Completing the square to find if quadratic form is positive definite.Determine if the quadratic form is positive definite, negative definite or undefiniteexample of quadratic form with nonunique +ve and -ve definite subspacesTransform $f(x_1,x_2,x_3)=2x_1^2+5x_2^2+5x_3^2+4x_1x_2-4x_1x_3-8x_2x_3$ to a diagonal form.Is there a systematic way of finding the matrix of a quadratic form?Quadratic forms - Completing squaresFind k for Positive Definite Quadratic FormUnitary Diagonalization of Quadratic FormHow to show that this matrix is positive semidefinite?Showing matrix $left[beginsmallmatrix 4 & 1 & 1\1 & 2 & -1 \ 1 & -1 & 3 endsmallmatrixright]$ is positive definiteWhat are the ways to determine whether quadratic form is positive definite?
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Completing the square to find if quadratic form is positive definite.
Determine if the quadratic form is positive definite, negative definite or undefiniteexample of quadratic form with nonunique +ve and -ve definite subspacesTransform $f(x_1,x_2,x_3)=2x_1^2+5x_2^2+5x_3^2+4x_1x_2-4x_1x_3-8x_2x_3$ to a diagonal form.Is there a systematic way of finding the matrix of a quadratic form?Quadratic forms - Completing squaresFind k for Positive Definite Quadratic FormUnitary Diagonalization of Quadratic FormHow to show that this matrix is positive semidefinite?Showing matrix $left[beginsmallmatrix 4 & 1 & 1\1 & 2 & -1 \ 1 & -1 & 3 endsmallmatrixright]$ is positive definiteWhat are the ways to determine whether quadratic form is positive definite?
$begingroup$
I have the quadratic form
$$g=x_1^2+6x_2^2+8x_3^2-4x_1x_2-6x_1x_3-x_2x_3$$
I have problems completing the square. I tried to rewrite the expression as follows
$$g=x_1^2-4x_1x_2+6x_2^2-x_2x_3+8x_3^2$$
Hence,
$$((x_1-2x_2)^2 -(2x_2)^2) + 6 left( x_2-fracx_312 right)^2 - left( fracx_32 right)^2 + 8x_3^2$$
Can I do it like this? On complete square calculator, it's said that you cannot complete the square for expression like this. Anyway, the reason I need to complete the square for this expression is because I need to determine whether the given quadratic form is positive definite. But I don't know how to proceed.
polynomials quadratic-forms positive-definite positive-semidefinite
edited 10 hours ago
Rodrigo de Azevedo
13.4k42065
asked 11 hours ago
Ieva BrakmaneIeva Brakmane
1389
$endgroup$
add a comment |
$begingroup$
I have the quadratic form
$$g=x_1^2+6x_2^2+8x_3^2-4x_1x_2-6x_1x_3-x_2x_3$$
I have problems completing the square. I tried to rewrite the expression as follows
$$g=x_1^2-4x_1x_2+6x_2^2-x_2x_3+8x_3^2$$
Hence,
$$((x_1-2x_2)^2 -(2x_2)^2) + 6 left( x_2-fracx_312 right)^2 - left( fracx_32 right)^2 + 8x_3^2$$
Can I do it like this? On complete square calculator, it's said that you cannot complete the square for expression like this. Anyway, the reason I need to complete the square for this expression is because I need to determine whether the given quadratic form is positive definite. But I don't know how to proceed.
polynomials quadratic-forms positive-definite positive-semidefinite
edited 10 hours ago
Rodrigo de Azevedo
13.4k42065
asked 11 hours ago
Ieva BrakmaneIeva Brakmane
1389
$endgroup$
$begingroup$
You can always complete the square using Lagrange's Method. In my answer you can see an example of this. Using the form's matrix is also a good method, many times easier.
$endgroup$
– DonAntonio
10 hours ago
add a comment |
$begingroup$
I have the quadratic form
$$g=x_1^2+6x_2^2+8x_3^2-4x_1x_2-6x_1x_3-x_2x_3$$
I have problems completing the square. I tried to rewrite the expression as follows
$$g=x_1^2-4x_1x_2+6x_2^2-x_2x_3+8x_3^2$$
Hence,
$$((x_1-2x_2)^2 -(2x_2)^2) + 6 left( x_2-fracx_312 right)^2 - left( fracx_32 right)^2 + 8x_3^2$$
Can I do it like this? On complete square calculator, it's said that you cannot complete the square for expression like this. Anyway, the reason I need to complete the square for this expression is because I need to determine whether the given quadratic form is positive definite. But I don't know how to proceed.
polynomials quadratic-forms positive-definite positive-semidefinite
edited 10 hours ago
Rodrigo de Azevedo
13.4k42065
asked 11 hours ago
Ieva BrakmaneIeva Brakmane
1389
$endgroup$
I have the quadratic form
$$g=x_1^2+6x_2^2+8x_3^2-4x_1x_2-6x_1x_3-x_2x_3$$
I have problems completing the square. I tried to rewrite the expression as follows
$$g=x_1^2-4x_1x_2+6x_2^2-x_2x_3+8x_3^2$$
Hence,
$$((x_1-2x_2)^2 -(2x_2)^2) + 6 left( x_2-fracx_312 right)^2 - left( fracx_32 right)^2 + 8x_3^2$$
Can I do it like this? On complete square calculator, it's said that you cannot complete the square for expression like this. Anyway, the reason I need to complete the square for this expression is because I need to determine whether the given quadratic form is positive definite. But I don't know how to proceed.
polynomials quadratic-forms positive-definite positive-semidefinite
polynomials quadratic-forms positive-definite positive-semidefinite
edited 10 hours ago
Rodrigo de Azevedo
13.4k42065
asked 11 hours ago
Ieva BrakmaneIeva Brakmane
1389
edited 10 hours ago
Rodrigo de Azevedo
13.4k42065
asked 11 hours ago
Ieva BrakmaneIeva Brakmane
1389
edited 10 hours ago
Rodrigo de Azevedo
13.4k42065
edited 10 hours ago
Rodrigo de Azevedo
13.4k42065
edited 10 hours ago
Rodrigo de Azevedo
13.4k42065
13.4k42065
asked 11 hours ago
Ieva BrakmaneIeva Brakmane
1389
asked 11 hours ago
Ieva BrakmaneIeva Brakmane
1389
asked 11 hours ago
Ieva BrakmaneIeva Brakmane
1389
1389
$begingroup$
You can always complete the square using Lagrange's Method. In my answer you can see an example of this. Using the form's matrix is also a good method, many times easier.
$endgroup$
– DonAntonio
10 hours ago
add a comment |
$begingroup$
You can always complete the square using Lagrange's Method. In my answer you can see an example of this. Using the form's matrix is also a good method, many times easier.
$endgroup$
– DonAntonio
10 hours ago
$begingroup$
You can always complete the square using Lagrange's Method. In my answer you can see an example of this. Using the form's matrix is also a good method, many times easier.
$endgroup$
– DonAntonio
10 hours ago
$begingroup$
You can always complete the square using Lagrange's Method. In my answer you can see an example of this. Using the form's matrix is also a good method, many times easier.
$endgroup$
– DonAntonio
10 hours ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Construct the coefficient matrix
$$A = beginbmatrix 1& -2& -3\ -2& 6& -1/2 \-3 &-1/2 & 8endbmatrix$$
Find the determinants
$D_1 = |a_11| = 1 > 0$
$D_2 = beginvmatrix1&-2 \ -2&6endvmatrix = 6 -4 =2 > 0$
$D_3 = det(A) = -177/4 < 0$
So, the given form is indefinite in nature.
