Find a diagonal form of the quadratic form $f(x) = sum_{i = 1}^nx_i^2 + sum_{i < j}x_ix_j$
$begingroup$
Find a diagonal form of the quadratic form
$$f(x) = sum_{i = 1}^nx_i^2 + sum_{i < j}x_ix_j.$$
It turned out to be such a problem:
How to change quadratic form $f(x) = sum_{i = 1}^nx_i^2 + sum_{i < j}x_ix_j$ into diagonal form?
We can solve it with congruent transformation, although it is a little complex.
Besides, I tried to transform it into a diagonal form by orthogonal transformation, since its quadratic matrix has the same diagonal elements and the same non-diagonal elements(I thought that I can find a simple way to find its eigenvalues and eigenvectors, but failed. Maybe you can try it).
Is there any other way?
linear-algebra quadratic-forms
$endgroup$
|
show 7 more comments
$begingroup$
Find a diagonal form of the quadratic form
$$f(x) = sum_{i = 1}^nx_i^2 + sum_{i < j}x_ix_j.$$
It turned out to be such a problem:
How to change quadratic form $f(x) = sum_{i = 1}^nx_i^2 + sum_{i < j}x_ix_j$ into diagonal form?
We can solve it with congruent transformation, although it is a little complex.
Besides, I tried to transform it into a diagonal form by orthogonal transformation, since its quadratic matrix has the same diagonal elements and the same non-diagonal elements(I thought that I can find a simple way to find its eigenvalues and eigenvectors, but failed. Maybe you can try it).
Is there any other way?
linear-algebra quadratic-forms
$endgroup$
1
$begingroup$
What is "standard form"? Is it $f(x) = (Ax mid x)$ for some simmetric linear transformation $A$?
$endgroup$
– Will M.
Dec 18 '18 at 6:10
$begingroup$
I'm sorry, I don't translate very well. The term may be diagonal form. such as this form $q(x)=a_1 x_1^2 + a_2 x_2^2+ ldots +a_n x_n^2$
$endgroup$
– congee
Dec 18 '18 at 6:39
$begingroup$
I see, you want a change of variables to make $f$ diagonal.
$endgroup$
– Will M.
Dec 18 '18 at 6:40
1
$begingroup$
yes, I try to solve the problem with congruent transformation, but it is a little complex. I wonder if there is any other way, such as orthogonal transformation
$endgroup$
– congee
Dec 18 '18 at 6:46
$begingroup$
By the way, I did not vote down. Some people do but they are not coming to give reasons for their votes. Such a shameful behaviour.
$endgroup$
– Will M.
Dec 18 '18 at 6:52
|
show 7 more comments
$begingroup$
Find a diagonal form of the quadratic form
$$f(x) = sum_{i = 1}^nx_i^2 + sum_{i < j}x_ix_j.$$
It turned out to be such a problem:
How to change quadratic form $f(x) = sum_{i = 1}^nx_i^2 + sum_{i < j}x_ix_j$ into diagonal form?
We can solve it with congruent transformation, although it is a little complex.
Besides, I tried to transform it into a diagonal form by orthogonal transformation, since its quadratic matrix has the same diagonal elements and the same non-diagonal elements(I thought that I can find a simple way to find its eigenvalues and eigenvectors, but failed. Maybe you can try it).
Is there any other way?
linear-algebra quadratic-forms
$endgroup$
Find a diagonal form of the quadratic form
$$f(x) = sum_{i = 1}^nx_i^2 + sum_{i < j}x_ix_j.$$
It turned out to be such a problem:
How to change quadratic form $f(x) = sum_{i = 1}^nx_i^2 + sum_{i < j}x_ix_j$ into diagonal form?
We can solve it with congruent transformation, although it is a little complex.
Besides, I tried to transform it into a diagonal form by orthogonal transformation, since its quadratic matrix has the same diagonal elements and the same non-diagonal elements(I thought that I can find a simple way to find its eigenvalues and eigenvectors, but failed. Maybe you can try it).
Is there any other way?
linear-algebra quadratic-forms
linear-algebra quadratic-forms
edited Dec 18 '18 at 16:16
congee
asked Dec 18 '18 at 6:07
congeecongee
133
133
1
$begingroup$
What is "standard form"? Is it $f(x) = (Ax mid x)$ for some simmetric linear transformation $A$?
$endgroup$
– Will M.
