Process of finding orthonormal basis of given bilinear form












2














Let $V$ be vector space of $ntimes n$ matrices .



$langle A,B rangle = {rm tr}(A^T B)$



I wanted to find orthonormal basis of it.



I know that if there small say 3 dimension vector say then I would have find first matrix of bilinear form



But here dimension are $n$ so how to tackle such problem






linear-algebra bilinear-form







share|cite|improve this question





























    2














    Let $V$ be vector space of $ntimes n$ matrices .



    $langle A,B rangle = {rm tr}(A^T B)$



    I wanted to find orthonormal basis of it.



    I know that if there small say 3 dimension vector say then I would have find first matrix of bilinear form



    But here dimension are $n$ so how to tackle such problem






    linear-algebra bilinear-form







    share|cite|improve this question



























      2












      2








      2







      Let $V$ be vector space of $ntimes n$ matrices .



      $langle A,B rangle = {rm tr}(A^T B)$



      I wanted to find orthonormal basis of it.



      I know that if there small say 3 dimension vector say then I would have find first matrix of bilinear form



      But here dimension are $n$ so how to tackle such problem






      linear-algebra bilinear-form







      share|cite|improve this question















      Let $V$ be vector space of $ntimes n$ matrices .



      $langle A,B rangle = {rm tr}(A^T B)$



      I wanted to find orthonormal basis of it.



      I know that if there small say 3 dimension vector say then I would have find first matrix of bilinear form



      But here dimension are $n$ so how to tackle such problem






      linear-algebra bilinear-form




      linear-algebra bilinear-form






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 25 at 15:19









      caverac

      13k21028




      13k21028










      asked Nov 25 at 15:07









      MathLover

      45710




      45710






















          1 Answer
          1






          active

          oldest

          votes


















          1














          There are $ntimes n$ elements in this basis, let's label each element as $M^{(k,l)}$, for $k,l = 1,cdots, n$



          $$
          M^{(k,l)}_{i,j} = delta^k_i delta^l_j
          $$



          So the matrix $M^{(k,l)}$ is full of zeros, except for the row $k$ and column $l$, where the value is $1$. The product of any two elements of this basis is



          begin{eqnarray}
          langle M^{(k,l)}, M^{(p,q)} rangle &=& sum_{i,j} M^{(k,l)}_{ji} M^{(p,q)}_{ji} \
          &=& sum_{i,j} delta^k_jdelta^l_i delta^p_j delta^q_i \
          &=& delta^k_p delta^l_q
          end{eqnarray}



          That is, the product is $1$ if $k = p $ and $l = q$, and is $0$ otherwise. So the set of matrices ${M^{(k,l)} }_{k,l = 1}^{n}$ is orthonormal






          share|cite|improve this answer





















          • Sir I could not understand how to algebra of delta is working? it is very new for me.Please can you tell me ?
            – MathLover
            Nov 25 at 15:42










          • @MathLover $$delta^i_j = begin{cases}1 &,& i = j \ 0 &,& {rm otherwise}end{cases}$$ That means that in a sum of the form $sum_i a_i delta^i_j$ all terms vanish, except the one for which $i = j$, $$sum_i a_i delta^i_j = a_j$$
            – caverac
            Nov 25 at 15:45













          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%2f3012952%2fprocess-of-finding-orthonormal-basis-of-given-bilinear-form%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














          There are $ntimes n$ elements in this basis, let's label each element as $M^{(k,l)}$, for $k,l = 1,cdots, n$



          $$
          M^{(k,l)}_{i,j} = delta^k_i delta^l_j
          $$



          So the matrix $M^{(k,l)}$ is full of zeros, except for the row $k$ and column $l$, where the value is $1$. The product of any two elements of this basis is



          begin{eqnarray}
          langle M^{(k,l)}, M^{(p,q)} rangle &=& sum_{i,j} M^{(k,l)}_{ji} M^{(p,q)}_{ji} \
          &=& sum_{i,j} delta^k_jdelta^l_i delta^p_j delta^q_i \
          &=& delta^k_p delta^l_q
          end{eqnarray}



          That is, the product is $1$ if $k = p $ and $l = q$, and is $0$ otherwise. So the set of matrices ${M^{(k,l)} }_{k,l = 1}^{n}$ is orthonormal






          share|cite|improve this answer





















          • Sir I could not understand how to algebra of delta is working? it is very new for me.Please can you tell me ?
            – MathLover
            Nov 25 at 15:42










          • @MathLover $$delta^i_j = begin{cases}1 &,& i = j \ 0 &,& {rm otherwise}end{cases}$$ That means that in a sum of the form $sum_i a_i delta^i_j$ all terms vanish, except the one for which $i = j$, $$sum_i a_i delta^i_j = a_j$$
            – caverac
            Nov 25 at 15:45


















          1














          There are $ntimes n$ elements in this basis, let's label each element as $M^{(k,l)}$, for $k,l = 1,cdots, n$



          $$
          M^{(k,l)}_{i,j} = delta^k_i delta^l_j
          $$



          So the matrix $M^{(k,l)}$ is full of zeros, except for the row $k$ and column $l$, where the value is $1$. The product of any two elements of this basis is



          begin{eqnarray}
          langle M^{(k,l)}, M^{(p,q)} rangle &=& sum_{i,j} M^{(k,l)}_{ji} M^{(p,q)}_{ji} \
          &=& sum_{i,j} delta^k_jdelta^l_i delta^p_j delta^q_i \
          &=& delta^k_p delta^l_q
          end{eqnarray}



