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
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          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













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          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




















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