Dec 18 '18 at 6:10
$begingroup$
I'm sorry, I don't translate very well. The term may be diagonal form. such as this form $q(x)=a_1 x_1^2 + a_2 x_2^2+ ldots +a_n x_n^2$
$endgroup$
– congee
Dec 18 '18 at 6:39
$begingroup$
I see, you want a change of variables to make $f$ diagonal.
$endgroup$
– Will M.
Dec 18 '18 at 6:40
1
$begingroup$
yes, I try to solve the problem with congruent transformation, but it is a little complex. I wonder if there is any other way, such as orthogonal transformation
$endgroup$
– congee
Dec 18 '18 at 6:46
$begingroup$
By the way, I did not vote down. Some people do but they are not coming to give reasons for their votes. Such a shameful behaviour.
$endgroup$
– Will M.
Dec 18 '18 at 6:52
|
show 7 more comments
1
$begingroup$
What is "standard form"? Is it $f(x) = (Ax mid x)$ for some simmetric linear transformation $A$?
$endgroup$
– Will M.
Dec 18 '18 at 6:10
$begingroup$
I'm sorry, I don't translate very well. The term may be diagonal form. such as this form $q(x)=a_1 x_1^2 + a_2 x_2^2+ ldots +a_n x_n^2$
$endgroup$
– congee
Dec 18 '18 at 6:39
$begingroup$
I see, you want a change of variables to make $f$ diagonal.
$endgroup$
– Will M.
Dec 18 '18 at 6:40
1
$begingroup$
yes, I try to solve the problem with congruent transformation, but it is a little complex. I wonder if there is any other way, such as orthogonal transformation
$endgroup$
– congee
Dec 18 '18 at 6:46
$begingroup$
By the way, I did not vote down. Some people do but they are not coming to give reasons for their votes. Such a shameful behaviour.
$endgroup$
– Will M.
Dec 18 '18 at 6:52
1
1
$begingroup$
What is "standard form"? Is it $f(x) = (Ax mid x)$ for some simmetric linear transformation $A$?
$endgroup$
– Will M.
Dec 18 '18 at 6:10
$begingroup$
What is "standard form"? Is it $f(x) = (Ax mid x)$ for some simmetric linear transformation $A$?
$endgroup$
– Will M.
Dec 18 '18 at 6:10
$begingroup$
I'm sorry, I don't translate very well. The term may be diagonal form. such as this form $q(x)=a_1 x_1^2 + a_2 x_2^2+ ldots +a_n x_n^2$
$endgroup$
– congee
Dec 18 '18 at 6:39
$begingroup$
I'm sorry, I don't translate very well. The term may be diagonal form. such as this form $q(x)=a_1 x_1^2 + a_2 x_2^2+ ldots +a_n x_n^2$
$endgroup$
– congee
Dec 18 '18 at 6:39
$begingroup$
I see, you want a change of variables to make $f$ diagonal.
$endgroup$
– Will M.
Dec 18 '18 at 6:40
$begingroup$
I see, you want a change of variables to make $f$ diagonal.
$endgroup$
– Will M.
Dec 18 '18 at 6:40
1
1
$begingroup$
yes, I try to solve the problem with congruent transformation, but it is a little complex. I wonder if there is any other way, such as orthogonal transformation
$endgroup$
– congee
Dec 18 '18 at 6:46
$begingroup$
yes, I try to solve the problem with congruent transformation, but it is a little complex. I wonder if there is any other way, such as orthogonal transformation
$endgroup$
– congee
Dec 18 '18 at 6:46
$begingroup$
By the way, I did not vote down. Some people do but they are not coming to give reasons for their votes. Such a shameful behaviour.
$endgroup$
– Will M.
Dec 18 '18 at 6:52
$begingroup$
By the way, I did not vote down. Some people do but they are not coming to give reasons for their votes. Such a shameful behaviour.
$endgroup$
– Will M.
Dec 18 '18 at 6:52
|
show 7 more comments
2 Answers
2
active
oldest
votes
$begingroup$
The matrix $M$ of this quadratic form has the form
$$
M=frac12(I_n+J_n),
$$
where $J_n$ is the $ntimes n$ matrix with all ones.
