Completing the square to decompose quadratic forms in three variables











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Given the quadratic form $Q(textbf{x}) = x^2 + 2xy - 4xz +2yz -4z^2$, the question asks to decompose $Q$ into sums of squares, first by eliminating terms in $z$, then terms in $y$, and then terms in $x$.



I know the signature of $Q$ is $(2,1)$ from the solution, but I'm at a loss for how to actually perform what the question is asking. For instance, I recognize that I can take the terms in $x$ and $z$ and form $(x-2z)^2$, but I don't really see what to do next. Any suggestions?










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  • Is there a typo? $(x-2z)^2$ doesn't give $-4z^2.$
    – user376343
    Nov 19 at 20:09















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Given the quadratic form $Q(textbf{x}) = x^2 + 2xy - 4xz +2yz -4z^2$, the question asks to decompose $Q$ into sums of squares, first by eliminating terms in $z$, then terms in $y$, and then terms in $x$.



I know the signature of $Q$ is $(2,1)$ from the solution, but I'm at a loss for how to actually perform what the question is asking. For instance, I recognize that I can take the terms in $x$ and $z$ and form $(x-2z)^2$, but I don't really see what to do next. Any suggestions?










share|cite|improve this question






















  • Is there a typo? $(x-2z)^2$ doesn't give $-4z^2.$
    – user376343
    Nov 19 at 20:09













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Given the quadratic form $Q(textbf{x}) = x^2 + 2xy - 4xz +2yz -4z^2$, the question asks to decompose $Q$ into sums of squares, first by eliminating terms in $z$, then terms in $y$, and then terms in $x$.



I know the signature of $Q$ is $(2,1)$ from the solution, but I'm at a loss for how to actually perform what the question is asking. For instance, I recognize that I can take the terms in $x$ and $z$ and form $(x-2z)^2$, but I don't really see what to do next. Any suggestions?










share|cite|improve this question













Given the quadratic form $Q(textbf{x}) = x^2 + 2xy - 4xz +2yz -4z^2$, the question asks to decompose $Q$ into sums of squares, first by eliminating terms in $z$, then terms in $y$, and then terms in $x$.



I know the signature of $Q$ is $(2,1)$ from the solution, but I'm at a loss for how to actually perform what the question is asking. For instance, I recognize that I can take the terms in $x$ and $z$ and form $(x-2z)^2$, but I don't really see what to do next. Any suggestions?







real-analysis quadratic-forms






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asked Nov 19 at 20:04









Matt Swanson

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132












  • Is there a typo? $(x-2z)^2$ doesn't give $-4z^2.$
    – user376343
    Nov 19 at 20:09


















  • Is there a typo? $(x-2z)^2$ doesn't give $-4z^2.$
    – user376343
    Nov 19 at 20:09
















Is there a typo? $(x-2z)^2$ doesn't give $-4z^2.$
– user376343
Nov 19 at 20:09




Is there a typo? $(x-2z)^2$ doesn't give $-4z^2.$
– user376343
Nov 19 at 20:09










1 Answer
1






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up vote
0
down vote













a method. Usually we take the matrix $H$ to be the Hessian matrix of second partial derivatives. In the special case of integer coefficients with all the mixed coefficent terms even, we can get $H$ all integers by taking half the Hessian matrix.



Outcome this time:



$$ P^T H P = D $$
$$left(
begin{array}{rrr}
1 & 0 & 0 \
- 1 & 1 & 0 \
- 1 & 3 & 1 \
end{array}
right)
left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
left(
begin{array}{rrr}
1 & - 1 & - 1 \
0 & 1 & 3 \
0 & 0 & 1 \
end{array}
right)
= left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$

$$ Q^T D Q = H $$
$$left(
begin{array}{rrr}
1 & 0 & 0 \
1 & 1 & 0 \
- 2 & - 3 & 1 \
end{array}
right)
left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 0 \
0 & 0 & 1 \
end{array}
right)
left(
begin{array}{rrr}
1 & 1 & - 2 \
0 & 1 & - 3 \
0 & 0 & 1 \
end{array}
right)
= left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
$$