For an nxn matrix if
$bullet D_1 , D_2, cdots D_n>0 implies$ Positive definite
$bullet D_1,D_3,D_5cdots<0 $ and $D_2,D_4,D_6cdots>0$ or $(-1)^kD_k >0 , k =0,1,cdots n implies$ Negative definite
$bullet D_1 , D_2, cdots D_n ge0$ or $D_k>0 $ with at least one value zero $implies$ Positive semi-definite
$bullet(-1)^kD_k ge0 , k =0,1,cdots n$ with at least one zero $implies$ Negative semi-definite
$bullet$All other cases are indefinite
answered 11 hours ago
Ak19Ak19
3,296215
$endgroup$
add a comment |
$begingroup$
Completing the square (Lagrange's Method):
$$x^2+6y^2+8z^2-4xy-6xz-yz=left(x-(2y+3z)right)^2-4y^2-12yz-9z^2+6y^2+8z^2-yz$$
$$=left(x-(2y+3z)right)^2+2y^2-13yz-z^2=left(x-(2y+3z)right)^2+2left(y-frac134zright)^2-frac1778z^2$$
Thus, for example, $;g(38,,13,,4)=-177cdot2=-354<0;$ and thus the form isn't positive definite.
BTW, $;g(1,0,0)=1;$ , so the form is actually indefinite.
answered 10 hours ago
DonAntonioDonAntonio
181k1497234
$endgroup$
$begingroup$
Yes, completing the square is possible! (+1)
$endgroup$
– Robert Z
10 hours ago
$begingroup$
@DonAntonio Can you explain why did you chose $2y+3z$ a the beginning to form square? Also why the last coefficient to $z^2$ is $frac1778$ not $frac1698$? Then I don't uderstand why do you chose such values $g(38,13,4)$? Why if $g(1,0,0)= 1$ is the form indefinite? Would it be also indefinite if $g=(0,1,0)=1$?
$endgroup$
– Ieva Brakmane
8 hours ago
$begingroup$
(1) The rule is: take half the coefficient of the linear term to complete square. For example, $,ax^2+bx=aleft(x^2+frac baxright)=aleft(x+frac b2aright)^2-fracb^24a,$ and etc. In the given quadratic from, the coefficient of $,x,$ is$,-(4x+6z),$ , so half this...and etc.
$endgroup$
– DonAntonio
8 hours ago
$begingroup$
Also, I chose a vector as simple as I could. With $,w:=(38,13,4),$ we get $,g(w)<0;$ . Then, we only need a vector for which its value is negative and thus the form is indefinite (perhaps you need to go over the definition of "definite positive, definite negative, indefinite and etc.)
$endgroup$
– DonAntonio
8 hours ago
$begingroup$
@DonAntonio Thank you, now I understand about the coefficients. 1) Did you manually by try and error found which value would make the expression zero? 2) What about that coefficient to $z^2$, why it's $frac1778$ instead of $frac1698$?
$endgroup$
– Ieva Brakmane
7 hours ago
|
show 1 more comment
$begingroup$
Note that if $x_2=0$ then the quadratic form $g$ is
$$x_1^2+8x_3^2-6x_1x_3=(x_1-2x_3)(x_1-4x_3)$$
whose sign is indefinite. So $g$ is NOT positive definite.
answered 10 hours ago
Robert ZRobert Z
103k1073147
$endgroup$
$begingroup$
What do you mean by "you cannot complete the square"? Lagrange's method for completing the square in quadratic forms is based precisely on the fact that you always can complete the square.
$endgroup$
– DonAntonio
10 hours ago
$begingroup$
But you always can ! Read my answer. It is just Lagrange's method applied. Of course, negative coefficients allowed
$endgroup$
– DonAntonio
10 hours ago
$begingroup$
If that were the case then completing the square would be pretty useless in many cases...
$endgroup$
– DonAntonio
10 hours ago
add a comment |
$begingroup$
Note that $H$ is the Hessian matrix of second partial derivatives of $g=x^2+6y^2+8z^2-4xy-6xz-yz$
This direction is what "completing the square" produces first:
$$ Q^T D Q = H $$
$$left(
beginarrayrrr
1 & 0 & 0 \
- 2 & 1 & 0 \
- 3 & - frac 13 4 & 1 \
endarray
right)
left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & 0 \
0 & 0 & - frac 177 4 \
endarray
right)
left(
beginarrayrrr
1 & - 2 & - 3 \
0 & 1 & - frac 13 4 \
0 & 0 & 1 \
endarray
right)
= left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
$$
Dividing the coefficients of $D$ by two gives
$$ (x-2y-3z)^2 + 2 left( y - frac134 zright)^2 - frac1778 z^2 = x^2+6y^2+8z^2-4xy-6xz-yz$$
and is exactly what DonAntonio got
The method actually produces the reverse first:
$$ P^T H P = D $$
$$left(
beginarrayrrr
1 & 0 & 0 \
2 & 1 & 0 \
frac 19 2 & frac 13 4 & 1 \
endarray
right)
left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
left(
beginarrayrrr
1 & 2 & frac 19 2 \
0 & 1 & frac 13 4 \
0 & 0 & 1 \
endarray
right)
= left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & 0 \
0 & 0 & - frac 177 4 \
endarray
right)
$$
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
$$ H = left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
$$
$$ D_0 = H $$
$$ E_j^T D_j-1 E_j = D_j $$
$$ P_j-1 E_j = P_j $$
$$ E_j^-1 Q_j-1 = Q_j $$
$$ P_j Q_j = Q_j P_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$
$$ H = left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
$$
==============================================
$$ E_1 = left(
beginarrayrrr
1 & 2 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
$$
$$ P_1 = left(
beginarrayrrr
1 & 2 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
, ; ; ; Q_1 = left(
beginarrayrrr
1 & - 2 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
, ; ; ; D_1 = left(
beginarrayrrr
2 & 0 & - 6 \
0 & 4 & - 13 \
- 6 & - 13 & 16 \
endarray
right)
$$
==============================================
$$ E_2 = left(
beginarrayrrr
1 & 0 & 3 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
$$
$$ P_2 = left(
beginarrayrrr
1 & 2 & 3 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
, ; ; ; Q_2 = left(
beginarrayrrr
1 & - 2 & - 3 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
, ; ; ; D_2 = left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & - 13 \
0 & - 13 & - 2 \
endarray
right)
$$
==============================================
$$ E_3 = left(
beginarrayrrr
1 & 0 & 0 \
0 & 1 & frac 13 4 \
0 & 0 & 1 \
endarray
right)
$$
$$ P_3 = left(
beginarrayrrr
1 & 2 & frac 19 2 \
0 & 1 & frac 13 4 \
0 & 0 & 1 \
endarray
right)
, ; ; ; Q_3 = left(
beginarrayrrr
1 & - 2 & - 3 \
0 & 1 & - frac 13 4 \
0 & 0 & 1 \
endarray
right)
, ; ; ; D_3 = left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & 0 \
0 & 0 & - frac 177 4 \
endarray
right)
$$
==============================================
$$ P^T H P = D $$
$$left(
beginarrayrrr
1 & 0 & 0 \
2 & 1 & 0 \
frac 19 2 & frac 13 4 & 1 \
endarray
right)
left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
left(
beginarrayrrr
1 & 2 & frac 19 2 \
0 & 1 & frac 13 4 \
0 & 0 & 1 \
endarray
right)
= left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & 0 \
0 & 0 & - frac 177 4 \
endarray
right)
$$
$$ Q^T D Q = H $$
$$left(
beginarrayrrr
1 & 0 & 0 \
- 2 & 1 & 0 \
- 3 & - frac 13 4 & 1 \
endarray
right)
left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & 0 \
0 & 0 & - frac 177 4 \
endarray
right)
left(
beginarrayrrr
1 & - 2 & - 3 \
0 & 1 & - frac 13 4 \
0 & 0 & 1 \
endarray
right)
= left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
$$
answered 8 hours ago
Will JagyWill Jagy
106k5105206
$endgroup$
$begingroup$
I am lost as to what you really tried to do here, though I suspect is something close to "elementary congruency" or "congruency transformation: carrying on row elementary operations and exactly the same column operations ...etc. until the symmetric* matrix is diagonalized. The term "Hessian matrix" is known to me from advanced calculus, but I can't see here how it pops up: if you meant the Hessian of $;g;$ as function $;Bbb R^3toBbb R;$ , then you get a diagonal one and not pretty useful, as far as I can see, matrix, as all the mixed partial derivatives of second order vanish...
