Mixing
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?
real-analysis quadratic-forms
asked 2 days ago
Matt Swanson
132
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Matt Swanson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
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?
real-analysis quadratic-forms
asked 2 days ago
Matt Swanson
132
New contributor
Matt Swanson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Is there a typo? $(x-2z)^2$ doesn't give $-4z^2.$
– user376343
2 days ago
add a comment |
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?
real-analysis quadratic-forms
asked 2 days ago
Matt Swanson
132
New contributor
Matt Swanson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
real-analysis quadratic-forms
asked 2 days ago
Matt Swanson
132
New contributor
Matt Swanson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Matt Swanson
132
New contributor
Matt Swanson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Matt Swanson
132
New contributor
Matt Swanson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Matt Swanson
132
asked 2 days ago
Matt Swanson
132
132
New contributor
Matt Swanson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Matt Swanson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Matt Swanson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Is there a typo? $(x-2z)^2$ doesn't give $-4z^2.$
– user376343
2 days ago
add a comment |
Is there a typo? $(x-2z)^2$ doesn't give $-4z^2.$
– user376343
2 days ago
Is there a typo? $(x-2z)^2$ doesn't give $-4z^2.$
– user376343
2 days ago
Is there a typo? $(x-2z)^2$ doesn't give $-4z^2.$
– user376343
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
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
$$ H = left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
$$
$$ D_0 = H $$
$$ E_j^T D_{j-1} E_j = D_j $$
$$ P_{j-1} E_j = P_j $$
$$ E_j^{-1} Q_{j-1} = Q_j $$
$$ P_j Q_j = Q_j P_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$
$$ H = left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
$$
==============================================
$$ E_{1} = left(
begin{array}{rrr}
1 & - 1 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{1} = left(
begin{array}{rrr}
1 & - 1 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{1} = left(
begin{array}{rrr}
1 & 1 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{1} = left(
begin{array}{rrr}
1 & 0 & - 2 \
0 & - 1 & 3 \
- 2 & 3 & - 4 \
end{array}
right)
$$
==============================================
$$ E_{2} = left(
begin{array}{rrr}
1 & 0 & 2 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{2} = left(
begin{array}{rrr}
1 & - 1 & 2 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{2} = left(
begin{array}{rrr}
1 & 1 & - 2 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{2} = left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 3 \
0 & 3 & - 8 \
end{array}
right)
$$
==============================================
$$ E_{3} = left(
begin{array}{rrr}
1 & 0 & 0 \
0 & 1 & 3 \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{3} = left(
begin{array}{rrr}
1 & - 1 & - 1 \
0 & 1 & 3 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{3} = left(
begin{array}{rrr}
1 & 1 & - 2 \
0 & 1 & - 3 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{3} = left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
==============================================
$$ P^T H P = D $$
$$left(
begin{array}{rrr}
1 & 0 & 0 \
- 1 & 1 & 0 \
- 1 & 3 & 1 \
end{array}
right)
left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
left(
begin{array}{rrr}
1 & - 1 & - 1 \
0 & 1 & 3 \
0 & 0 & 1 \
end{array}
right)
= left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
$$ Q^T D Q = H $$
$$left(
begin{array}{rrr}
1 & 0 & 0 \
1 & 1 & 0 \
- 2 & - 3 & 1 \
end{array}
right)
left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 0 \
0 & 0 & 1 \
end{array}
right)
left(
begin{array}{rrr}
1 & 1 & - 2 \
0 & 1 & - 3 \
0 & 0 & 1 \
end{array}
right)
= left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
$$
answered 2 days ago
Will Jagy
100k597198
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
$$ H = left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
$$
$$ D_0 = H $$
$$ E_j^T D_{j-1} E_j = D_j $$
$$ P_{j-1} E_j = P_j $$
$$ E_j^{-1} Q_{j-1} = Q_j $$
$$ P_j Q_j = Q_j P_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$
$$ H = left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
$$
==============================================
$$ E_{1} = left(
begin{array}{rrr}
1 & - 1 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{1} = left(
begin{array}{rrr}