It is the diagonal elements of $D$ and the rows of $Q$ that give the expression,
$$ color{red}{ (x+y-2z)^2 - (y-3z)^2 + z^2 } $$



Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
$$

$$ D_0 = H $$
$$ E_j^T D_{j-1} E_j = D_j $$
$$ P_{j-1} E_j = P_j $$
$$ E_j^{-1} Q_{j-1} = Q_j $$
$$ P_j Q_j = Q_j P_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$



$$ H = left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
$$



==============================================



$$ E_{1} = left(
begin{array}{rrr}
1 & - 1 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$

$$ P_{1} = left(
begin{array}{rrr}
1 & - 1 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{1} = left(
begin{array}{rrr}
1 & 1 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{1} = left(
begin{array}{rrr}
1 & 0 & - 2 \
0 & - 1 & 3 \
- 2 & 3 & - 4 \
end{array}
right)
$$



==============================================



$$ E_{2} = left(
begin{array}{rrr}
1 & 0 & 2 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$

$$ P_{2} = left(
begin{array}{rrr}
1 & - 1 & 2 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{2} = left(
begin{array}{rrr}
1 & 1 & - 2 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{2} = left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 3 \
0 & 3 & - 8 \
end{array}
right)
$$



==============================================



$$ E_{3} = left(
begin{array}{rrr}
1 & 0 & 0 \
0 & 1 & 3 \
0 & 0 & 1 \
end{array}
right)
$$

$$ P_{3} = left(
begin{array}{rrr}
1 & - 1 & - 1 \
0 & 1 & 3 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{3} = left(
begin{array}{rrr}
1 & 1 & - 2 \
0 & 1 & - 3 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{3} = left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$



==============================================



$$ P^T H P = D $$
$$left(
begin{array}{rrr}
1 & 0 & 0 \
- 1 & 1 & 0 \
- 1 & 3 & 1 \
end{array}
right)
left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
left(
begin{array}{rrr}
1 & - 1 & - 1 \
0 & 1 & 3 \
0 & 0 & 1 \
end{array}
right)
= left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$

$$ Q^T D Q = H $$
$$left(
begin{array}{rrr}
1 & 0 & 0 \
1 & 1 & 0 \
- 2 & - 3 & 1 \
end{array}
right)
left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 0 \
0 & 0 & 1 \
end{array}
right)
left(
begin{array}{rrr}
1 & 1 & - 2 \
0 & 1 & - 3 \
0 & 0 & 1 \
end{array}
right)
= left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
$$






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    1 Answer
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    oldest

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    up vote
    0
    down vote













    a method. Usually we take the matrix $H$ to be the Hessian matrix of second partial derivatives. In the special case of integer coefficients with all the mixed coefficent terms even, we can get $H$ all integers by taking half the Hessian matrix.



    Outcome this time:



    $$ P^T H P = D $$
    $$left(
    begin{array}{rrr}
    1 & 0 & 0 \
    - 1 & 1 & 0 \
    - 1 & 3 & 1 \
    end{array}
    right)
    left(
    begin{array}{rrr}
    1 & 1 & - 2 \
    1 & 0 & 1 \
    - 2 & 1 & - 4 \
    end{array}
    right)
    left(
    begin{array}{rrr}
    1 & - 1 & - 1 \
    0 & 1 & 3 \
    0 & 0 & 1 \
    end{array}
    right)
    = left(
    begin{array}{rrr}
    1 & 0 & 0 \
    0 & - 1 & 0 \
    0 & 0 & 1 \
    end{array}
    right)
    $$

    $$ Q^T D Q = H $$
    $$left(
    begin{array}{rrr}
    1 & 0 & 0 \
    1 & 1 & 0 \
    - 2 & - 3 & 1 \
    end{array}
    right)
    left(
    begin{array}{rrr}
    1 & 0 & 0 \
    0 & - 1 & 0 \
    0 & 0 & 1 \
    end{array}
    right)
    left(
    begin{array}{rrr}
    1 & 1 & - 2 \
    0 & 1 & - 3 \
    0 & 0 & 1 \
    end{array}
    right)
    = left(
    begin{array}{rrr}
    1 & 1 & - 2 \
    1 & 0 & 1 \
    - 2 & 1 & - 4 \
    end{array}
    right)
    $$