The eigenvalues of $J_n$ are obviously $lambda_1=n$ (multiplicity one) and
$lambda_2=lambda_3=cdots=lambda_n=0$ (multiplicity $n-1$). Therefore the eigenvalues of $M$ are $(n+1)/2$ (multiplicity one) and $1/2$ (multiplicity $n-1$). Implying that there exists an orthogonal coordinate transformation $(x_1,x_2,ldots,x_n)mapsto (x_1',x_2',ldots,x_n')$ such that in the primed coordinates the quadratic form takes the form
$$
Q(x_1',x_2',ldots,x_n')=frac12left((n+1)x_1'^2+x_2'^2+x_3'^2+cdots+x_n'^2right).
$$
An orthonormal basis corresponding to the primed coordinates can be found by orthonormalizing any basis of the $(n-1)$-dimensional subspace
$$V={(x_1,x_2,ldots,x_n)inBbb{R}^nmid x_1+x_2+ldots+x_n=0}$$
which is equal to the eigenspace of $J_n$ belonging to the eigenvalue zero. The missing basis vector is the unit vector $dfrac1{sqrt n}(1,1,ldots,1)$ spanning the orthogonal complement of $V$.
It is (very likely) possible to describe an orthonormal basis of $V$ with a lot of symmetries. I'm not sure I want to go there, though. I would start with a complex basis consisting of roots of unity of order $n$, and go from there.
$endgroup$
$begingroup$
As you said, this quadratic form is positive definite. So, $I_n$ is a diagonal form of it according to Sylvester's law of inertia
$endgroup$
– congee
Dec 20 '18 at 5:07
$begingroup$
Yeah, @congee. $J_n$ is positive semi-definite, and we add a positive multiple of $I_n$ to it.
$endgroup$
– Jyrki Lahtonen
Dec 20 '18 at 6:18
add a comment |
$begingroup$
This is too long for the comment box.
The essence of your exercise is given in the following example.
Suppose we have $f(x, y, z) = x^2 + y^2 + z^2 + xy + xz + yz.$ We complete the square by getting rid of one of the variables, say we pick the first one ($x$ in my notation). So you get,
$$begin{align*}
f(x, y, z) &= (x^2 + xy + xz) + y^2 + z^2 + yz \
&= left( x + dfrac{y}{2} + dfrac{z}{2} right)^2 - frac{y^2}{4} - frac{z^2}{4} -frac{yz}{2} + y^2 + z^2 + yz\
&=u^2 + dfrac{3y^2}{4} + dfrac{yz}{2} + dfrac{3z^2}{4}
end{align*},$$
where $u = x + dfrac{y}{2} + dfrac{z}{2}.$ Now, we can get rid of the second variable, $y$ in my case, by, again, completing the square
$$begin{align*}
f(x, y, z) &= u^2 + left( dfrac{sqrt{3}y}{2} + frac{z}{2sqrt{3}} right)^2-dfrac{z^2}{12}+dfrac{3z^2}{4} = u^2 + v^2+dfrac{5}{6}z^2
end{align*},$$
which is a required representation of $f.$
The case with $x_1, ldots, x_p$ is just guessing a pattern and using induction (I am guessing).
$endgroup$
$begingroup$
I've tried this method to deal directly with situation that has $n$ variables, but it seems hard to handle for me. It seems that it's easy to find the rule from a simple point of view. Thank you.
$endgroup$
– congee
Dec 18 '18 at 7:11
$begingroup$
@WillM. When you complete the square, you should put $-frac{yz}2$ instead of $-frac{yz}4$, because $left(x+frac y2+frac z2right)^2=x^2+frac{y^2}4+frac{z^2}4+xy+xz+frac{yz}2$. But then, I don't see how you group together a common factor $frac34$ in the passage that follows.
$endgroup$
– Saucy O'Path
Dec 18 '18 at 7:11
$begingroup$
(correction: I do see how to deal with it, but the error is still there)
$endgroup$
– Saucy O'Path
Dec 18 '18 at 7:17
$begingroup$
Yes, I didn't notice it just now.
$endgroup$
– congee
Dec 18 '18 at 7:19
1
$begingroup$
@SaucyO'Path this brute force completing the square method always work... provided you can do the arithmetic. I usually cannot.
$endgroup$
– Will M.
Dec 18 '18 at 7:52
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
The matrix $M$ of this quadratic form has the form
$$
M=frac12(I_n+J_n),
$$
where $J_n$ is the $ntimes n$ matrix with all ones.