$endgroup$
– DonAntonio
7 hours ago
$begingroup$
@DonAntonio $H$ is the Hessian of $g=x^2+6y^2+8z^2-4xy-6xz-yz ; . ;$ Sometimes this is called congruence diagonalization. It tells us the signs of the eigenvalues of $H,$ according to Sylvester's Law of Inertia. I give a fair amount of detail at math.stackexchange.com/questions/1388421/… Note that the matrix version $Q^T D Q = H$ is almost exactly what you gave.
$endgroup$
– Will Jagy
7 hours ago
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Construct the coefficient matrix
$$A = beginbmatrix 1& -2& -3\ -2& 6& -1/2 \-3 &-1/2 & 8endbmatrix$$
Find the determinants
$D_1 = |a_11| = 1 > 0$
$D_2 = beginvmatrix1&-2 \ -2&6endvmatrix = 6 -4 =2 > 0$
$D_3 = det(A) = -177/4 < 0$
So, the given form is indefinite in nature.
For an nxn matrix if
$bullet D_1 , D_2, cdots D_n>0 implies$ Positive definite
$bullet D_1,D_3,D_5cdots<0 $ and $D_2,D_4,D_6cdots>0$ or $(-1)^kD_k >0 , k =0,1,cdots n implies$ Negative definite
$bullet D_1 , D_2, cdots D_n ge0$ or $D_k>0 $ with at least one value zero $implies$ Positive semi-definite
$bullet(-1)^kD_k ge0 , k =0,1,cdots n$ with at least one zero $implies$ Negative semi-definite
$bullet$All other cases are indefinite
answered 11 hours ago
Ak19Ak19
3,296215
$endgroup$
add a comment |
$begingroup$
Construct the coefficient matrix
$$A = beginbmatrix 1& -2& -3\ -2& 6& -1/2 \-3 &-1/2 & 8endbmatrix$$
Find the determinants
$D_1 = |a_11| = 1 > 0$
$D_2 = beginvmatrix1&-2 \ -2&6endvmatrix = 6 -4 =2 > 0$
$D_3 = det(A) = -177/4 < 0$
So, the given form is indefinite in nature.
For an nxn matrix if
$bullet D_1 , D_2, cdots D_n>0 implies$ Positive definite
$bullet D_1,D_3,D_5cdots<0 $ and $D_2,D_4,D_6cdots>0$ or $(-1)^kD_k >0 , k =0,1,cdots n implies$ Negative definite
$bullet D_1 , D_2, cdots D_n ge0$ or $D_k>0 $ with at least one value zero $implies$ Positive semi-definite
$bullet(-1)^kD_k ge0 , k =0,1,cdots n$ with at least one zero $implies$ Negative semi-definite
$bullet$All other cases are indefinite
answered 11 hours ago
Ak19Ak19
3,296215
$endgroup$
add a comment |
$begingroup$
Construct the coefficient matrix
$$A = beginbmatrix 1& -2& -3\ -2& 6& -1/2 \-3 &-1/2 & 8endbmatrix$$
Find the determinants
$D_1 = |a_11| = 1 > 0$
$D_2 = beginvmatrix1&-2 \ -2&6endvmatrix = 6 -4 =2 > 0$
$D_3 = det(A) = -177/4 < 0$
So, the given form is indefinite in nature.
For an nxn matrix if
$bullet D_1 , D_2, cdots D_n>0 implies$ Positive definite
$bullet D_1,D_3,D_5cdots<0 $ and $D_2,D_4,D_6cdots>0$ or $(-1)^kD_k >0 , k =0,1,cdots n implies$ Negative definite
$bullet D_1 , D_2, cdots D_n ge0$ or $D_k>0 $ with at least one value zero $implies$ Positive semi-definite
$bullet(-1)^kD_k ge0 , k =0,1,cdots n$ with at least one zero $implies$ Negative semi-definite
$bullet$All other cases are indefinite
answered 11 hours ago
Ak19Ak19
3,296215
$endgroup$
Construct the coefficient matrix
$$A = beginbmatrix 1& -2& -3\ -2& 6& -1/2 \-3 &-1/2 & 8endbmatrix$$
Find the determinants
$D_1 = |a_11| = 1 > 0$
$D_2 = beginvmatrix1&-2 \ -2&6endvmatrix = 6 -4 =2 > 0$
$D_3 = det(A) = -177/4 < 0$
So, the given form is indefinite in nature.
For an nxn matrix if
$bullet D_1 , D_2, cdots D_n>0 implies$ Positive definite
$bullet D_1,D_3,D_5cdots<0 $ and $D_2,D_4,D_6cdots>0$ or $(-1)^kD_k >0 , k =0,1,cdots n implies$ Negative definite
$bullet D_1 , D_2, cdots D_n ge0$ or $D_k>0 $ with at least one value zero $implies$ Positive semi-definite
$bullet(-1)^kD_k ge0 , k =0,1,cdots n$ with at least one zero $implies$ Negative semi-definite
$bullet$All other cases are indefinite
answered 11 hours ago
Ak19Ak19
3,296215
edited 10 hours ago
answered 11 hours ago
Ak19Ak19
3,296215
answered 11 hours ago
Ak19Ak19
3,296215
answered 11 hours ago
Ak19Ak19
3,296215
3,296215
add a comment |
add a comment |
$begingroup$
Completing the square (Lagrange's Method):
$$x^2+6y^2+8z^2-4xy-6xz-yz=left(x-(2y+3z)right)^2-4y^2-12yz-9z^2+6y^2+8z^2-yz$$
$$=left(x-(2y+3z)right)^2+2y^2-13yz-z^2=left(x-(2y+3z)right)^2+2left(y-frac134zright)^2-frac1778z^2$$
Thus, for example, $;g(38,,13,,4)=-177cdot2=-354<0;$ and thus the form isn't positive definite.
BTW, $;g(1,0,0)=1;$ , so the form is actually indefinite.
answered 10 hours ago
DonAntonioDonAntonio
181k1497234
$endgroup$
$begingroup$
Yes, completing the square is possible! (+1)
$endgroup$
– Robert Z
10 hours ago
$begingroup$
@DonAntonio Can you explain why did you chose $2y+3z$ a the beginning to form square? Also why the last coefficient to $z^2$ is $frac1778$ not $frac1698$? Then I don't uderstand why do you chose such values $g(38,13,4)$? Why if $g(1,0,0)= 1$ is the form indefinite? Would it be also indefinite if $g=(0,1,0)=1$?
$endgroup$
– Ieva Brakmane
8 hours ago
$begingroup$
(1) The rule is: take half the coefficient of the linear term to complete square. For example, $,ax^2+bx=aleft(x^2+frac baxright)=aleft(x+frac b2aright)^2-fracb^24a,$ and etc. In the given quadratic from, the coefficient of $,x,$ is$,-(4x+6z),$ , so half this...and etc.
$endgroup$
– DonAntonio
8 hours ago
$begingroup$
Also, I chose a vector as simple as I could. With $,w:=(38,13,4),$ we get $,g(w)<0;$ . Then, we only need a vector for which its value is negative and thus the form is indefinite (perhaps you need to go over the definition of "definite positive, definite negative, indefinite and etc.)
