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Put quadratic form into sum of squares

0

$begingroup$

Is there a method or process that doesn't require a matrix to put quadratic forms into a sum of squares ?

Two examples that I find extremely challenging.

i)

$q(x, y, z) = (x − y)
^2 + (y − z)
^2 − (z − x)
^2$

ii)
Finding the linear forms that constitute the orthogonal matrix that diagonalizes the symmetric matrix:

$ M= beginbmatrix3&1&-3/2\1&1&0\-3/2&0&0 endbmatrix$

the quadratic form should look something like this:

$q(x,y,z) =3x^2 + 2xy -3xz+y^2$

Not looking for the solution, but for a method I can apply in most cases.

linear-algebra matrices quadratic-forms sums-of-squares

share|cite|improve this question

asked Sep 4 '17 at 2:07

LeroyLeroy

1267

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Isn’t (i) already a sum of squares?
  $endgroup$
  – amd
  Sep 4 '17 at 3:20
  

  
  
  
  

add a comment | 

0

$begingroup$

Is there a method or process that doesn't require a matrix to put quadratic forms into a sum of squares ?

Two examples that I find extremely challenging.

i)

$q(x, y, z) = (x − y)
^2 + (y − z)
^2 − (z − x)
^2$

ii)
Finding the linear forms that constitute the orthogonal matrix that diagonalizes the symmetric matrix:

$ M= beginbmatrix3&1&-3/2\1&1&0\-3/2&0&0 endbmatrix$

the quadratic form should look something like this:

$q(x,y,z) =3x^2 + 2xy -3xz+y^2$

Not looking for the solution, but for a method I can apply in most cases.

linear-algebra matrices quadratic-forms sums-of-squares

share|cite|improve this question

asked Sep 4 '17 at 2:07

LeroyLeroy

1267

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Isn’t (i) already a sum of squares?
  $endgroup$
  – amd
  Sep 4 '17 at 3:20
  

  
  
  
  

add a comment | 

$begingroup$

Is there a method or process that doesn't require a matrix to put quadratic forms into a sum of squares ?

Two examples that I find extremely challenging.

i)

$q(x, y, z) = (x − y)
^2 + (y − z)
^2 − (z − x)
^2$

ii)
Finding the linear forms that constitute the orthogonal matrix that diagonalizes the symmetric matrix:

$ M= beginbmatrix3&1&-3/2\1&1&0\-3/2&0&0 endbmatrix$

the quadratic form should look something like this:

$q(x,y,z) =3x^2 + 2xy -3xz+y^2$

Not looking for the solution, but for a method I can apply in most cases.

linear-algebra matrices quadratic-forms sums-of-squares

share|cite|improve this question

asked Sep 4 '17 at 2:07

LeroyLeroy

1267

$endgroup$

share|cite|improve this question

asked Sep 4 '17 at 2:07

LeroyLeroy

1267

share|cite|improve this question

asked Sep 4 '17 at 2:07

LeroyLeroy

1267

asked Sep 4 '17 at 2:07

LeroyLeroy

1267

asked Sep 4 '17 at 2:07

LeroyLeroy

1267

1 Answer
1

active

oldest

votes

0

$begingroup$

Let's tackle your second quadratic form. Start by choosing one of the variables which appears as a square (if there isn't such a variable, you can always make a change of variables to get one). Let's start with $y$. This variable appears only with $x,y$ so we should try and complete the square to get all the terms that occur. In our case,

$$ 3x^2 + 2xy - 3xz + y^2 = (y + x)^2 + 2x^2 - 3xz. $$

This way, we have eliminated $y$ from the problem and now we handle only $2x^2 - 3xz$ in which $y$ doesn't appear and repeat the process. Then

$$ 2x^2 - 3xz = left( sqrt2x - frac32sqrt2z right)^2 - frac98z^2$$

so the final answer can be

$$ q(x,y,z) = (y + x)^2 + left( sqrt2x - frac32sqrt2z right)^2 - frac98z^2.$$

If we would have started with $x$, the process would look like this:

