatorname{trace}X^*X
                $$
                ($X^*$ is the Hermitian conjugate of $X$)

              2. To know the cyclic trace property
                $$
                operatorname{trace}(AB)=operatorname{trace}(BA).
                $$
                Then everything comes nicely: for any unitary matrix $U$ (the columns are an orthonormal basis $u_1ldots,u_n$) we have $UU^*=I$, so
                begin{align}
                |X|^2&=operatorname{trace}X^*X=operatorname{trace}X^*XUU^*=\
                &=operatorname{trace}U^*X^*XU=operatorname{trace}U^*X^*UU^*XU=|U^*XU|^2.
                end{align}
                Here any element of $U^*XU$ is
                $$
                u_j^*Xu_k=langle u_j,Xu_krangle.
                $$






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                To complete you may use two things:




                1. To figure out that
                  $$
                  |X|^2=bigg(sum_{j,k=1}^n|X_{jk}|^2bigg)=operatorname{trace}X^*X
                  $$
                  ($X^*$ is the Hermitian conjugate of $X$)

                2. To know the cyclic trace property
                  $$
                  operatorname{trace}(AB)=operatorname{trace}(BA).
                  $$
                  Then everything comes nicely: for any unitary matrix $U$ (the columns are an orthonormal basis $u_1ldots,u_n$) we have $UU^*=I$, so
                  begin{align}
                  |X|^2&=operatorname{trace}X^*X=operatorname{trace}X^*XUU^*=\
                  &=operatorname{trace}U^*X^*XU=operatorname{trace}U^*X^*UU^*XU=|U^*XU|^2.
                  end{align}
                  Here any element of $U^*XU$ is
                  $$
                  u_j^*Xu_k=langle u_j,Xu_krangle.
                  $$






                share|cite|improve this answer











                $endgroup$



                To complete you may use two things:




                1. To figure out that
                  $$
                  |X|^2=bigg(sum_{j,k=1}^n|X_{jk}|^2bigg)=operatorname{trace}X^*X
                  $$
                  ($X^*$ is the Hermitian conjugate of $X$)

                2. To know the cyclic trace property
                  $$
                  operatorname{trace}(AB)=operatorname{trace}(BA).
                  $$
                  Then everything comes nicely: for any unitary matrix $U$ (the columns are an orthonormal basis $u_1ldots,u_n$) we have $UU^*=I$, so
                  begin{align}
                  |X|^2&=operatorname{trace}X^*X=operatorname{trace}X^*XUU^*=\
                  &=operatorname{trace}U^*X^*XU=operatorname{trace}U^*X^*UU^*XU=|U^*XU|^2.
                  end{align}
                  Here any element of $U^*XU$ is
                  $$
                  u_j^*Xu_k=langle u_j,Xu_krangle.
                  $$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 2 at 22:47

























                answered Jan 2 at 22:36









                A.Γ.A.Γ.

                22.7k32656




                22.7k32656






























                    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%2f3060044%2ffinding-the-norm-of-a-matrix-given-an-orthonormal-basis%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

                    'app-layout' is not a known element: how to share Component with different Modules

                    Image
                    0 My application was working fine until I decided to go with lazy loading: So, my shared component looks like this: import { Component, Renderer2 } from '@angular/core'; export interface FormModel { captcha?: string; } @Component({ selector: 'app-layout', templateUrl: './app-layout.component.html', styleUrls: ['./app-layout.css'] }) export class FullLayoutComponent { } app-layout.component.ts I am using this component in my Another component say school-home.component.html like this: <app-layout> </app-layout> and it was working fine until I created a Module school-home.module.ts like this: import { NgModule } from '@angular/core'; import { CommonModule } from '@angular/common'; import { SchoolhomeRoutingModule ...
                    Read more

                    android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 This is a simple project showing cubes on Unity, exported to android project and imported on Android studio. On maifest file it showed following warning: Expecting <uses-feature android:name="android.software.leanback" android:required="false" /> tag I have no information about it. And also on default emptyview android project does not use/mention this feature. I know it is required by TV's and such but why its not added by default? Why is this? And Should I concern about it? android unity3d ...
                    Read more

                    WPF add header to Image with URL pettitions [duplicate]

                    Image
                    up vote 0 down vote favorite This question already has an answer here: Image source URI requiring authorization 3 answers Add header to all* HTTP requests on a machine 1 answer first of all sorry for my english. Recently I added some security in the backend of my project, and now I need to send a validation code(apiKey) inside the headers to retrieve the images. I have a problem with ALL images in my project that I had attached in my UI. For example, before I had: <Image Height="90" Source="{Binding Video.Thumbnail}"/> Now I need to change my code to create a CUSTOM petition add the apiKey in the header:...
                    Read more