$endgroup$
– DonAntonio
8 hours ago
$begingroup$
@DonAntonio Thank you, now I understand about the coefficients. 1) Did you manually by try and error found which value would make the expression zero? 2) What about that coefficient to $z^2$, why it's $frac1778$ instead of $frac1698$?
$endgroup$
– Ieva Brakmane
7 hours ago
|
show 1 more comment
$begingroup$
Completing the square (Lagrange's Method):
$$x^2+6y^2+8z^2-4xy-6xz-yz=left(x-(2y+3z)right)^2-4y^2-12yz-9z^2+6y^2+8z^2-yz$$
$$=left(x-(2y+3z)right)^2+2y^2-13yz-z^2=left(x-(2y+3z)right)^2+2left(y-frac134zright)^2-frac1778z^2$$
Thus, for example, $;g(38,,13,,4)=-177cdot2=-354<0;$ and thus the form isn't positive definite.
BTW, $;g(1,0,0)=1;$ , so the form is actually indefinite.
answered 10 hours ago
DonAntonioDonAntonio
181k1497234
$endgroup$
$begingroup$
Yes, completing the square is possible! (+1)
$endgroup$
– Robert Z
10 hours ago
$begingroup$
@DonAntonio Can you explain why did you chose $2y+3z$ a the beginning to form square? Also why the last coefficient to $z^2$ is $frac1778$ not $frac1698$? Then I don't uderstand why do you chose such values $g(38,13,4)$? Why if $g(1,0,0)= 1$ is the form indefinite? Would it be also indefinite if $g=(0,1,0)=1$?
$endgroup$
– Ieva Brakmane
8 hours ago
$begingroup$
(1) The rule is: take half the coefficient of the linear term to complete square. For example, $,ax^2+bx=aleft(x^2+frac baxright)=aleft(x+frac b2aright)^2-fracb^24a,$ and etc. In the given quadratic from, the coefficient of $,x,$ is$,-(4x+6z),$ , so half this...and etc.
$endgroup$
– DonAntonio
8 hours ago
$begingroup$
Also, I chose a vector as simple as I could. With $,w:=(38,13,4),$ we get $,g(w)<0;$ . Then, we only need a vector for which its value is negative and thus the form is indefinite (perhaps you need to go over the definition of "definite positive, definite negative, indefinite and etc.)
$endgroup$
– DonAntonio
8 hours ago
$begingroup$
@DonAntonio Thank you, now I understand about the coefficients. 1) Did you manually by try and error found which value would make the expression zero? 2) What about that coefficient to $z^2$, why it's $frac1778$ instead of $frac1698$?
$endgroup$
– Ieva Brakmane
7 hours ago
|
show 1 more comment
$begingroup$
Completing the square (Lagrange's Method):
$$x^2+6y^2+8z^2-4xy-6xz-yz=left(x-(2y+3z)right)^2-4y^2-12yz-9z^2+6y^2+8z^2-yz$$
$$=left(x-(2y+3z)right)^2+2y^2-13yz-z^2=left(x-(2y+3z)right)^2+2left(y-frac134zright)^2-frac1778z^2$$
Thus, for example, $;g(38,,13,,4)=-177cdot2=-354<0;$ and thus the form isn't positive definite.
BTW, $;g(1,0,0)=1;$ , so the form is actually indefinite.
answered 10 hours ago
DonAntonioDonAntonio
181k1497234
$endgroup$
Completing the square (Lagrange's Method):
$$x^2+6y^2+8z^2-4xy-6xz-yz=left(x-(2y+3z)right)^2-4y^2-12yz-9z^2+6y^2+8z^2-yz$$
$$=left(x-(2y+3z)right)^2+2y^2-13yz-z^2=left(x-(2y+3z)right)^2+2left(y-frac134zright)^2-frac1778z^2$$
Thus, for example, $;g(38,,13,,4)=-177cdot2=-354<0;$ and thus the form isn't positive definite.
BTW, $;g(1,0,0)=1;$ , so the form is actually indefinite.
answered 10 hours ago
DonAntonioDonAntonio
181k1497234
edited 10 hours ago
answered 10 hours ago
DonAntonioDonAntonio
181k1497234
answered 10 hours ago
DonAntonioDonAntonio
181k1497234
answered 10 hours ago
DonAntonioDonAntonio
181k1497234
181k1497234
$begingroup$
Yes, completing the square is possible! (+1)
$endgroup$
– Robert Z
10 hours ago
$begingroup$
@DonAntonio Can you explain why did you chose $2y+3z$ a the beginning to form square? Also why the last coefficient to $z^2$ is $frac1778$ not $frac1698$? Then I don't uderstand why do you chose such values $g(38,13,4)$? Why if $g(1,0,0)= 1$ is the form indefinite? Would it be also indefinite if $g=(0,1,0)=1$?
$endgroup$
– Ieva Brakmane
8 hours ago
$begingroup$
(1) The rule is: take half the coefficient of the linear term to complete square. For example, $,ax^2+bx=aleft(x^2+frac baxright)=aleft(x+frac b2aright)^2-fracb^24a,$ and etc. In the given quadratic from, the coefficient of $,x,$ is$,-(4x+6z),$ , so half this...and etc.
$endgroup$
– DonAntonio
8 hours ago
$begingroup$
Also, I chose a vector as simple as I could. With $,w:=(38,13,4),$ we get $,g(w)<0;$ . Then, we only need a vector for which its value is negative and thus the form is indefinite (perhaps you need to go over the definition of "definite positive, definite negative, indefinite and etc.)
$endgroup$
– DonAntonio
8 hours ago
$begingroup$
@DonAntonio Thank you, now I understand about the coefficients. 1) Did you manually by try and error found which value would make the expression zero? 2) What about that coefficient to $z^2$, why it's $frac1778$ instead of $frac1698$?
$endgroup$
– Ieva Brakmane
7 hours ago
|
show 1 more comment
$begingroup$
Yes, completing the square is possible! (+1)
$endgroup$
– Robert Z
10 hours ago
$begingroup$
@DonAntonio Can you explain why did you chose $2y+3z$ a the beginning to form square? Also why the last coefficient to $z^2$ is $frac1778$ not $frac1698$? Then I don't uderstand why do you chose such values $g(38,13,4)$? Why if $g(1,0,0)= 1$ is the form indefinite? Would it be also indefinite if $g=(0,1,0)=1$?
$endgroup$
– Ieva Brakmane
8 hours ago
$begingroup$
(1) The rule is: take half the coefficient of the linear term to complete square. For example, $,ax^2+bx=aleft(x^2+frac baxright)=aleft(x+frac b2aright)^2-fracb^24a,$ and etc. In the given quadratic from, the coefficient of $,x,$ is$,-(4x+6z),$ , so half this...and etc.
$endgroup$
– DonAntonio
8 hours ago
$begingroup$
Also, I chose a vector as simple as I could. With $,w:=(38,13,4),$ we get $,g(w)<0;$ . Then, we only need a vector for which its value is negative and thus the form is indefinite (perhaps you need to go over the definition of "definite positive, definite negative, indefinite and etc.)
$endgroup$
– DonAntonio
8 hours ago
$begingroup$
@DonAntonio Thank you, now I understand about the coefficients. 1) Did you manually by try and error found which value would make the expression zero? 2) What about that coefficient to $z^2$, why it's $frac1778$ instead of $frac1698$?
$endgroup$
– Ieva Brakmane
7 hours ago
$begingroup$
Yes, completing the square is possible! (+1)
$endgroup$
– Robert Z
10 hours ago
$begingroup$
Yes, completing the square is possible! (+1)
$endgroup$
– Robert Z
10 hours ago
$begingroup$
@DonAntonio Can you explain why did you chose $2y+3z$ a the beginning to form square? Also why the last coefficient to $z^2$ is $frac1778$ not $frac1698$? Then I don't uderstand why do you chose such values $g(38,13,4)$? Why if $g(1,0,0)= 1$ is the form indefinite? Would it be also indefinite if $g=(0,1,0)=1$?