          That is, the product is $1$ if $k = p $ and $l = q$, and is $0$ otherwise. So the set of matrices ${M^{(k,l)} }_{k,l = 1}^{n}$ is orthonormal






          share|cite|improve this answer





















          • Sir I could not understand how to algebra of delta is working? it is very new for me.Please can you tell me ?
            – MathLover
            Nov 25 at 15:42










          • @MathLover $$delta^i_j = begin{cases}1 &,& i = j \ 0 &,& {rm otherwise}end{cases}$$ That means that in a sum of the form $sum_i a_i delta^i_j$ all terms vanish, except the one for which $i = j$, $$sum_i a_i delta^i_j = a_j$$
            – caverac
            Nov 25 at 15:45
















          1












          1








          1






          There are $ntimes n$ elements in this basis, let's label each element as $M^{(k,l)}$, for $k,l = 1,cdots, n$



          $$
          M^{(k,l)}_{i,j} = delta^k_i delta^l_j
          $$



          So the matrix $M^{(k,l)}$ is full of zeros, except for the row $k$ and column $l$, where the value is $1$. The product of any two elements of this basis is



          begin{eqnarray}
          langle M^{(k,l)}, M^{(p,q)} rangle &=& sum_{i,j} M^{(k,l)}_{ji} M^{(p,q)}_{ji} \
          &=& sum_{i,j} delta^k_jdelta^l_i delta^p_j delta^q_i \
          &=& delta^k_p delta^l_q
          end{eqnarray}



          That is, the product is $1$ if $k = p $ and $l = q$, and is $0$ otherwise. So the set of matrices ${M^{(k,l)} }_{k,l = 1}^{n}$ is orthonormal






          share|cite|improve this answer












          There are $ntimes n$ elements in this basis, let's label each element as $M^{(k,l)}$, for $k,l = 1,cdots, n$



          $$
          M^{(k,l)}_{i,j} = delta^k_i delta^l_j
          $$



          So the matrix $M^{(k,l)}$ is full of zeros, except for the row $k$ and column $l$, where the value is $1$. The product of any two elements of this basis is



          begin{eqnarray}
          langle M^{(k,l)}, M^{(p,q)} rangle &=& sum_{i,j} M^{(k,l)}_{ji} M^{(p,q)}_{ji} \
          &=& sum_{i,j} delta^k_jdelta^l_i delta^p_j delta^q_i \
          &=& delta^k_p delta^l_q
          end{eqnarray}



          That is, the product is $1$ if $k = p $ and $l = q$, and is $0$ otherwise. So the set of matrices ${M^{(k,l)} }_{k,l = 1}^{n}$ is orthonormal







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 25 at 15:33









          caverac

          13k21028




          13k21028












          • Sir I could not understand how to algebra of delta is working? it is very new for me.Please can you tell me ?
            – MathLover
            Nov 25 at 15:42










          • @MathLover $$delta^i_j = begin{cases}1 &,& i = j \ 0 &,& {rm otherwise}end{cases}$$ That means that in a sum of the form $sum_i a_i delta^i_j$ all terms vanish, except the one for which $i = j$, $$sum_i a_i delta^i_j = a_j$$
            – caverac
            Nov 25 at 15:45




















          • Sir I could not understand how to algebra of delta is working? it is very new for me.Please can you tell me ?
            – MathLover
            Nov 25 at 15:42










          • @MathLover $$delta^i_j = begin{cases}1 &,& i = j \ 0 &,& {rm otherwise}end{cases}$$ That means that in a sum of the form $sum_i a_i delta^i_j$ all terms vanish, except the one for which $i = j$, $$sum_i a_i delta^i_j = a_j$$
            – caverac
            Nov 25 at 15:45


















          Sir I could not understand how to algebra of delta is working? it is very new for me.Please can you tell me ?
          – MathLover
          Nov 25 at 15:42




          Sir I could not understand how to algebra of delta is working? it is very new for me.Please can you tell me ?
          – MathLover
          Nov 25 at 15:42












          @MathLover $$delta^i_j = begin{cases}1 &,& i = j \ 0 &,& {rm otherwise}end{cases}$$ That means that in a sum of the form $sum_i a_i delta^i_j$ all terms vanish, except the one for which $i = j$, $$sum_i a_i delta^i_j = a_j$$
          – caverac
          Nov 25 at 15:45






          @MathLover $$delta^i_j = begin{cases}1 &,& i = j \ 0 &,& {rm otherwise}end{cases}$$ That means that in a sum of the form $sum_i a_i delta^i_j$ all terms vanish, except the one for which $i = j$, $$sum_i a_i delta^i_j = a_j$$
          – caverac
          Nov 25 at 15:45




















          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.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • 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.


          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%2f3012952%2fprocess-of-finding-orthonormal-basis-of-given-bilinear-form%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

          Probability when a professor distributes a quiz and homework assignment to a class of n students.

          Image
          0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
          Read more

          Aardman Animations

          Image
          Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
          Read more

          Are they similar matrix

          Image
          0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
          Read more