    It is the diagonal elements of $D$ and the rows of $Q$ that give the expression,
    $$ color{red}{ (x+y-2z)^2 - (y-3z)^2 + z^2 } $$



    Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
    https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
    $$ H = left(
    begin{array}{rrr}
    1 & 1 & - 2 \
    1 & 0 & 1 \
    - 2 & 1 & - 4 \
    end{array}
    right)
    $$

    $$ D_0 = H $$
    $$ E_j^T D_{j-1} E_j = D_j $$
    $$ P_{j-1} E_j = P_j $$
    $$ E_j^{-1} Q_{j-1} = Q_j $$
    $$ P_j Q_j = Q_j P_j = I $$
    $$ P_j^T H P_j = D_j $$
    $$ Q_j^T D_j Q_j = H $$



    $$ H = left(
    begin{array}{rrr}
    1 & 1 & - 2 \
    1 & 0 & 1 \
    - 2 & 1 & - 4 \
    end{array}
    right)
    $$



    ==============================================



    $$ E_{1} = left(
    begin{array}{rrr}
    1 & - 1 & 0 \
    0 & 1 & 0 \
    0 & 0 & 1 \
    end{array}
    right)
    $$

    $$ P_{1} = left(
    begin{array}{rrr}
    1 & - 1 & 0 \
    0 & 1 & 0 \
    0 & 0 & 1 \
    end{array}
    right)
    , ; ; ; Q_{1} = left(
    begin{array}{rrr}
    1 & 1 & 0 \
    0 & 1 & 0 \
    0 & 0 & 1 \
    end{array}
    right)
    , ; ; ; D_{1} = left(
    begin{array}{rrr}
    1 & 0 & - 2 \
    0 & - 1 & 3 \
    - 2 & 3 & - 4 \
    end{array}
    right)
    $$



    ==============================================



    $$ E_{2} = left(
    begin{array}{rrr}
    1 & 0 & 2 \
    0 & 1 & 0 \
    0 & 0 & 1 \
    end{array}
    right)
    $$

    $$ P_{2} = left(
    begin{array}{rrr}
    1 & - 1 & 2 \
    0 & 1 & 0 \
    0 & 0 & 1 \
    end{array}
    right)
    , ; ; ; Q_{2} = left(
    begin{array}{rrr}
    1 & 1 & - 2 \
    0 & 1 & 0 \
    0 & 0 & 1 \
    end{array}
    right)
    , ; ; ; D_{2} = left(
    begin{array}{rrr}
    1 & 0 & 0 \
    0 & - 1 & 3 \
    0 & 3 & - 8 \
    end{array}
    right)
    $$



    ==============================================



    $$ E_{3} = left(
    begin{array}{rrr}
    1 & 0 & 0 \
    0 & 1 & 3 \
    0 & 0 & 1 \
    end{array}
    right)
    $$

    $$ P_{3} = left(
    begin{array}{rrr}
    1 & - 1 & - 1 \
    0 & 1 & 3 \
    0 & 0 & 1 \
    end{array}
    right)
    , ; ; ; Q_{3} = left(
    begin{array}{rrr}
    1 & 1 & - 2 \
    0 & 1 & - 3 \
    0 & 0 & 1 \
    end{array}
    right)
    , ; ; ; D_{3} = left(
    begin{array}{rrr}
    1 & 0 & 0 \
    0 & - 1 & 0 \
    0 & 0 & 1 \
    end{array}
    right)
    $$



    ==============================================



    $$ P^T H P = D $$
    $$left(
    begin{array}{rrr}
    1 & 0 & 0 \
    - 1 & 1 & 0 \
    - 1 & 3 & 1 \
    end{array}
    right)
    left(
    begin{array}{rrr}
    1 & 1 & - 2 \
    1 & 0 & 1 \
    - 2 & 1 & - 4 \
    end{array}
    right)
    left(
    begin{array}{rrr}
    1 & - 1 & - 1 \
    0 & 1 & 3 \
    0 & 0 & 1 \
    end{array}
    right)
    = left(
    begin{array}{rrr}
    1 & 0 & 0 \
    0 & - 1 & 0 \
    0 & 0 & 1 \
    end{array}
    right)
    $$