1 & - 1 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{1} = left(
begin{array}{rrr}
1 & 1 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{1} = left(
begin{array}{rrr}
1 & 0 & - 2 \
0 & - 1 & 3 \
- 2 & 3 & - 4 \
end{array}
right)
$$
==============================================
$$ E_{2} = left(
begin{array}{rrr}
1 & 0 & 2 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{2} = left(
begin{array}{rrr}
1 & - 1 & 2 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{2} = left(
begin{array}{rrr}
1 & 1 & - 2 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{2} = left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 3 \
0 & 3 & - 8 \
end{array}
right)
$$
==============================================
$$ E_{3} = left(
begin{array}{rrr}
1 & 0 & 0 \
0 & 1 & 3 \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{3} = left(
begin{array}{rrr}
1 & - 1 & - 1 \
0 & 1 & 3 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{3} = left(
begin{array}{rrr}
1 & 1 & - 2 \
0 & 1 & - 3 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{3} = left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
==============================================
$$ P^T H P = D $$
$$left(
begin{array}{rrr}
1 & 0 & 0 \
- 1 & 1 & 0 \
- 1 & 3 & 1 \
end{array}
right)
left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
left(
begin{array}{rrr}
1 & - 1 & - 1 \
0 & 1 & 3 \
0 & 0 & 1 \
end{array}
right)
= left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
$$ Q^T D Q = H $$
$$left(
begin{array}{rrr}
1 & 0 & 0 \
1 & 1 & 0 \
- 2 & - 3 & 1 \
end{array}
right)
left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 0 \
0 & 0 & 1 \
end{array}
right)
left(
begin{array}{rrr}
1 & 1 & - 2 \
0 & 1 & - 3 \
0 & 0 & 1 \
end{array}
right)
= left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
$$
answered 2 days ago
Will Jagy
100k597198
add a comment |
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
$$ H = left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
$$
$$ D_0 = H $$
$$ E_j^T D_{j-1} E_j = D_j $$
$$ P_{j-1} E_j = P_j $$
$$ E_j^{-1} Q_{j-1} = Q_j $$
$$ P_j Q_j = Q_j P_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$
$$ H = left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
$$
==============================================
$$ E_{1} = left(
begin{array}{rrr}
1 & - 1 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{1} = left(
begin{array}{rrr}
1 & - 1 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{1} = left(
begin{array}{rrr}
1 & 1 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{1} = left(
begin{array}{rrr}
1 & 0 & - 2 \
0 & - 1 & 3 \
- 2 & 3 & - 4 \
end{array}
right)
$$
==============================================
$$ E_{2} = left(
begin{array}{rrr}
1 & 0 & 2 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{2} = left(
begin{array}{rrr}
1 & - 1 & 2 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{2} = left(
begin{array}{rrr}
1 & 1 & - 2 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{2} = left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 3 \
0 & 3 & - 8 \
end{array}
right)
$$
==============================================
$$ E_{3} = left(
begin{array}{rrr}
1 & 0 & 0 \
0 & 1 & 3 \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{3} = left(
begin{array}{rrr}
1 & - 1 & - 1 \
0 & 1 & 3 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{3} = left(
begin{array}{rrr}
1 & 1 & - 2 \
0 & 1 & - 3 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{3} = left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
==============================================
$$ P^T H P = D $$
$$left(
begin{array}{rrr}
1 & 0 & 0 \
- 1 & 1 & 0 \
- 1 & 3 & 1 \
end{array}
right)
left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
left(
begin{array}{rrr}
1 & - 1 & - 1 \
0 & 1 & 3 \
0 & 0 & 1 \
end{array}
right)
= left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
$$ Q^T D Q = H $$
$$left(
begin{array}{rrr}
1 & 0 & 0 \
1 & 1 & 0 \
- 2 & - 3 & 1 \
end{array}
right)
left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 0 \
0 & 0 & 1 \
end{array}
right)
left(
begin{array}{rrr}
1 & 1 & - 2 \
0 & 1 & - 3 \
0 & 0 & 1 \
end{array}
right)
= left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
$$
answered 2 days ago
Will Jagy
100k597198
add a comment |
up vote
0
down vote
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
$$ H = left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
$$
$$ D_0 = H $$
$$ E_j^T D_{j-1} E_j = D_j $$
$$ P_{j-1} E_j = P_j $$
$$ E_j^{-1} Q_{j-1} = Q_j $$
$$ P_j Q_j = Q_j P_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$
$$ H = left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
$$
==============================================
