Finding the norm of a matrix given an orthonormal basis












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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







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    $begingroup$


    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











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      asked Jan 2 at 22:25









      Ian B.Ian B.

      898




      898






















          1 Answer
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          active

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          $begingroup$

          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











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            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






























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