    $$ Q^T D Q = H $$
    $$left(
    begin{array}{rrr}
    1 & 0 & 0 \
    1 & 1 & 0 \
    - 2 & - 3 & 1 \
    end{array}
    right)
    left(
    begin{array}{rrr}
    1 & 0 & 0 \
    0 & - 1 & 0 \
    0 & 0 & 1 \
    end{array}
    right)
    left(
    begin{array}{rrr}
    1 & 1 & - 2 \
    0 & 1 & - 3 \
    0 & 0 & 1 \
    end{array}
    right)
    = left(
    begin{array}{rrr}
    1 & 1 & - 2 \
    1 & 0 & 1 \
    - 2 & 1 & - 4 \
    end{array}
    right)
    $$






    share|cite|improve this answer

























      up vote
      0
      down vote













      a method. Usually we take the matrix $H$ to be the Hessian matrix of second partial derivatives. In the special case of integer coefficients with all the mixed coefficent terms even, we can get $H$ all integers by taking half the Hessian matrix.



      Outcome this time:



      $$ P^T H P = D $$
      $$left(
      begin{array}{rrr}
      1 & 0 & 0 \
      - 1 & 1 & 0 \
      - 1 & 3 & 1 \
      end{array}
      right)
      left(
      begin{array}{rrr}
      1 & 1 & - 2 \
      1 & 0 & 1 \
      - 2 & 1 & - 4 \
      end{array}
      right)
      left(
      begin{array}{rrr}
      1 & - 1 & - 1 \
      0 & 1 & 3 \
      0 & 0 & 1 \
      end{array}
      right)
      = left(
      begin{array}{rrr}
      1 & 0 & 0 \
      0 & - 1 & 0 \
      0 & 0 & 1 \
      end{array}
      right)
      $$

      $$ Q^T D Q = H $$
      $$left(
      begin{array}{rrr}
      1 & 0 & 0 \
      1 & 1 & 0 \
      - 2 & - 3 & 1 \
      end{array}
      right)
      left(
      begin{array}{rrr}
      1 & 0 & 0 \
      0 & - 1 & 0 \
      0 & 0 & 1 \
      end{array}
      right)
      left(
      begin{array}{rrr}
      1 & 1 & - 2 \
      0 & 1 & - 3 \
      0 & 0 & 1 \
      end{array}
      right)
      = left(
      begin{array}{rrr}
      1 & 1 & - 2 \
      1 & 0 & 1 \
      - 2 & 1 & - 4 \
      end{array}
      right)
      $$



      It is the diagonal elements of $D$ and the rows of $Q$ that give the expression,
      $$ color{red}{ (x+y-2z)^2 - (y-3z)^2 + z^2 } $$



      Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
      https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
      $$ H = left(
      begin{array}{rrr}
      1 & 1 & - 2 \
      1 & 0 & 1 \
      - 2 & 1 & - 4 \
      end{array}
      right)
      $$

      $$ D_0 = H $$
      $$ E_j^T D_{j-1} E_j = D_j $$
      $$ P_{j-1} E_j = P_j $$
      $$ E_j^{-1} Q_{j-1} = Q_j $$
      $$ P_j Q_j = Q_j P_j = I $$
      $$ P_j^T H P_j = D_j $$
      $$ Q_j^T D_j Q_j = H $$



      $$ H = left(
      begin{array}{rrr}
      1 & 1 & - 2 \
      1 & 0 & 1 \
      - 2 & 1 & - 4 \
      end{array}
      right)
      $$



      ==============================================



      $$ E_{1} = left(
      begin{array}{rrr}
      1 & - 1 & 0 \
      0 & 1 & 0 \
      0 & 0 & 1 \
      end{array}
      right)
      $$

      $$ P_{1} = left(
      begin{array}{rrr}
      1 & - 1 & 0 \
      0 & 1 & 0 \
      0 & 0 & 1 \
      end{array}
      right)
      , ; ; ; Q_{1} = left(
      begin{array}{rrr}
      1 & 1 & 0 \
      0 & 1 & 0 \
      0 & 0 & 1 \
      end{array}
      right)
      , ; ; ; D_{1} = left(
      begin{array}{rrr}
      1 & 0 & - 2 \
      0 & - 1 & 3 \
      - 2 & 3 & - 4 \
      end{array}
      right)
      $$