$$ q(x,y,z) = 3x^2 + 2xy - 3xz + y^2 = left( sqrt3x + frac1sqrt3y - fracsqrt32z right)^2 + frac23y^2 - frac34z^2.$$

share|cite|improve this answer

answered Sep 4 '17 at 8:13

levaplevap

48k33274

$endgroup$

add a comment | 

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0

$begingroup$

Let's tackle your second quadratic form. Start by choosing one of the variables which appears as a square (if there isn't such a variable, you can always make a change of variables to get one). Let's start with $y$. This variable appears only with $x,y$ so we should try and complete the square to get all the terms that occur. In our case,

$$ 3x^2 + 2xy - 3xz + y^2 = (y + x)^2 + 2x^2 - 3xz. $$

This way, we have eliminated $y$ from the problem and now we handle only $2x^2 - 3xz$ in which $y$ doesn't appear and repeat the process. Then

$$ 2x^2 - 3xz = left( sqrt2x - frac32sqrt2z right)^2 - frac98z^2$$

so the final answer can be

$$ q(x,y,z) = (y + x)^2 + left( sqrt2x - frac32sqrt2z right)^2 - frac98z^2.$$

If we would have started with $x$, the process would look like this:

$$ q(x,y,z) = 3x^2 + 2xy - 3xz + y^2 = left( sqrt3x + frac1sqrt3y - fracsqrt32z right)^2 + frac23y^2 - frac34z^2.$$

share|cite|improve this answer

answered Sep 4 '17 at 8:13

levaplevap

48k33274

$endgroup$

add a comment | 

0

$begingroup$

Let's tackle your second quadratic form. Start by choosing one of the variables which appears as a square (if there isn't such a variable, you can always make a change of variables to get one). Let's start with $y$. This variable appears only with $x,y$ so we should try and complete the square to get all the terms that occur. In our case,

$$ 3x^2 + 2xy - 3xz + y^2 = (y + x)^2 + 2x^2 - 3xz. $$

This way, we have eliminated $y$ from the problem and now we handle only $2x^2 - 3xz$ in which $y$ doesn't appear and repeat the process. Then

$$ 2x^2 - 3xz = left( sqrt2x - frac32sqrt2z right)^2 - frac98z^2$$

so the final answer can be

$$ q(x,y,z) = (y + x)^2 + left( sqrt2x - frac32sqrt2z right)^2 - frac98z^2.$$

If we would have started with $x$, the process would look like this:

$$ q(x,y,z) = 3x^2 + 2xy - 3xz + y^2 = left( sqrt3x + frac1sqrt3y - fracsqrt32z right)^2 + frac23y^2 - frac34z^2.$$

share|cite|improve this answer

answered Sep 4 '17 at 8:13

levaplevap

48k33274

$endgroup$

add a comment | 

$begingroup$

Let's tackle your second quadratic form. Start by choosing one of the variables which appears as a square (if there isn't such a variable, you can always make a change of variables to get one). Let's start with $y$. This variable appears only with $x,y$ so we should try and complete the square to get all the terms that occur. In our case,

$$ 3x^2 + 2xy - 3xz + y^2 = (y + x)^2 + 2x^2 - 3xz. $$

This way, we have eliminated $y$ from the problem and now we handle only $2x^2 - 3xz$ in which $y$ doesn't appear and repeat the process. Then

$$ 2x^2 - 3xz = left( sqrt2x - frac32sqrt2z right)^2 - frac98z^2$$

so the final answer can be

$$ q(x,y,z) = (y + x)^2 + left( sqrt2x - frac32sqrt2z right)^2 - frac98z^2.$$

If we would have started with $x$, the process would look like this:

$$ q(x,y,z) = 3x^2 + 2xy - 3xz + y^2 = left( sqrt3x + frac1sqrt3y - fracsqrt32z right)^2 + frac23y^2 - frac34z^2.$$

share|cite|improve this answer

answered Sep 4 '17 at 8:13

levaplevap

48k33274

$endgroup$

share|cite|improve this answer

answered Sep 4 '17 at 8:13

levaplevap

48k33274

answered Sep 4 '17 at 8:13

levaplevap

48k33274

answered Sep 4 '17 at 8:13

levaplevap

48k33274