prove Gram's determinant invariant under orthogonalisation












0














let $mathcal H$ be a Hilbert space, and let $g_1,...g_nin mathcal H$. let $G(g_1,...g_n)$ be Gram's determinant:



$left|{begin{array}{ccc}
left(g_{1},g_{1}right) & dots & left(g_{1},g_{n}right)\
vdots & ddots & vdots\
left(g_{n},g_{1}right) & dots & left(g_{n},g_{n}right)
end{array}}right|$



write $(h_1,...h_n)$ to be the result of orthogonalisating $(g_1,...g_n)$, without normalizing the vectors. I want to show that
$G(g_1,...g_n)=G(h_1,...h_n)$



It's not hard to see that suffices to show $G(g_1,...g_n)=G(g_1,h_2,g_3,...g_n)$ when $h_2=g_2+alpha g_1$, such that $g_1perp h_2$. but I didn't manage to finish that up. I tried to calculate it directly, but it gets mass.






functional-analysis hilbert-spaces







share|cite|improve this question





























    0














    let $mathcal H$ be a Hilbert space, and let $g_1,...g_nin mathcal H$. let $G(g_1,...g_n)$ be Gram's determinant:



    $left|{begin{array}{ccc}
    left(g_{1},g_{1}right) & dots & left(g_{1},g_{n}right)\
    vdots & ddots & vdots\
    left(g_{n},g_{1}right) & dots & left(g_{n},g_{n}right)
    end{array}}right|$



    write $(h_1,...h_n)$ to be the result of orthogonalisating $(g_1,...g_n)$, without normalizing the vectors. I want to show that
    $G(g_1,...g_n)=G(h_1,...h_n)$



    It's not hard to see that suffices to show $G(g_1,...g_n)=G(g_1,h_2,g_3,...g_n)$ when $h_2=g_2+alpha g_1$, such that $g_1perp h_2$. but I didn't manage to finish that up. I tried to calculate it directly, but it gets mass.






    functional-analysis hilbert-spaces







    share|cite|improve this question



























      0












      0








      0







      let $mathcal H$ be a Hilbert space, and let $g_1,...g_nin mathcal H$. let $G(g_1,...g_n)$ be Gram's determinant:



      $left|{begin{array}{ccc}
      left(g_{1},g_{1}right) & dots & left(g_{1},g_{n}right)\
      vdots & ddots & vdots\
      left(g_{n},g_{1}right) & dots & left(g_{n},g_{n}right)
      end{array}}right|$



      write $(h_1,...h_n)$ to be the result of orthogonalisating $(g_1,...g_n)$, without normalizing the vectors. I want to show that
      $G(g_1,...g_n)=G(h_1,...h_n)$



      It's not hard to see that suffices to show $G(g_1,...g_n)=G(g_1,h_2,g_3,...g_n)$ when $h_2=g_2+alpha g_1$, such that $g_1perp h_2$. but I didn't manage to finish that up. I tried to calculate it directly, but it gets mass.






      functional-analysis hilbert-spaces







      share|cite|improve this question















      let $mathcal H$ be a Hilbert space, and let $g_1,...g_nin mathcal H$. let $G(g_1,...g_n)$ be Gram's determinant:



      $left|{begin{array}{ccc}
      left(g_{1},g_{1}right) & dots & left(g_{1},g_{n}right)\
      vdots & ddots & vdots\
      left(g_{n},g_{1}right) & dots & left(g_{n},g_{n}right)
      end{array}}right|$



      write $(h_1,...h_n)$ to be the result of orthogonalisating $(g_1,...g_n)$, without normalizing the vectors. I want to show that
      $G(g_1,...g_n)=G(h_1,...h_n)$



      It's not hard to see that suffices to show $G(g_1,...g_n)=G(g_1,h_2,g_3,...g_n)$ when $h_2=g_2+alpha g_1$, such that $g_1perp h_2$. but I didn't manage to finish that up. I tried to calculate it directly, but it gets mass.