      ==============================================



      $$ E_{2} = left(
      begin{array}{rrr}
      1 & 0 & 2 \
      0 & 1 & 0 \
      0 & 0 & 1 \
      end{array}
      right)
      $$

      $$ P_{2} = left(
      begin{array}{rrr}
      1 & - 1 & 2 \
      0 & 1 & 0 \
      0 & 0 & 1 \
      end{array}
      right)
      , ; ; ; Q_{2} = left(
      begin{array}{rrr}
      1 & 1 & - 2 \
      0 & 1 & 0 \
      0 & 0 & 1 \
      end{array}
      right)
      , ; ; ; D_{2} = left(
      begin{array}{rrr}
      1 & 0 & 0 \
      0 & - 1 & 3 \
      0 & 3 & - 8 \
      end{array}
      right)
      $$



      ==============================================



      $$ E_{3} = left(
      begin{array}{rrr}
      1 & 0 & 0 \
      0 & 1 & 3 \
      0 & 0 & 1 \
      end{array}
      right)
      $$

      $$ P_{3} = left(
      begin{array}{rrr}
      1 & - 1 & - 1 \
      0 & 1 & 3 \
      0 & 0 & 1 \
      end{array}
      right)
      , ; ; ; Q_{3} = left(
      begin{array}{rrr}
      1 & 1 & - 2 \
      0 & 1 & - 3 \
      0 & 0 & 1 \
      end{array}
      right)
      , ; ; ; D_{3} = left(
      begin{array}{rrr}
      1 & 0 & 0 \
      0 & - 1 & 0 \
      0 & 0 & 1 \
      end{array}
      right)
      $$



      ==============================================



      $$ P^T H P = D $$
      $$left(
      begin{array}{rrr}
      1 & 0 & 0 \
      - 1 & 1 & 0 \
      - 1 & 3 & 1 \
      end{array}
      right)
      left(
      begin{array}{rrr}
      1 & 1 & - 2 \
      1 & 0 & 1 \
      - 2 & 1 & - 4 \
      end{array}
      right)
      left(
      begin{array}{rrr}
      1 & - 1 & - 1 \
      0 & 1 & 3 \
      0 & 0 & 1 \
      end{array}
      right)
      = left(
      begin{array}{rrr}
      1 & 0 & 0 \
      0 & - 1 & 0 \
      0 & 0 & 1 \
      end{array}
      right)
      $$

      $$ Q^T D Q = H $$
      $$left(
      begin{array}{rrr}
      1 & 0 & 0 \
      1 & 1 & 0 \
      - 2 & - 3 & 1 \
      end{array}
      right)
      left(
      begin{array}{rrr}
      1 & 0 & 0 \
      0 & - 1 & 0 \
      0 & 0 & 1 \
      end{array}
      right)
      left(
      begin{array}{rrr}
      1 & 1 & - 2 \
      0 & 1 & - 3 \
      0 & 0 & 1 \
      end{array}
      right)
      = left(
      begin{array}{rrr}
      1 & 1 & - 2 \
      1 & 0 & 1 \
      - 2 & 1 & - 4 \
      end{array}
      right)
      $$






      share|cite|improve this answer























        up vote
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        a method. Usually we take the matrix $H$ to be the Hessian matrix of second partial derivatives. In the special case of integer coefficients with all the mixed coefficent terms even, we can get $H$ all integers by taking half the Hessian matrix.



        Outcome this time:



        $$ P^T H P = D $$
        $$left(
        begin{array}{rrr}
        1 & 0 & 0 \
        - 1 & 1 & 0 \
        - 1 & 3 & 1 \
        end{array}
        right)
        left(
        begin{array}{rrr}
        1 & 1 & - 2 \
        1 & 0 & 1 \
        - 2 & 1 & - 4 \
        end{array}
        right)
        left(
        begin{array}{rrr}
        1 & - 1 & - 1 \
        0 & 1 & 3 \
        0 & 0 & 1 \
        end{array}
        right)
        = left(
        begin{array}{rrr}
        1 & 0 & 0 \
        0 & - 1 & 0 \
        0 & 0 & 1 \
        end{array}
        right)
        $$