The eigenvalues of $J_n$ are obviously $lambda_1=n$ (multiplicity one) and
$lambda_2=lambda_3=cdots=lambda_n=0$ (multiplicity $n-1$). Therefore the eigenvalues of $M$ are $(n+1)/2$ (multiplicity one) and $1/2$ (multiplicity $n-1$). Implying that there exists an orthogonal coordinate transformation $(x_1,x_2,ldots,x_n)mapsto (x_1',x_2',ldots,x_n')$ such that in the primed coordinates the quadratic form takes the form
$$
Q(x_1',x_2',ldots,x_n')=frac12left((n+1)x_1'^2+x_2'^2+x_3'^2+cdots+x_n'^2right).
$$
An orthonormal basis corresponding to the primed coordinates can be found by orthonormalizing any basis of the $(n-1)$-dimensional subspace
$$V={(x_1,x_2,ldots,x_n)inBbb{R}^nmid x_1+x_2+ldots+x_n=0}$$
which is equal to the eigenspace of $J_n$ belonging to the eigenvalue zero. The missing basis vector is the unit vector $dfrac1{sqrt n}(1,1,ldots,1)$ spanning the orthogonal complement of $V$.
It is (very likely) possible to describe an orthonormal basis of $V$ with a lot of symmetries. I'm not sure I want to go there, though. I would start with a complex basis consisting of roots of unity of order $n$, and go from there.
$endgroup$
$begingroup$
As you said, this quadratic form is positive definite. So, $I_n$ is a diagonal form of it according to Sylvester's law of inertia
$endgroup$
– congee
Dec 20 '18 at 5:07
$begingroup$
Yeah, @congee. $J_n$ is positive semi-definite, and we add a positive multiple of $I_n$ to it.
$endgroup$
– Jyrki Lahtonen
Dec 20 '18 at 6:18
add a comment |
$begingroup$
The matrix $M$ of this quadratic form has the form
$$
M=frac12(I_n+J_n),
$$
where $J_n$ is the $ntimes n$ matrix with all ones.
The eigenvalues of $J_n$ are obviously $lambda_1=n$ (multiplicity one) and
$lambda_2=lambda_3=cdots=lambda_n=0$ (multiplicity $n-1$). Therefore the eigenvalues of $M$ are $(n+1)/2$ (multiplicity one) and $1/2$ (multiplicity $n-1$). Implying that there exists an orthogonal coordinate transformation $(x_1,x_2,ldots,x_n)mapsto (x_1',x_2',ldots,x_n')$ such that in the primed coordinates the quadratic form takes the form
$$
Q(x_1',x_2',ldots,x_n')=frac12left((n+1)x_1'^2+x_2'^2+x_3'^2+cdots+x_n'^2right).
$$
An orthonormal basis corresponding to the primed coordinates can be found by orthonormalizing any basis of the $(n-1)$-dimensional subspace
$$V={(x_1,x_2,ldots,x_n)inBbb{R}^nmid x_1+x_2+ldots+x_n=0}$$
which is equal to the eigenspace of $J_n$ belonging to the eigenvalue zero. The missing basis vector is the unit vector $dfrac1{sqrt n}(1,1,ldots,1)$ spanning the orthogonal complement of $V$.
It is (very likely) possible to describe an orthonormal basis of $V$ with a lot of symmetries. I'm not sure I want to go there, though. I would start with a complex basis consisting of roots of unity of order $n$, and go from there.
$endgroup$
$begingroup$
As you said, this quadratic form is positive definite. So, $I_n$ is a diagonal form of it according to Sylvester's law of inertia
$endgroup$
– congee
Dec 20 '18 at 5:07
$begingroup$
Yeah, @congee. $J_n$ is positive semi-definite, and we add a positive multiple of $I_n$ to it.
$endgroup$
– Jyrki Lahtonen
Dec 20 '18 at 6:18
add a comment |
$begingroup$
The matrix $M$ of this quadratic form has the form
$$
M=frac12(I_n+J_n),
$$
where $J_n$ is the $ntimes n$ matrix with all ones.