      functional-analysis hilbert-spaces




      functional-analysis hilbert-spaces






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 22 at 15:57

























      asked Nov 22 at 14:46









      S. R

      886




      886






















          1 Answer
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          0














          Try in 2D first. We have $h_2 = g_2 - (g_1,g_2)g_1$. Let $t = (g_1,g_2)$ and lets say its an inner product over $mathbb R$. Let $mathcal G$ be the matrix whose determinant is $G$.
          $$
          mathcal G(g_1,h_2)=
          begin{bmatrix}
          |g_1|^2 & (g_1,h_2)\
          (h_2,g_1) & |g_2|^2
          end{bmatrix}
          =
          begin{bmatrix}
          |g_1|^2 & t-t|g_1|^2\
          t-t|g_1|^2 & |g_2|^2-2t^2 + t^2|g_1|^2
          end{bmatrix}
          $$
          This matrix can be obtained from
          $$ mathcal G (g_1,g_2) =
          begin{bmatrix}
          |g_1|^2 & t\
          t & |g_2|^2
          end{bmatrix}
          $$
          by subtracting $t$ times row 1 from row 2 and then from this new matrix subtracting $t$ times column 1 from column 2. Both operations used are valid elementary operations represented by matrices of determinant 1, so the determinant is unchanged.



          For higher dimensions you see the emergence of terms
          begin{align} |h_2|^2 &= |g_2|^2 - 2t^2 +t^2|g_1|^2 quad &(text{ same as before }) \
          (h_2,g_i) &= (g_2,g_i) - t(g_1,g_i) & ineq 2end{align}
          The original entries of the second row of $mathcal G(g_i)$ would have been $(g_2,g_i)$. If we subtract $t$ times the first row we get
          $$ (g_2,g_i) - t(g_1,g_i)$$
          as needed, except the diagonal term.
          If we then subtract $t$ times the first column, again the same terms appear, and the diagonal term is now exactly as in the $2times 2$ case.






          share|cite|improve this answer























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            1 Answer
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            0














            Try in 2D first. We have $h_2 = g_2 - (g_1,g_2)g_1$. Let $t = (g_1,g_2)$ and lets say its an inner product over $mathbb R$. Let $mathcal G$ be the matrix whose determinant is $G$.
            $$
            mathcal G(g_1,h_2)=
            begin{bmatrix}
            |g_1|^2 & (g_1,h_2)\
            (h_2,g_1) & |g_2|^2
            end{bmatrix}
            =
            begin{bmatrix}
            |g_1|^2 & t-t|g_1|^2\
            t-t|g_1|^2 & |g_2|^2-2t^2 + t^2|g_1|^2
            end{bmatrix}
            $$
            This matrix can be obtained from
            $$ mathcal G (g_1,g_2) =
            begin{bmatrix}
            |g_1|^2 & t\
            t & |g_2|^2
            end{bmatrix}
            $$
            by subtracting $t$ times row 1 from row 2 and then from this new matrix subtracting $t$ times column 1 from column 2. Both operations used are valid elementary operations represented by matrices of determinant 1, so the determinant is unchanged.



            For higher dimensions you see the emergence of terms
            begin{align} |h_2|^2 &= |g_2|^2 - 2t^2 +t^2|g_1|^2 quad &(text{ same as before }) \
            (h_2,g_i) &= (g_2,g_i) - t(g_1,g_i) & ineq 2end{align}
            The original entries of the second row of $mathcal G(g_i)$ would have been $(g_2,g_i)$. If we subtract $t$ times the first row we get
            $$ (g_2,g_i) - t(g_1,g_i)$$
            as needed, except the diagonal term.
            If we then subtract $t$ times the first column, again the same terms appear, and the diagonal term is now exactly as in the $2times 2$ case.






            share|cite|improve this answer




























              0














              Try in 2D first. We have $h_2 = g_2 - (g_1,g_2)g_1$. Let $t = (g_1,g_2)$ and lets say its an inner product over $mathbb R$. Let $mathcal G$ be the matrix whose determinant is $G$.
              $$
              mathcal G(g_1,h_2)=
              begin{bmatrix}
              |g_1|^2 & (g_1,h_2)\
              (h_2,g_1) & |g_2|^2
              end{bmatrix}
              =
              begin{bmatrix}
              |g_1|^2 & t-t|g_1|^2\
              t-t|g_1|^2 & |g_2|^2-2t^2 + t^2|g_1|^2
              end{bmatrix}
              $$
              This matrix can be obtained from
              $$ mathcal G (g_1,g_2) =
              begin{bmatrix}
              |g_1|^2 & t\
              t & |g_2|^2
              end{bmatrix}
              $$
              by subtracting $t$ times row 1 from row 2 and then from this new matrix subtracting $t$ times column 1 from column 2. Both operations used are valid elementary operations represented by matrices of determinant 1, so the determinant is unchanged.