        $$ Q^T D Q = H $$
        $$left(
        begin{array}{rrr}
        1 & 0 & 0 \
        1 & 1 & 0 \
        - 2 & - 3 & 1 \
        end{array}
        right)
        left(
        begin{array}{rrr}
        1 & 0 & 0 \
        0 & - 1 & 0 \
        0 & 0 & 1 \
        end{array}
        right)
        left(
        begin{array}{rrr}
        1 & 1 & - 2 \
        0 & 1 & - 3 \
        0 & 0 & 1 \
        end{array}
        right)
        = left(
        begin{array}{rrr}
        1 & 1 & - 2 \
        1 & 0 & 1 \
        - 2 & 1 & - 4 \
        end{array}
        right)
        $$



        It is the diagonal elements of $D$ and the rows of $Q$ that give the expression,
        $$ color{red}{ (x+y-2z)^2 - (y-3z)^2 + z^2 } $$



        Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
        https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
        $$ H = left(
        begin{array}{rrr}
        1 & 1 & - 2 \
        1 & 0 & 1 \
        - 2 & 1 & - 4 \
        end{array}
        right)
        $$

        $$ D_0 = H $$
        $$ E_j^T D_{j-1} E_j = D_j $$
        $$ P_{j-1} E_j = P_j $$
        $$ E_j^{-1} Q_{j-1} = Q_j $$
        $$ P_j Q_j = Q_j P_j = I $$
        $$ P_j^T H P_j = D_j $$
        $$ Q_j^T D_j Q_j = H $$



        $$ H = left(
        begin{array}{rrr}
        1 & 1 & - 2 \
        1 & 0 & 1 \
        - 2 & 1 & - 4 \
        end{array}
        right)
        $$



        ==============================================



        $$ E_{1} = left(
        begin{array}{rrr}
        1 & - 1 & 0 \
        0 & 1 & 0 \
        0 & 0 & 1 \
        end{array}
        right)
        $$

        $$ P_{1} = left(
        begin{array}{rrr}
        1 & - 1 & 0 \
        0 & 1 & 0 \
        0 & 0 & 1 \
        end{array}
        right)
        , ; ; ; Q_{1} = left(
        begin{array}{rrr}
        1 & 1 & 0 \
        0 & 1 & 0 \
        0 & 0 & 1 \
        end{array}
        right)
        , ; ; ; D_{1} = left(
        begin{array}{rrr}
        1 & 0 & - 2 \
        0 & - 1 & 3 \
        - 2 & 3 & - 4 \
        end{array}
        right)
        $$



        ==============================================



        $$ E_{2} = left(
        begin{array}{rrr}
        1 & 0 & 2 \
        0 & 1 & 0 \
        0 & 0 & 1 \
        end{array}
        right)
        $$

        $$ P_{2} = left(
        begin{array}{rrr}
        1 & - 1 & 2 \
        0 & 1 & 0 \
        0 & 0 & 1 \
        end{array}
        right)
        , ; ; ; Q_{2} = left(
        begin{array}{rrr}
        1 & 1 & - 2 \
        0 & 1 & 0 \
        0 & 0 & 1 \
        end{array}
        right)
        , ; ; ; D_{2} = left(
        begin{array}{rrr}
        1 & 0 & 0 \
        0 & - 1 & 3 \
        0 & 3 & - 8 \
        end{array}
        right)
        $$



        ==============================================



        $$ E_{3} = left(
        begin{array}{rrr}
        1 & 0 & 0 \
        0 & 1 & 3 \
        0 & 0 & 1 \
        end{array}
        right)
        $$

        $$ P_{3} = left(
        begin{array}{rrr}
        1 & - 1 & - 1 \
        0 & 1 & 3 \
        0 & 0 & 1 \
        end{array}
        right)
        , ; ; ; Q_{3} = left(
        begin{array}{rrr}
        1 & 1 & - 2 \
        0 & 1 & - 3 \
        0 & 0 & 1 \
        end{array}
        right)
        , ; ; ; D_{3} = left(
        begin{array}{rrr}
        1 & 0 & 0 \
        0 & - 1 & 0 \
        0 & 0 & 1 \
        end{array}
        right)
        $$