The eigenvalues of $J_n$ are obviously $lambda_1=n$ (multiplicity one) and
$lambda_2=lambda_3=cdots=lambda_n=0$ (multiplicity $n-1$). Therefore the eigenvalues of $M$ are $(n+1)/2$ (multiplicity one) and $1/2$ (multiplicity $n-1$). Implying that there exists an orthogonal coordinate transformation $(x_1,x_2,ldots,x_n)mapsto (x_1',x_2',ldots,x_n')$ such that in the primed coordinates the quadratic form takes the form
$$
Q(x_1',x_2',ldots,x_n')=frac12left((n+1)x_1'^2+x_2'^2+x_3'^2+cdots+x_n'^2right).
$$
An orthonormal basis corresponding to the primed coordinates can be found by orthonormalizing any basis of the $(n-1)$-dimensional subspace
$$V={(x_1,x_2,ldots,x_n)inBbb{R}^nmid x_1+x_2+ldots+x_n=0}$$
which is equal to the eigenspace of $J_n$ belonging to the eigenvalue zero. The missing basis vector is the unit vector $dfrac1{sqrt n}(1,1,ldots,1)$ spanning the orthogonal complement of $V$.
It is (very likely) possible to describe an orthonormal basis of $V$ with a lot of symmetries. I'm not sure I want to go there, though. I would start with a complex basis consisting of roots of unity of order $n$, and go from there.
$endgroup$
The matrix $M$ of this quadratic form has the form
$$
M=frac12(I_n+J_n),
$$
where $J_n$ is the $ntimes n$ matrix with all ones.
The eigenvalues of $J_n$ are obviously $lambda_1=n$ (multiplicity one) and
$lambda_2=lambda_3=cdots=lambda_n=0$ (multiplicity $n-1$). Therefore the eigenvalues of $M$ are $(n+1)/2$ (multiplicity one) and $1/2$ (multiplicity $n-1$). Implying that there exists an orthogonal coordinate transformation $(x_1,x_2,ldots,x_n)mapsto (x_1',x_2',ldots,x_n')$ such that in the primed coordinates the quadratic form takes the form
$$
Q(x_1',x_2',ldots,x_n')=frac12left((n+1)x_1'^2+x_2'^2+x_3'^2+cdots+x_n'^2right).
$$
An orthonormal basis corresponding to the primed coordinates can be found by orthonormalizing any basis of the $(n-1)$-dimensional subspace
$$V={(x_1,x_2,ldots,x_n)inBbb{R}^nmid x_1+x_2+ldots+x_n=0}$$
which is equal to the eigenspace of $J_n$ belonging to the eigenvalue zero. The missing basis vector is the unit vector $dfrac1{sqrt n}(1,1,ldots,1)$ spanning the orthogonal complement of $V$.
It is (very likely) possible to describe an orthonormal basis of $V$ with a lot of symmetries. I'm not sure I want to go there, though. I would start with a complex basis consisting of roots of unity of order $n$, and go from there.
answered Dec 18 '18 at 7:39
Jyrki LahtonenJyrki Lahtonen
109k13170376
109k13170376
$begingroup$
As you said, this quadratic form is positive definite. So, $I_n$ is a diagonal form of it according to Sylvester's law of inertia
$endgroup$
– congee
Dec 20 '18 at 5:07
$begingroup$
Yeah, @congee. $J_n$ is positive semi-definite, and we add a positive multiple of $I_n$ to it.
$endgroup$
– Jyrki Lahtonen
Dec 20 '18 at 6:18
add a comment |
$begingroup$
As you said, this quadratic form is positive definite. So, $I_n$ is a diagonal form of it according to Sylvester's law of inertia
$endgroup$
– congee
Dec 20 '18 at 5:07
$begingroup$
Yeah, @congee. $J_n$ is positive semi-definite, and we add a positive multiple of $I_n$ to it.
$endgroup$
– Jyrki Lahtonen
Dec 20 '18 at 6:18
$begingroup$
As you said, this quadratic form is positive definite. So, $I_n$ is a diagonal form of it according to Sylvester's law of inertia
$endgroup$
– congee
Dec 20 '18 at 5:07
$begingroup$
As you said, this quadratic form is positive definite. So, $I_n$ is a diagonal form of it according to Sylvester's law of inertia
$endgroup$
– congee
Dec 20 '18 at 5:07
$begingroup$
Yeah, @congee. $J_n$ is positive semi-definite, and we add a positive multiple of $I_n$ to it.
$endgroup$
– Jyrki Lahtonen
Dec 20 '18 at 6:18
$begingroup$
Yeah, @congee. $J_n$ is positive semi-definite, and we add a positive multiple of $I_n$ to it.
$endgroup$
– Jyrki Lahtonen
Dec 20 '18 at 6:18
add a comment |
$begingroup$
This is too long for the comment box.
The essence of your exercise is given in the following example.