              For higher dimensions you see the emergence of terms
              begin{align} |h_2|^2 &= |g_2|^2 - 2t^2 +t^2|g_1|^2 quad &(text{ same as before }) \
              (h_2,g_i) &= (g_2,g_i) - t(g_1,g_i) & ineq 2end{align}
              The original entries of the second row of $mathcal G(g_i)$ would have been $(g_2,g_i)$. If we subtract $t$ times the first row we get
              $$ (g_2,g_i) - t(g_1,g_i)$$
              as needed, except the diagonal term.
              If we then subtract $t$ times the first column, again the same terms appear, and the diagonal term is now exactly as in the $2times 2$ case.






              share|cite|improve this answer


























                0












                0








                0






                Try in 2D first. We have $h_2 = g_2 - (g_1,g_2)g_1$. Let $t = (g_1,g_2)$ and lets say its an inner product over $mathbb R$. Let $mathcal G$ be the matrix whose determinant is $G$.
                $$
                mathcal G(g_1,h_2)=
                begin{bmatrix}
                |g_1|^2 & (g_1,h_2)\
                (h_2,g_1) & |g_2|^2
                end{bmatrix}
                =
                begin{bmatrix}
                |g_1|^2 & t-t|g_1|^2\
                t-t|g_1|^2 & |g_2|^2-2t^2 + t^2|g_1|^2
                end{bmatrix}
                $$
                This matrix can be obtained from
                $$ mathcal G (g_1,g_2) =
                begin{bmatrix}
                |g_1|^2 & t\
                t & |g_2|^2
                end{bmatrix}
                $$
                by subtracting $t$ times row 1 from row 2 and then from this new matrix subtracting $t$ times column 1 from column 2. Both operations used are valid elementary operations represented by matrices of determinant 1, so the determinant is unchanged.



                For higher dimensions you see the emergence of terms
                begin{align} |h_2|^2 &= |g_2|^2 - 2t^2 +t^2|g_1|^2 quad &(text{ same as before }) \
                (h_2,g_i) &= (g_2,g_i) - t(g_1,g_i) & ineq 2end{align}
                The original entries of the second row of $mathcal G(g_i)$ would have been $(g_2,g_i)$. If we subtract $t$ times the first row we get
                $$ (g_2,g_i) - t(g_1,g_i)$$
                as needed, except the diagonal term.
                If we then subtract $t$ times the first column, again the same terms appear, and the diagonal term is now exactly as in the $2times 2$ case.






                share|cite|improve this answer














                Try in 2D first. We have $h_2 = g_2 - (g_1,g_2)g_1$. Let $t = (g_1,g_2)$ and lets say its an inner product over $mathbb R$. Let $mathcal G$ be the matrix whose determinant is $G$.
                $$
                mathcal G(g_1,h_2)=
                begin{bmatrix}
                |g_1|^2 & (g_1,h_2)\
                (h_2,g_1) & |g_2|^2
                end{bmatrix}
                =
                begin{bmatrix}
                |g_1|^2 & t-t|g_1|^2\
                t-t|g_1|^2 & |g_2|^2-2t^2 + t^2|g_1|^2
                end{bmatrix}
                $$
                This matrix can be obtained from
                $$ mathcal G (g_1,g_2) =
                begin{bmatrix}
                |g_1|^2 & t\
                t & |g_2|^2
                end{bmatrix}
                $$
                by subtracting $t$ times row 1 from row 2 and then from this new matrix subtracting $t$ times column 1 from column 2. Both operations used are valid elementary operations represented by matrices of determinant 1, so the determinant is unchanged.



                For higher dimensions you see the emergence of terms
                begin{align} |h_2|^2 &= |g_2|^2 - 2t^2 +t^2|g_1|^2 quad &(text{ same as before }) \
                (h_2,g_i) &= (g_2,g_i) - t(g_1,g_i) & ineq 2end{align}
                The original entries of the second row of $mathcal G(g_i)$ would have been $(g_2,g_i)$. If we subtract $t$ times the first row we get
                $$ (g_2,g_i) - t(g_1,g_i)$$
                as needed, except the diagonal term.
                If we then subtract $t$ times the first column, again the same terms appear, and the diagonal term is now exactly as in the $2times 2$ case.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 22 at 17:40

























                answered Nov 22 at 17:27









                Calvin Khor

                11.2k21438




                11.2k21438






























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