$endgroup$
– Ieva Brakmane
8 hours ago
$begingroup$
@DonAntonio Can you explain why did you chose $2y+3z$ a the beginning to form square? Also why the last coefficient to $z^2$ is $frac1778$ not $frac1698$? Then I don't uderstand why do you chose such values $g(38,13,4)$? Why if $g(1,0,0)= 1$ is the form indefinite? Would it be also indefinite if $g=(0,1,0)=1$?
$endgroup$
– Ieva Brakmane
8 hours ago
$begingroup$
(1) The rule is: take half the coefficient of the linear term to complete square. For example, $,ax^2+bx=aleft(x^2+frac baxright)=aleft(x+frac b2aright)^2-fracb^24a,$ and etc. In the given quadratic from, the coefficient of $,x,$ is$,-(4x+6z),$ , so half this...and etc.
$endgroup$
– DonAntonio
8 hours ago
$begingroup$
(1) The rule is: take half the coefficient of the linear term to complete square. For example, $,ax^2+bx=aleft(x^2+frac baxright)=aleft(x+frac b2aright)^2-fracb^24a,$ and etc. In the given quadratic from, the coefficient of $,x,$ is$,-(4x+6z),$ , so half this...and etc.
$endgroup$
– DonAntonio
8 hours ago
$begingroup$
Also, I chose a vector as simple as I could. With $,w:=(38,13,4),$ we get $,g(w)<0;$ . Then, we only need a vector for which its value is negative and thus the form is indefinite (perhaps you need to go over the definition of "definite positive, definite negative, indefinite and etc.)
$endgroup$
– DonAntonio
8 hours ago
$begingroup$
Also, I chose a vector as simple as I could. With $,w:=(38,13,4),$ we get $,g(w)<0;$ . Then, we only need a vector for which its value is negative and thus the form is indefinite (perhaps you need to go over the definition of "definite positive, definite negative, indefinite and etc.)
$endgroup$
– DonAntonio
8 hours ago
$begingroup$
@DonAntonio Thank you, now I understand about the coefficients. 1) Did you manually by try and error found which value would make the expression zero? 2) What about that coefficient to $z^2$, why it's $frac1778$ instead of $frac1698$?
$endgroup$
– Ieva Brakmane
7 hours ago
$begingroup$
@DonAntonio Thank you, now I understand about the coefficients. 1) Did you manually by try and error found which value would make the expression zero? 2) What about that coefficient to $z^2$, why it's $frac1778$ instead of $frac1698$?
$endgroup$
– Ieva Brakmane
7 hours ago
|
show 1 more comment
$begingroup$
Note that if $x_2=0$ then the quadratic form $g$ is
$$x_1^2+8x_3^2-6x_1x_3=(x_1-2x_3)(x_1-4x_3)$$
whose sign is indefinite. So $g$ is NOT positive definite.
answered 10 hours ago
Robert ZRobert Z
103k1073147
$endgroup$
$begingroup$
What do you mean by "you cannot complete the square"? Lagrange's method for completing the square in quadratic forms is based precisely on the fact that you always can complete the square.
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– DonAntonio
10 hours ago
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But you always can ! Read my answer. It is just Lagrange's method applied. Of course, negative coefficients allowed
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– DonAntonio
10 hours ago
$begingroup$
If that were the case then completing the square would be pretty useless in many cases...
$endgroup$
– DonAntonio
10 hours ago
add a comment |
$begingroup$
Note that if $x_2=0$ then the quadratic form $g$ is
$$x_1^2+8x_3^2-6x_1x_3=(x_1-2x_3)(x_1-4x_3)$$
whose sign is indefinite. So $g$ is NOT positive definite.
answered 10 hours ago
Robert ZRobert Z
103k1073147
$endgroup$
$begingroup$
What do you mean by "you cannot complete the square"? Lagrange's method for completing the square in quadratic forms is based precisely on the fact that you always can complete the square.
$endgroup$
– DonAntonio
10 hours ago
$begingroup$
But you always can ! Read my answer. It is just Lagrange's method applied. Of course, negative coefficients allowed
$endgroup$
– DonAntonio
10 hours ago
$begingroup$
If that were the case then completing the square would be pretty useless in many cases...
$endgroup$
– DonAntonio
10 hours ago
add a comment |
$begingroup$
Note that if $x_2=0$ then the quadratic form $g$ is
$$x_1^2+8x_3^2-6x_1x_3=(x_1-2x_3)(x_1-4x_3)$$
whose sign is indefinite. So $g$ is NOT positive definite.
answered 10 hours ago
Robert ZRobert Z
103k1073147
$endgroup$
Note that if $x_2=0$ then the quadratic form $g$ is
$$x_1^2+8x_3^2-6x_1x_3=(x_1-2x_3)(x_1-4x_3)$$
whose sign is indefinite. So $g$ is NOT positive definite.
answered 10 hours ago
Robert ZRobert Z
103k1073147
edited 10 hours ago
answered 10 hours ago
Robert ZRobert Z
103k1073147
answered 10 hours ago
Robert ZRobert Z
103k1073147
answered 10 hours ago
Robert ZRobert Z
103k1073147
103k1073147
$begingroup$
What do you mean by "you cannot complete the square"? Lagrange's method for completing the square in quadratic forms is based precisely on the fact that you always can complete the square.
$endgroup$
– DonAntonio
10 hours ago
$begingroup$
But you always can ! Read my answer. It is just Lagrange's method applied. Of course, negative coefficients allowed
$endgroup$
– DonAntonio
10 hours ago
$begingroup$
If that were the case then completing the square would be pretty useless in many cases...
$endgroup$
– DonAntonio
10 hours ago
add a comment |
$begingroup$
What do you mean by "you cannot complete the square"? Lagrange's method for completing the square in quadratic forms is based precisely on the fact that you always can complete the square.
$endgroup$
– DonAntonio
10 hours ago
$begingroup$
But you always can ! Read my answer. It is just Lagrange's method applied. Of course, negative coefficients allowed
$endgroup$
– DonAntonio
10 hours ago
$begingroup$
If that were the case then completing the square would be pretty useless in many cases...
$endgroup$
– DonAntonio
10 hours ago
$begingroup$
What do you mean by "you cannot complete the square"? Lagrange's method for completing the square in quadratic forms is based precisely on the fact that you always can complete the square.
$endgroup$
– DonAntonio
10 hours ago
$begingroup$
What do you mean by "you cannot complete the square"? Lagrange's method for completing the square in quadratic forms is based precisely on the fact that you always can complete the square.
$endgroup$
– DonAntonio
10 hours ago
$begingroup$
But you always can ! Read my answer. It is just Lagrange's method applied. Of course, negative coefficients allowed
$endgroup$
– DonAntonio
10 hours ago
$begingroup$
But you always can ! Read my answer. It is just Lagrange's method applied. Of course, negative coefficients allowed
$endgroup$
– DonAntonio
10 hours ago
$begingroup$
If that were the case then completing the square would be pretty useless in many cases...
$endgroup$
– DonAntonio
10 hours ago
$begingroup$
If that were the case then completing the square would be pretty useless in many cases...