$$ E_{1} = left(
begin{array}{rrr}
1 & - 1 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{1} = left(
begin{array}{rrr}
1 & - 1 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{1} = left(
begin{array}{rrr}
1 & 1 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{1} = left(
begin{array}{rrr}
1 & 0 & - 2 \
0 & - 1 & 3 \
- 2 & 3 & - 4 \
end{array}
right)
$$
==============================================
$$ E_{2} = left(
begin{array}{rrr}
1 & 0 & 2 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{2} = left(
begin{array}{rrr}
1 & - 1 & 2 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{2} = left(
begin{array}{rrr}
1 & 1 & - 2 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{2} = left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 3 \
0 & 3 & - 8 \
end{array}
right)
$$
==============================================
$$ E_{3} = left(
begin{array}{rrr}
1 & 0 & 0 \
0 & 1 & 3 \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{3} = left(
begin{array}{rrr}
1 & - 1 & - 1 \
0 & 1 & 3 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{3} = left(
begin{array}{rrr}
1 & 1 & - 2 \
0 & 1 & - 3 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{3} = left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
==============================================
$$ P^T H P = D $$
$$left(
begin{array}{rrr}
1 & 0 & 0 \
- 1 & 1 & 0 \
- 1 & 3 & 1 \
end{array}
right)
left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
left(
begin{array}{rrr}
1 & - 1 & - 1 \
0 & 1 & 3 \
0 & 0 & 1 \
end{array}
right)
= left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
$$ Q^T D Q = H $$
$$left(
begin{array}{rrr}
1 & 0 & 0 \
1 & 1 & 0 \
- 2 & - 3 & 1 \
end{array}
right)
left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 0 \
0 & 0 & 1 \
end{array}
right)
left(
begin{array}{rrr}
1 & 1 & - 2 \
0 & 1 & - 3 \
0 & 0 & 1 \
end{array}
right)
= left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
$$
answered 2 days ago
Will Jagy
100k597198
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
$$ H = left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
$$
$$ D_0 = H $$
$$ E_j^T D_{j-1} E_j = D_j $$
$$ P_{j-1} E_j = P_j $$
$$ E_j^{-1} Q_{j-1} = Q_j $$
$$ P_j Q_j = Q_j P_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$
$$ H = left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
$$
==============================================
$$ E_{1} = left(
begin{array}{rrr}
1 & - 1 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{1} = left(
begin{array}{rrr}
1 & - 1 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{1} = left(
begin{array}{rrr}
1 & 1 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{1} = left(
begin{array}{rrr}
1 & 0 & - 2 \
0 & - 1 & 3 \
- 2 & 3 & - 4 \
end{array}
right)
$$
==============================================
$$ E_{2} = left(
begin{array}{rrr}
1 & 0 & 2 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{2} = left(
begin{array}{rrr}
1 & - 1 & 2 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{2} = left(
begin{array}{rrr}
1 & 1 & - 2 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{2} = left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 3 \
0 & 3 & - 8 \
end{array}
right)
$$
==============================================
$$ E_{3} = left(
begin{array}{rrr}
1 & 0 & 0 \
0 & 1 & 3 \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{3} = left(
begin{array}{rrr}
1 & - 1 & - 1 \
0 & 1 & 3 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{3} = left(
begin{array}{rrr}
1 & 1 & - 2 \
0 & 1 & - 3 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{3} = left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
==============================================
$$ P^T H P = D $$
$$left(
begin{array}{rrr}
1 & 0 & 0 \
- 1 & 1 & 0 \
- 1 & 3 & 1 \
end{array}
right)
left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
left(
begin{array}{rrr}
1 & - 1 & - 1 \
0 & 1 & 3 \
0 & 0 & 1 \
end{array}
right)
= left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
$$ Q^T D Q = H $$
$$left(
begin{array}{rrr}
1 & 0 & 0 \
1 & 1 & 0 \
- 2 & - 3 & 1 \
end{array}
right)
left(
begin{array}{rrr}
1 & 0 & 0 \
0 & - 1 & 0 \
0 & 0 & 1 \
end{array}
right)
left(
begin{array}{rrr}
1 & 1 & - 2 \
0 & 1 & - 3 \
0 & 0 & 1 \
end{array}
right)
= left(
begin{array}{rrr}
1 & 1 & - 2 \
1 & 0 & 1 \
- 2 & 1 & - 4 \
end{array}
right)
$$
answered 2 days ago
Will Jagy
100k597198
answered 2 days ago
Will Jagy
100k597198
answered 2 days ago
Will Jagy
100k597198
answered 2 days ago
Will Jagy
100k597198
100k597198
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Is there a typo? $(x-2z)^2$ doesn't give $-4z^2.$
– user376343
2 days ago