Suppose we have $f(x, y, z) = x^2 + y^2 + z^2 + xy + xz + yz.$ We complete the square by getting rid of one of the variables, say we pick the first one ($x$ in my notation). So you get,
$$begin{align*}
f(x, y, z) &= (x^2 + xy + xz) + y^2 + z^2 + yz \
&= left( x + dfrac{y}{2} + dfrac{z}{2} right)^2 - frac{y^2}{4} - frac{z^2}{4} -frac{yz}{2} + y^2 + z^2 + yz\
&=u^2 + dfrac{3y^2}{4} + dfrac{yz}{2} + dfrac{3z^2}{4}
end{align*},$$
where $u = x + dfrac{y}{2} + dfrac{z}{2}.$ Now, we can get rid of the second variable, $y$ in my case, by, again, completing the square
$$begin{align*}
f(x, y, z) &= u^2 + left( dfrac{sqrt{3}y}{2} + frac{z}{2sqrt{3}} right)^2-dfrac{z^2}{12}+dfrac{3z^2}{4} = u^2 + v^2+dfrac{5}{6}z^2
end{align*},$$
which is a required representation of $f.$
The case with $x_1, ldots, x_p$ is just guessing a pattern and using induction (I am guessing).
$endgroup$
$begingroup$
I've tried this method to deal directly with situation that has $n$ variables, but it seems hard to handle for me. It seems that it's easy to find the rule from a simple point of view. Thank you.
$endgroup$
– congee
Dec 18 '18 at 7:11
$begingroup$
@WillM. When you complete the square, you should put $-frac{yz}2$ instead of $-frac{yz}4$, because $left(x+frac y2+frac z2right)^2=x^2+frac{y^2}4+frac{z^2}4+xy+xz+frac{yz}2$. But then, I don't see how you group together a common factor $frac34$ in the passage that follows.
$endgroup$
– Saucy O'Path
Dec 18 '18 at 7:11
$begingroup$
(correction: I do see how to deal with it, but the error is still there)
$endgroup$
– Saucy O'Path
Dec 18 '18 at 7:17
$begingroup$
Yes, I didn't notice it just now.
$endgroup$
– congee
Dec 18 '18 at 7:19
1
$begingroup$
@SaucyO'Path this brute force completing the square method always work... provided you can do the arithmetic. I usually cannot.
$endgroup$
– Will M.
Dec 18 '18 at 7:52
add a comment |
$begingroup$
This is too long for the comment box.
The essence of your exercise is given in the following example.
Suppose we have $f(x, y, z) = x^2 + y^2 + z^2 + xy + xz + yz.$ We complete the square by getting rid of one of the variables, say we pick the first one ($x$ in my notation). So you get,
$$begin{align*}
f(x, y, z) &= (x^2 + xy + xz) + y^2 + z^2 + yz \
&= left( x + dfrac{y}{2} + dfrac{z}{2} right)^2 - frac{y^2}{4} - frac{z^2}{4} -frac{yz}{2} + y^2 + z^2 + yz\
&=u^2 + dfrac{3y^2}{4} + dfrac{yz}{2} + dfrac{3z^2}{4}
end{align*},$$
where $u = x + dfrac{y}{2} + dfrac{z}{2}.$ Now, we can get rid of the second variable, $y$ in my case, by, again, completing the square
$$begin{align*}
f(x, y, z) &= u^2 + left( dfrac{sqrt{3}y}{2} + frac{z}{2sqrt{3}} right)^2-dfrac{z^2}{12}+dfrac{3z^2}{4} = u^2 + v^2+dfrac{5}{6}z^2
end{align*},$$
which is a required representation of $f.$
The case with $x_1, ldots, x_p$ is just guessing a pattern and using induction (I am guessing).
$endgroup$
$begingroup$
I've tried this method to deal directly with situation that has $n$ variables, but it seems hard to handle for me. It seems that it's easy to find the rule from a simple point of view. Thank you.
$endgroup$
– congee
Dec 18 '18 at 7:11
$begingroup$
@WillM. When you complete the square, you should put $-frac{yz}2$ instead of $-frac{yz}4$, because $left(x+frac y2+frac z2right)^2=x^2+frac{y^2}4+frac{z^2}4+xy+xz+frac{yz}2$. But then, I don't see how you group together a common factor $frac34$ in the passage that follows.