$endgroup$
– DonAntonio
10 hours ago
add a comment |
$begingroup$
Note that $H$ is the Hessian matrix of second partial derivatives of $g=x^2+6y^2+8z^2-4xy-6xz-yz$
This direction is what "completing the square" produces first:
$$ Q^T D Q = H $$
$$left(
beginarrayrrr
1 & 0 & 0 \
- 2 & 1 & 0 \
- 3 & - frac 13 4 & 1 \
endarray
right)
left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & 0 \
0 & 0 & - frac 177 4 \
endarray
right)
left(
beginarrayrrr
1 & - 2 & - 3 \
0 & 1 & - frac 13 4 \
0 & 0 & 1 \
endarray
right)
= left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
$$
Dividing the coefficients of $D$ by two gives
$$ (x-2y-3z)^2 + 2 left( y - frac134 zright)^2 - frac1778 z^2 = x^2+6y^2+8z^2-4xy-6xz-yz$$
and is exactly what DonAntonio got
The method actually produces the reverse first:
$$ P^T H P = D $$
$$left(
beginarrayrrr
1 & 0 & 0 \
2 & 1 & 0 \
frac 19 2 & frac 13 4 & 1 \
endarray
right)
left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
left(
beginarrayrrr
1 & 2 & frac 19 2 \
0 & 1 & frac 13 4 \
0 & 0 & 1 \
endarray
right)
= left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & 0 \
0 & 0 & - frac 177 4 \
endarray
right)
$$
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
$$ H = left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
$$
$$ D_0 = H $$
$$ E_j^T D_j-1 E_j = D_j $$
$$ P_j-1 E_j = P_j $$
$$ E_j^-1 Q_j-1 = Q_j $$
$$ P_j Q_j = Q_j P_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$
$$ H = left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
$$
==============================================
$$ E_1 = left(
beginarrayrrr
1 & 2 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
$$
$$ P_1 = left(
beginarrayrrr
1 & 2 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
, ; ; ; Q_1 = left(
beginarrayrrr
1 & - 2 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
, ; ; ; D_1 = left(
beginarrayrrr
2 & 0 & - 6 \
0 & 4 & - 13 \
- 6 & - 13 & 16 \
endarray
right)
$$
==============================================
$$ E_2 = left(
beginarrayrrr
1 & 0 & 3 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
$$
$$ P_2 = left(
beginarrayrrr
1 & 2 & 3 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
, ; ; ; Q_2 = left(
beginarrayrrr
1 & - 2 & - 3 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
, ; ; ; D_2 = left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & - 13 \
0 & - 13 & - 2 \
endarray
right)
$$
==============================================
$$ E_3 = left(
beginarrayrrr
1 & 0 & 0 \
0 & 1 & frac 13 4 \
0 & 0 & 1 \
endarray
right)
$$
$$ P_3 = left(
beginarrayrrr
1 & 2 & frac 19 2 \
0 & 1 & frac 13 4 \
0 & 0 & 1 \
endarray
right)
, ; ; ; Q_3 = left(
beginarrayrrr
1 & - 2 & - 3 \
0 & 1 & - frac 13 4 \
0 & 0 & 1 \
endarray
right)
, ; ; ; D_3 = left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & 0 \
0 & 0 & - frac 177 4 \
endarray
right)
$$
==============================================
$$ P^T H P = D $$
$$left(
beginarrayrrr
1 & 0 & 0 \
2 & 1 & 0 \
frac 19 2 & frac 13 4 & 1 \
endarray
right)
left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
left(
beginarrayrrr
1 & 2 & frac 19 2 \
0 & 1 & frac 13 4 \
0 & 0 & 1 \
endarray
right)
= left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & 0 \
0 & 0 & - frac 177 4 \
endarray
right)
$$
$$ Q^T D Q = H $$
$$left(
beginarrayrrr
1 & 0 & 0 \
- 2 & 1 & 0 \
- 3 & - frac 13 4 & 1 \
endarray
right)
left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & 0 \
0 & 0 & - frac 177 4 \
endarray
right)
left(
beginarrayrrr
1 & - 2 & - 3 \
0 & 1 & - frac 13 4 \
0 & 0 & 1 \
endarray
right)
= left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
$$
answered 8 hours ago
Will JagyWill Jagy
106k5105206
$endgroup$
$begingroup$
I am lost as to what you really tried to do here, though I suspect is something close to "elementary congruency" or "congruency transformation: carrying on row elementary operations and exactly the same column operations ...etc. until the symmetric* matrix is diagonalized. The term "Hessian matrix" is known to me from advanced calculus, but I can't see here how it pops up: if you meant the Hessian of $;g;$ as function $;Bbb R^3toBbb R;$ , then you get a diagonal one and not pretty useful, as far as I can see, matrix, as all the mixed partial derivatives of second order vanish...
$endgroup$
– DonAntonio
7 hours ago
$begingroup$
@DonAntonio $H$ is the Hessian of $g=x^2+6y^2+8z^2-4xy-6xz-yz ; . ;$ Sometimes this is called congruence diagonalization. It tells us the signs of the eigenvalues of $H,$ according to Sylvester's Law of Inertia. I give a fair amount of detail at math.stackexchange.com/questions/1388421/… Note that the matrix version $Q^T D Q = H$ is almost exactly what you gave.
$endgroup$
– Will Jagy
7 hours ago
add a comment |
$begingroup$
Note that $H$ is the Hessian matrix of second partial derivatives of $g=x^2+6y^2+8z^2-4xy-6xz-yz$
This direction is what "completing the square" produces first:
$$ Q^T D Q = H $$
$$left(
beginarrayrrr
1 & 0 & 0 \
- 2 & 1 & 0 \
- 3 & - frac 13 4 & 1 \
endarray
right)
left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & 0 \
0 & 0 & - frac 177 4 \
endarray
right)
left(
beginarrayrrr
1 & - 2 & - 3 \
0 & 1 & - frac 13 4 \
0 & 0 & 1 \
endarray
right)
= left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
$$
Dividing the coefficients of $D$ by two gives
$$ (x-2y-3z)^2 + 2 left( y - frac134 zright)^2 - frac1778 z^2 = x^2+6y^2+8z^2-4xy-6xz-yz$$
and is exactly what DonAntonio got
The method actually produces the reverse first:
$$ P^T H P = D $$
$$left(
beginarrayrrr
1 & 0 & 0 \
2 & 1 & 0 \
frac 19 2 & frac 13 4 & 1 \
endarray
right)
left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
left(
beginarrayrrr
1 & 2 & frac 19 2 \
0 & 1 & frac 13 4 \
0 & 0 & 1 \
endarray
right)
= left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & 0 \
0 & 0 & - frac 177 4 \
endarray
right)
$$
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
$$ H = left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
$$
$$ D_0 = H $$
$$ E_j^T D_j-1 E_j = D_j $$
$$ P_j-1 E_j = P_j $$
$$ E_j^-1 Q_j-1 = Q_j $$
$$ P_j Q_j = Q_j P_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$
$$ H = left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
$$
==============================================
$$ E_1 = left(
beginarrayrrr
1 & 2 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
$$
$$ P_1 = left(
beginarrayrrr
1 & 2 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
, ; ; ; Q_1 = left(
beginarrayrrr
1 & - 2 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
, ; ; ; D_1 = left(
beginarrayrrr
2 & 0 & - 6 \
0 & 4 & - 13 \
- 6 & - 13 & 16 \
endarray
right)
$$
==============================================
$$ E_2 = left(
beginarrayrrr
1 & 0 & 3 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
$$
$$ P_2 = left(
beginarrayrrr
1 & 2 & 3 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
, ; ; ; Q_2 = left(
beginarrayrrr
1 & - 2 & - 3 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
, ; ; ; D_2 = left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & - 13 \
0 & - 13 & - 2 \
endarray
right)
$$
==============================================
$$ E_3 = left(
beginarrayrrr
1 & 0 & 0 \
0 & 1 & frac 13 4 \
0 & 0 & 1 \
endarray
right)
$$
$$ P_3 = left(
beginarrayrrr
1 & 2 & frac 19 2 \
0 & 1 & frac 13 4 \
0 & 0 & 1 \
endarray
right)
, ; ; ; Q_3 = left(
beginarrayrrr
1 & - 2 & - 3 \
0 & 1 & - frac 13 4 \
0 & 0 & 1 \
endarray
right)
, ; ; ; D_3 = left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & 0 \
0 & 0 & - frac 177 4 \
endarray
right)
$$
==============================================
$$ P^T H P = D $$
$$left(
beginarrayrrr
1 & 0 & 0 \
2 & 1 & 0 \
frac 19 2 & frac 13 4 & 1 \
endarray
right)
left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
left(
beginarrayrrr
1 & 2 & frac 19 2 \
0 & 1 & frac 13 4 \
0 & 0 & 1 \
endarray
right)
= left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & 0 \
0 & 0 & - frac 177 4 \
endarray
right)
$$
$$ Q^T D Q = H $$
$$left(
beginarrayrrr
1 & 0 & 0 \
- 2 & 1 & 0 \
- 3 & - frac 13 4 & 1 \
endarray
right)
left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & 0 \
0 & 0 & - frac 177 4 \
endarray
right)
left(
beginarrayrrr
1 & - 2 & - 3 \
0 & 1 & - frac 13 4 \
0 & 0 & 1 \
endarray
right)
= left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
$$
answered 8 hours ago
Will JagyWill Jagy
106k5105206
$endgroup$
$begingroup$
I am lost as to what you really tried to do here, though I suspect is something close to "elementary congruency" or "congruency transformation: carrying on row elementary operations and exactly the same column operations ...etc. until the symmetric* matrix is diagonalized. The term "Hessian matrix" is known to me from advanced calculus, but I can't see here how it pops up: if you meant the Hessian of $;g;$ as function $;Bbb R^3toBbb R;$ , then you get a diagonal one and not pretty useful, as far as I can see, matrix, as all the mixed partial derivatives of second order vanish...