        ==============================================



        $$ P^T H P = D $$
        $$left(
        begin{array}{rrr}
        1 & 0 & 0 \
        - 1 & 1 & 0 \
        - 1 & 3 & 1 \
        end{array}
        right)
        left(
        begin{array}{rrr}
        1 & 1 & - 2 \
        1 & 0 & 1 \
        - 2 & 1 & - 4 \
        end{array}
        right)
        left(
        begin{array}{rrr}
        1 & - 1 & - 1 \
        0 & 1 & 3 \
        0 & 0 & 1 \
        end{array}
        right)
        = left(
        begin{array}{rrr}
        1 & 0 & 0 \
        0 & - 1 & 0 \
        0 & 0 & 1 \
        end{array}
        right)
        $$

        $$ Q^T D Q = H $$
        $$left(
        begin{array}{rrr}
        1 & 0 & 0 \
        1 & 1 & 0 \
        - 2 & - 3 & 1 \
        end{array}
        right)
        left(
        begin{array}{rrr}
        1 & 0 & 0 \
        0 & - 1 & 0 \
        0 & 0 & 1 \
        end{array}
        right)
        left(
        begin{array}{rrr}
        1 & 1 & - 2 \
        0 & 1 & - 3 \
        0 & 0 & 1 \
        end{array}
        right)
        = left(
        begin{array}{rrr}
        1 & 1 & - 2 \
        1 & 0 & 1 \
        - 2 & 1 & - 4 \
        end{array}
        right)
        $$






        share|cite|improve this answer












        a method. Usually we take the matrix $H$ to be the Hessian matrix of second partial derivatives. In the special case of integer coefficients with all the mixed coefficent terms even, we can get $H$ all integers by taking half the Hessian matrix.



        Outcome this time:



        $$ P^T H P = D $$
        $$left(
        begin{array}{rrr}
        1 & 0 & 0 \
        - 1 & 1 & 0 \
        - 1 & 3 & 1 \
        end{array}
        right)
        left(
        begin{array}{rrr}
        1 & 1 & - 2 \
        1 & 0 & 1 \
        - 2 & 1 & - 4 \
        end{array}
        right)
        left(
        begin{array}{rrr}
        1 & - 1 & - 1 \
        0 & 1 & 3 \
        0 & 0 & 1 \
        end{array}
        right)
        = left(
        begin{array}{rrr}
        1 & 0 & 0 \
        0 & - 1 & 0 \
        0 & 0 & 1 \
        end{array}
        right)
        $$

        $$ Q^T D Q = H $$
        $$left(
        begin{array}{rrr}
        1 & 0 & 0 \
        1 & 1 & 0 \
        - 2 & - 3 & 1 \
        end{array}
        right)
        left(
        begin{array}{rrr}
        1 & 0 & 0 \
        0 & - 1 & 0 \
        0 & 0 & 1 \
        end{array}
        right)
        left(
        begin{array}{rrr}
        1 & 1 & - 2 \
        0 & 1 & - 3 \
        0 & 0 & 1 \
        end{array}
        right)
        = left(
        begin{array}{rrr}
        1 & 1 & - 2 \
        1 & 0 & 1 \
        - 2 & 1 & - 4 \
        end{array}
        right)
        $$



        It is the diagonal elements of $D$ and the rows of $Q$ that give the expression,
        $$ color{red}{ (x+y-2z)^2 - (y-3z)^2 + z^2 } $$



        Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
        https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
        $$ H = left(
        begin{array}{rrr}
        1 & 1 & - 2 \
        1 & 0 & 1 \
        - 2 & 1 & - 4 \
        end{array}
        right)
        $$

        $$ D_0 = H $$
        $$ E_j^T D_{j-1} E_j = D_j $$
        $$ P_{j-1} E_j = P_j $$
        $$ E_j^{-1} Q_{j-1} = Q_j $$
        $$ P_j Q_j = Q_j P_j = I $$
        $$ P_j^T H P_j = D_j $$
        $$ Q_j^T D_j Q_j = H $$



        $$ H = left(
        begin{array}{rrr}
        1 & 1 & - 2 \
        1 & 0 & 1 \
        - 2 & 1 & - 4 \
        end{array}
        right)
        $$