$endgroup$
– Saucy O'Path
Dec 18 '18 at 7:11
$begingroup$
(correction: I do see how to deal with it, but the error is still there)
$endgroup$
– Saucy O'Path
Dec 18 '18 at 7:17
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Yes, I didn't notice it just now.
$endgroup$
– congee
Dec 18 '18 at 7:19
1
$begingroup$
@SaucyO'Path this brute force completing the square method always work... provided you can do the arithmetic. I usually cannot.
$endgroup$
– Will M.
Dec 18 '18 at 7:52
add a comment |
$begingroup$
This is too long for the comment box.
The essence of your exercise is given in the following example.
Suppose we have $f(x, y, z) = x^2 + y^2 + z^2 + xy + xz + yz.$ We complete the square by getting rid of one of the variables, say we pick the first one ($x$ in my notation). So you get,
$$begin{align*}
f(x, y, z) &= (x^2 + xy + xz) + y^2 + z^2 + yz \
&= left( x + dfrac{y}{2} + dfrac{z}{2} right)^2 - frac{y^2}{4} - frac{z^2}{4} -frac{yz}{2} + y^2 + z^2 + yz\
&=u^2 + dfrac{3y^2}{4} + dfrac{yz}{2} + dfrac{3z^2}{4}
end{align*},$$
where $u = x + dfrac{y}{2} + dfrac{z}{2}.$ Now, we can get rid of the second variable, $y$ in my case, by, again, completing the square
$$begin{align*}
f(x, y, z) &= u^2 + left( dfrac{sqrt{3}y}{2} + frac{z}{2sqrt{3}} right)^2-dfrac{z^2}{12}+dfrac{3z^2}{4} = u^2 + v^2+dfrac{5}{6}z^2
end{align*},$$
which is a required representation of $f.$
The case with $x_1, ldots, x_p$ is just guessing a pattern and using induction (I am guessing).
$endgroup$
This is too long for the comment box.
The essence of your exercise is given in the following example.
Suppose we have $f(x, y, z) = x^2 + y^2 + z^2 + xy + xz + yz.$ We complete the square by getting rid of one of the variables, say we pick the first one ($x$ in my notation). So you get,
$$begin{align*}
f(x, y, z) &= (x^2 + xy + xz) + y^2 + z^2 + yz \
&= left( x + dfrac{y}{2} + dfrac{z}{2} right)^2 - frac{y^2}{4} - frac{z^2}{4} -frac{yz}{2} + y^2 + z^2 + yz\
&=u^2 + dfrac{3y^2}{4} + dfrac{yz}{2} + dfrac{3z^2}{4}
end{align*},$$
where $u = x + dfrac{y}{2} + dfrac{z}{2}.$ Now, we can get rid of the second variable, $y$ in my case, by, again, completing the square
$$begin{align*}
f(x, y, z) &= u^2 + left( dfrac{sqrt{3}y}{2} + frac{z}{2sqrt{3}} right)^2-dfrac{z^2}{12}+dfrac{3z^2}{4} = u^2 + v^2+dfrac{5}{6}z^2
end{align*},$$
which is a required representation of $f.$
The case with $x_1, ldots, x_p$ is just guessing a pattern and using induction (I am guessing).
edited Dec 18 '18 at 7:51
answered Dec 18 '18 at 6:51
Will M.Will M.
2,725315
2,725315
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I've tried this method to deal directly with situation that has $n$ variables, but it seems hard to handle for me. It seems that it's easy to find the rule from a simple point of view. Thank you.
$endgroup$
– congee
Dec 18 '18 at 7:11
$begingroup$
@WillM. When you complete the square, you should put $-frac{yz}2$ instead of $-frac{yz}4$, because $left(x+frac y2+frac z2right)^2=x^2+frac{y^2}4+frac{z^2}4+xy+xz+frac{yz}2$. But then, I don't see how you group together a common factor $frac34$ in the passage that follows.
$endgroup$
– Saucy O'Path
Dec 18 '18 at 7:11
$begingroup$
(correction: I do see how to deal with it, but the error is still there)
$endgroup$
– Saucy O'Path
Dec 18 '18 at 7:17
$begingroup$
Yes, I didn't notice it just now.
$endgroup$
– congee
Dec 18 '18 at 7:19
1
$begingroup$
@SaucyO'Path this brute force completing the square method always work... provided you can do the arithmetic. I usually cannot.
$endgroup$
– Will M.