$endgroup$
– DonAntonio
7 hours ago
$begingroup$
@DonAntonio $H$ is the Hessian of $g=x^2+6y^2+8z^2-4xy-6xz-yz ; . ;$ Sometimes this is called congruence diagonalization. It tells us the signs of the eigenvalues of $H,$ according to Sylvester's Law of Inertia. I give a fair amount of detail at math.stackexchange.com/questions/1388421/… Note that the matrix version $Q^T D Q = H$ is almost exactly what you gave.
$endgroup$
– Will Jagy
7 hours ago
add a comment |
$begingroup$
Note that $H$ is the Hessian matrix of second partial derivatives of $g=x^2+6y^2+8z^2-4xy-6xz-yz$
This direction is what "completing the square" produces first:
$$ Q^T D Q = H $$
$$left(
beginarrayrrr
1 & 0 & 0 \
- 2 & 1 & 0 \
- 3 & - frac 13 4 & 1 \
endarray
right)
left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & 0 \
0 & 0 & - frac 177 4 \
endarray
right)
left(
beginarrayrrr
1 & - 2 & - 3 \
0 & 1 & - frac 13 4 \
0 & 0 & 1 \
endarray
right)
= left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
$$
Dividing the coefficients of $D$ by two gives
$$ (x-2y-3z)^2 + 2 left( y - frac134 zright)^2 - frac1778 z^2 = x^2+6y^2+8z^2-4xy-6xz-yz$$
and is exactly what DonAntonio got
The method actually produces the reverse first:
$$ P^T H P = D $$
$$left(
beginarrayrrr
1 & 0 & 0 \
2 & 1 & 0 \
frac 19 2 & frac 13 4 & 1 \
endarray
right)
left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
left(
beginarrayrrr
1 & 2 & frac 19 2 \
0 & 1 & frac 13 4 \
0 & 0 & 1 \
endarray
right)
= left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & 0 \
0 & 0 & - frac 177 4 \
endarray
right)
$$
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
$$ H = left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
$$
$$ D_0 = H $$
$$ E_j^T D_j-1 E_j = D_j $$
$$ P_j-1 E_j = P_j $$
$$ E_j^-1 Q_j-1 = Q_j $$
$$ P_j Q_j = Q_j P_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$
$$ H = left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
$$
==============================================
$$ E_1 = left(
beginarrayrrr
1 & 2 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
$$
$$ P_1 = left(
beginarrayrrr
1 & 2 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
, ; ; ; Q_1 = left(
beginarrayrrr
1 & - 2 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
, ; ; ; D_1 = left(
beginarrayrrr
2 & 0 & - 6 \
0 & 4 & - 13 \
- 6 & - 13 & 16 \
endarray
right)
$$
==============================================
$$ E_2 = left(
beginarrayrrr
1 & 0 & 3 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
$$
$$ P_2 = left(
beginarrayrrr
1 & 2 & 3 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
, ; ; ; Q_2 = left(
beginarrayrrr
1 & - 2 & - 3 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
, ; ; ; D_2 = left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & - 13 \
0 & - 13 & - 2 \
endarray
right)
$$
==============================================
$$ E_3 = left(
beginarrayrrr
1 & 0 & 0 \
0 & 1 & frac 13 4 \
0 & 0 & 1 \
endarray
right)
$$
$$ P_3 = left(
beginarrayrrr
1 & 2 & frac 19 2 \
0 & 1 & frac 13 4 \
0 & 0 & 1 \
endarray
right)
, ; ; ; Q_3 = left(
beginarrayrrr
1 & - 2 & - 3 \
0 & 1 & - frac 13 4 \
0 & 0 & 1 \
endarray
right)
, ; ; ; D_3 = left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & 0 \
0 & 0 & - frac 177 4 \
endarray
right)
$$
==============================================
$$ P^T H P = D $$
$$left(
beginarrayrrr
1 & 0 & 0 \
2 & 1 & 0 \
frac 19 2 & frac 13 4 & 1 \
endarray
right)
left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
left(
beginarrayrrr
1 & 2 & frac 19 2 \
0 & 1 & frac 13 4 \
0 & 0 & 1 \
endarray
right)
= left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & 0 \
0 & 0 & - frac 177 4 \
endarray
right)
$$
$$ Q^T D Q = H $$
$$left(
beginarrayrrr
1 & 0 & 0 \
- 2 & 1 & 0 \
- 3 & - frac 13 4 & 1 \
endarray
right)
left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & 0 \
0 & 0 & - frac 177 4 \
endarray
right)
left(
beginarrayrrr
1 & - 2 & - 3 \
0 & 1 & - frac 13 4 \
0 & 0 & 1 \
endarray
right)
= left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
$$
answered 8 hours ago
Will JagyWill Jagy
106k5105206
$endgroup$
Note that $H$ is the Hessian matrix of second partial derivatives of $g=x^2+6y^2+8z^2-4xy-6xz-yz$
This direction is what "completing the square" produces first:
$$ Q^T D Q = H $$
$$left(
beginarrayrrr
1 & 0 & 0 \
- 2 & 1 & 0 \
- 3 & - frac 13 4 & 1 \
endarray
right)
left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & 0 \
0 & 0 & - frac 177 4 \
endarray
right)
left(
beginarrayrrr
1 & - 2 & - 3 \
0 & 1 & - frac 13 4 \
0 & 0 & 1 \
endarray
right)
= left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
$$
Dividing the coefficients of $D$ by two gives
$$ (x-2y-3z)^2 + 2 left( y - frac134 zright)^2 - frac1778 z^2 = x^2+6y^2+8z^2-4xy-6xz-yz$$
and is exactly what DonAntonio got
The method actually produces the reverse first:
$$ P^T H P = D $$
$$left(
beginarrayrrr
1 & 0 & 0 \
2 & 1 & 0 \
frac 19 2 & frac 13 4 & 1 \
endarray
right)
left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
left(
beginarrayrrr
1 & 2 & frac 19 2 \
0 & 1 & frac 13 4 \
0 & 0 & 1 \
endarray
right)
= left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & 0 \
0 & 0 & - frac 177 4 \
endarray
right)
$$
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
$$ H = left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
$$
$$ D_0 = H $$
$$ E_j^T D_j-1 E_j = D_j $$
$$ P_j-1 E_j = P_j $$
$$ E_j^-1 Q_j-1 = Q_j $$
$$ P_j Q_j = Q_j P_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$
$$ H = left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
$$
==============================================
$$ E_1 = left(
beginarrayrrr
1 & 2 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
$$
$$ P_1 = left(
beginarrayrrr
1 & 2 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
, ; ; ; Q_1 = left(
beginarrayrrr
1 & - 2 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
, ; ; ; D_1 = left(
beginarrayrrr
2 & 0 & - 6 \
0 & 4 & - 13 \
- 6 & - 13 & 16 \
endarray
right)
$$
==============================================
$$ E_2 = left(
beginarrayrrr
1 & 0 & 3 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
$$
$$ P_2 = left(
beginarrayrrr
1 & 2 & 3 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