        ==============================================



        $$ E_{1} = left(
        begin{array}{rrr}
        1 & - 1 & 0 \
        0 & 1 & 0 \
        0 & 0 & 1 \
        end{array}
        right)
        $$

        $$ P_{1} = left(
        begin{array}{rrr}
        1 & - 1 & 0 \
        0 & 1 & 0 \
        0 & 0 & 1 \
        end{array}
        right)
        , ; ; ; Q_{1} = left(
        begin{array}{rrr}
        1 & 1 & 0 \
        0 & 1 & 0 \
        0 & 0 & 1 \
        end{array}
        right)
        , ; ; ; D_{1} = left(
        begin{array}{rrr}
        1 & 0 & - 2 \
        0 & - 1 & 3 \
        - 2 & 3 & - 4 \
        end{array}
        right)
        $$



        ==============================================



        $$ E_{2} = left(
        begin{array}{rrr}
        1 & 0 & 2 \
        0 & 1 & 0 \
        0 & 0 & 1 \
        end{array}
        right)
        $$

        $$ P_{2} = left(
        begin{array}{rrr}
        1 & - 1 & 2 \
        0 & 1 & 0 \
        0 & 0 & 1 \
        end{array}
        right)
        , ; ; ; Q_{2} = left(
        begin{array}{rrr}
        1 & 1 & - 2 \
        0 & 1 & 0 \
        0 & 0 & 1 \
        end{array}
        right)
        , ; ; ; D_{2} = left(
        begin{array}{rrr}
        1 & 0 & 0 \
        0 & - 1 & 3 \
        0 & 3 & - 8 \
        end{array}
        right)
        $$



        ==============================================



        $$ E_{3} = left(
        begin{array}{rrr}
        1 & 0 & 0 \
        0 & 1 & 3 \
        0 & 0 & 1 \
        end{array}
        right)
        $$

        $$ P_{3} = left(
        begin{array}{rrr}
        1 & - 1 & - 1 \
        0 & 1 & 3 \
        0 & 0 & 1 \
        end{array}
        right)
        , ; ; ; Q_{3} = left(
        begin{array}{rrr}
        1 & 1 & - 2 \
        0 & 1 & - 3 \
        0 & 0 & 1 \
        end{array}
        right)
        , ; ; ; D_{3} = left(
        begin{array}{rrr}
        1 & 0 & 0 \
        0 & - 1 & 0 \
        0 & 0 & 1 \
        end{array}
        right)
        $$



        ==============================================



        $$ P^T H P = D $$
        $$left(
        begin{array}{rrr}
        1 & 0 & 0 \
        - 1 & 1 & 0 \
        - 1 & 3 & 1 \
        end{array}
        right)
        left(
        begin{array}{rrr}
        1 & 1 & - 2 \
        1 & 0 & 1 \
        - 2 & 1 & - 4 \
        end{array}
        right)
        left(
        begin{array}{rrr}
        1 & - 1 & - 1 \
        0 & 1 & 3 \
        0 & 0 & 1 \
        end{array}
        right)
        = left(
        begin{array}{rrr}
        1 & 0 & 0 \
        0 & - 1 & 0 \
        0 & 0 & 1 \
        end{array}
        right)
        $$

        $$ Q^T D Q = H $$
        $$left(
        begin{array}{rrr}
        1 & 0 & 0 \
        1 & 1 & 0 \
        - 2 & - 3 & 1 \
        end{array}
        right)
        left(
        begin{array}{rrr}
        1 & 0 & 0 \
        0 & - 1 & 0 \
        0 & 0 & 1 \
        end{array}
        right)
        left(
        begin{array}{rrr}
        1 & 1 & - 2 \
        0 & 1 & - 3 \
        0 & 0 & 1 \
        end{array}
        right)
        = left(
        begin{array}{rrr}
        1 & 1 & - 2 \
        1 & 0 & 1 \
        - 2 & 1 & - 4 \
        end{array}
        right)
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 19 at 21:41









        Will Jagy

        101k597198




        101k597198






























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