Dec 18 '18 at 7:52
add a comment |
$begingroup$
I've tried this method to deal directly with situation that has $n$ variables, but it seems hard to handle for me. It seems that it's easy to find the rule from a simple point of view. Thank you.
$endgroup$
– congee
Dec 18 '18 at 7:11
$begingroup$
@WillM. When you complete the square, you should put $-frac{yz}2$ instead of $-frac{yz}4$, because $left(x+frac y2+frac z2right)^2=x^2+frac{y^2}4+frac{z^2}4+xy+xz+frac{yz}2$. But then, I don't see how you group together a common factor $frac34$ in the passage that follows.
$endgroup$
– Saucy O'Path
Dec 18 '18 at 7:11
$begingroup$
(correction: I do see how to deal with it, but the error is still there)
$endgroup$
– Saucy O'Path
Dec 18 '18 at 7:17
$begingroup$
Yes, I didn't notice it just now.
$endgroup$
– congee
Dec 18 '18 at 7:19
1
$begingroup$
@SaucyO'Path this brute force completing the square method always work... provided you can do the arithmetic. I usually cannot.
$endgroup$
– Will M.
Dec 18 '18 at 7:52
$begingroup$
I've tried this method to deal directly with situation that has $n$ variables, but it seems hard to handle for me. It seems that it's easy to find the rule from a simple point of view. Thank you.
$endgroup$
– congee
Dec 18 '18 at 7:11
$begingroup$
I've tried this method to deal directly with situation that has $n$ variables, but it seems hard to handle for me. It seems that it's easy to find the rule from a simple point of view. Thank you.
$endgroup$
– congee
Dec 18 '18 at 7:11
$begingroup$
@WillM. When you complete the square, you should put $-frac{yz}2$ instead of $-frac{yz}4$, because $left(x+frac y2+frac z2right)^2=x^2+frac{y^2}4+frac{z^2}4+xy+xz+frac{yz}2$. But then, I don't see how you group together a common factor $frac34$ in the passage that follows.
$endgroup$
– Saucy O'Path
Dec 18 '18 at 7:11
$begingroup$
@WillM. When you complete the square, you should put $-frac{yz}2$ instead of $-frac{yz}4$, because $left(x+frac y2+frac z2right)^2=x^2+frac{y^2}4+frac{z^2}4+xy+xz+frac{yz}2$. But then, I don't see how you group together a common factor $frac34$ in the passage that follows.
$endgroup$
– Saucy O'Path
Dec 18 '18 at 7:11
$begingroup$
(correction: I do see how to deal with it, but the error is still there)
$endgroup$
– Saucy O'Path
Dec 18 '18 at 7:17
$begingroup$
(correction: I do see how to deal with it, but the error is still there)
$endgroup$
– Saucy O'Path
Dec 18 '18 at 7:17
$begingroup$
Yes, I didn't notice it just now.
$endgroup$
– congee
Dec 18 '18 at 7:19
$begingroup$
Yes, I didn't notice it just now.
$endgroup$
– congee
Dec 18 '18 at 7:19
1
1
$begingroup$
@SaucyO'Path this brute force completing the square method always work... provided you can do the arithmetic. I usually cannot.
$endgroup$
– Will M.
Dec 18 '18 at 7:52
$begingroup$
@SaucyO'Path this brute force completing the square method always work... provided you can do the arithmetic. I usually cannot.
$endgroup$
– Will M.
Dec 18 '18 at 7:52
add a comment |
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What is "standard form"? Is it $f(x) = (Ax mid x)$ for some simmetric linear transformation $A$?
$endgroup$
– Will M.
Dec 18 '18 at 6:10
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I'm sorry, I don't translate very well. The term may be diagonal form. such as this form $q(x)=a_1 x_1^2 + a_2 x_2^2+ ldots +a_n x_n^2$
$endgroup$
– congee
Dec 18 '18 at 6:39
$begingroup$
I see, you want a change of variables to make $f$ diagonal.
$endgroup$
– Will M.
Dec 18 '18 at 6:40
1
$begingroup$
yes, I try to solve the problem with congruent transformation, but it is a little complex. I wonder if there is any other way, such as orthogonal transformation
$endgroup$
– congee
Dec 18 '18 at 6:46
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By the way, I did not vote down. Some people do but they are not coming to give reasons for their votes. Such a shameful behaviour.
$endgroup$
– Will M.
Dec 18 '18 at 6:52