, ; ; ; Q_2 = left(
beginarrayrrr
1 & - 2 & - 3 \
0 & 1 & 0 \
0 & 0 & 1 \
endarray
right)
, ; ; ; D_2 = left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & - 13 \
0 & - 13 & - 2 \
endarray
right)
$$
==============================================
$$ E_3 = left(
beginarrayrrr
1 & 0 & 0 \
0 & 1 & frac 13 4 \
0 & 0 & 1 \
endarray
right)
$$
$$ P_3 = left(
beginarrayrrr
1 & 2 & frac 19 2 \
0 & 1 & frac 13 4 \
0 & 0 & 1 \
endarray
right)
, ; ; ; Q_3 = left(
beginarrayrrr
1 & - 2 & - 3 \
0 & 1 & - frac 13 4 \
0 & 0 & 1 \
endarray
right)
, ; ; ; D_3 = left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & 0 \
0 & 0 & - frac 177 4 \
endarray
right)
$$
==============================================
$$ P^T H P = D $$
$$left(
beginarrayrrr
1 & 0 & 0 \
2 & 1 & 0 \
frac 19 2 & frac 13 4 & 1 \
endarray
right)
left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
left(
beginarrayrrr
1 & 2 & frac 19 2 \
0 & 1 & frac 13 4 \
0 & 0 & 1 \
endarray
right)
= left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & 0 \
0 & 0 & - frac 177 4 \
endarray
right)
$$
$$ Q^T D Q = H $$
$$left(
beginarrayrrr
1 & 0 & 0 \
- 2 & 1 & 0 \
- 3 & - frac 13 4 & 1 \
endarray
right)
left(
beginarrayrrr
2 & 0 & 0 \
0 & 4 & 0 \
0 & 0 & - frac 177 4 \
endarray
right)
left(
beginarrayrrr
1 & - 2 & - 3 \
0 & 1 & - frac 13 4 \
0 & 0 & 1 \
endarray
right)
= left(
beginarrayrrr
2 & - 4 & - 6 \
- 4 & 12 & - 1 \
- 6 & - 1 & 16 \
endarray
right)
$$
answered 8 hours ago
Will JagyWill Jagy
106k5105206
edited 7 hours ago
answered 8 hours ago
Will JagyWill Jagy
106k5105206
answered 8 hours ago
Will JagyWill Jagy
106k5105206
answered 8 hours ago
Will JagyWill Jagy
106k5105206
106k5105206
$begingroup$
I am lost as to what you really tried to do here, though I suspect is something close to "elementary congruency" or "congruency transformation: carrying on row elementary operations and exactly the same column operations ...etc. until the symmetric* matrix is diagonalized. The term "Hessian matrix" is known to me from advanced calculus, but I can't see here how it pops up: if you meant the Hessian of $;g;$ as function $;Bbb R^3toBbb R;$ , then you get a diagonal one and not pretty useful, as far as I can see, matrix, as all the mixed partial derivatives of second order vanish...
$endgroup$
– DonAntonio
7 hours ago
$begingroup$
@DonAntonio $H$ is the Hessian of $g=x^2+6y^2+8z^2-4xy-6xz-yz ; . ;$ Sometimes this is called congruence diagonalization. It tells us the signs of the eigenvalues of $H,$ according to Sylvester's Law of Inertia. I give a fair amount of detail at math.stackexchange.com/questions/1388421/… Note that the matrix version $Q^T D Q = H$ is almost exactly what you gave.
$endgroup$
– Will Jagy
7 hours ago
add a comment |
$begingroup$
I am lost as to what you really tried to do here, though I suspect is something close to "elementary congruency" or "congruency transformation: carrying on row elementary operations and exactly the same column operations ...etc. until the symmetric* matrix is diagonalized. The term "Hessian matrix" is known to me from advanced calculus, but I can't see here how it pops up: if you meant the Hessian of $;g;$ as function $;Bbb R^3toBbb R;$ , then you get a diagonal one and not pretty useful, as far as I can see, matrix, as all the mixed partial derivatives of second order vanish...
$endgroup$
– DonAntonio
7 hours ago
$begingroup$
@DonAntonio $H$ is the Hessian of $g=x^2+6y^2+8z^2-4xy-6xz-yz ; . ;$ Sometimes this is called congruence diagonalization. It tells us the signs of the eigenvalues of $H,$ according to Sylvester's Law of Inertia. I give a fair amount of detail at math.stackexchange.com/questions/1388421/… Note that the matrix version $Q^T D Q = H$ is almost exactly what you gave.
$endgroup$
– Will Jagy
7 hours ago
$begingroup$
I am lost as to what you really tried to do here, though I suspect is something close to "elementary congruency" or "congruency transformation: carrying on row elementary operations and exactly the same column operations ...etc. until the symmetric* matrix is diagonalized. The term "Hessian matrix" is known to me from advanced calculus, but I can't see here how it pops up: if you meant the Hessian of $;g;$ as function $;Bbb R^3toBbb R;$ , then you get a diagonal one and not pretty useful, as far as I can see, matrix, as all the mixed partial derivatives of second order vanish...
$endgroup$
– DonAntonio
7 hours ago
$begingroup$
I am lost as to what you really tried to do here, though I suspect is something close to "elementary congruency" or "congruency transformation: carrying on row elementary operations and exactly the same column operations ...etc. until the symmetric* matrix is diagonalized. The term "Hessian matrix" is known to me from advanced calculus, but I can't see here how it pops up: if you meant the Hessian of $;g;$ as function $;Bbb R^3toBbb R;$ , then you get a diagonal one and not pretty useful, as far as I can see, matrix, as all the mixed partial derivatives of second order vanish...
$endgroup$
– DonAntonio
7 hours ago
$begingroup$
@DonAntonio $H$ is the Hessian of $g=x^2+6y^2+8z^2-4xy-6xz-yz ; . ;$ Sometimes this is called congruence diagonalization. It tells us the signs of the eigenvalues of $H,$ according to Sylvester's Law of Inertia. I give a fair amount of detail at math.stackexchange.com/questions/1388421/… Note that the matrix version $Q^T D Q = H$ is almost exactly what you gave.
$endgroup$
– Will Jagy
7 hours ago
$begingroup$
@DonAntonio $H$ is the Hessian of $g=x^2+6y^2+8z^2-4xy-6xz-yz ; . ;$ Sometimes this is called congruence diagonalization. It tells us the signs of the eigenvalues of $H,$ according to Sylvester's Law of Inertia. I give a fair amount of detail at math.stackexchange.com/questions/1388421/… Note that the matrix version $Q^T D Q = H$ is almost exactly what you gave.
$endgroup$
– Will Jagy
7 hours ago
add a comment |
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$begingroup$
You can always complete the square using Lagrange's Method. In my answer you can see an example of this. Using the form's matrix is also a good method, many times easier.
$endgroup$
– DonAntonio
10 hours ago