How to find image of a quadratic form?
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Suppose the quadratic form $f: mathbb R^3to mathbb R$ with $$f(x_1,x_2,x_3) = x_1^2 - x_2^2 - 11x_3^2 - 2x_1x_2 + 4x_1x_3 + 8x_2x_3.$$
By using Lagrange's Reduction, we have the canonical expression of $f,$
$$g(y_1,y_2,y_3) = y_1^2 - 2y_2^2 + 3y_3^2,$$ where
$$y_1 = x_1 - x_2 + 2x_3,\ y_2 = x_2 - 3x_3,\ y_3 = x_3.$$
My question is:
How to find the sets $f(mathbb R^3)$ and $f(mathbb R^3setminus {(0,0,0)})$?
Thank you for your help!
linear-algebra quadratic-forms
asked Nov 22 at 23:16
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Suppose the quadratic form $f: mathbb R^3to mathbb R$ with $$f(x_1,x_2,x_3) = x_1^2 - x_2^2 - 11x_3^2 - 2x_1x_2 + 4x_1x_3 + 8x_2x_3.$$
By using Lagrange's Reduction, we have the canonical expression of $f,$
$$g(y_1,y_2,y_3) = y_1^2 - 2y_2^2 + 3y_3^2,$$ where
$$y_1 = x_1 - x_2 + 2x_3,\ y_2 = x_2 - 3x_3,\ y_3 = x_3.$$
My question is:
How to find the sets $f(mathbb R^3)$ and $f(mathbb R^3setminus {(0,0,0)})$?
Thank you for your help!
linear-algebra quadratic-forms
asked Nov 22 at 23:16
Success
112
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose the quadratic form $f: mathbb R^3to mathbb R$ with $$f(x_1,x_2,x_3) = x_1^2 - x_2^2 - 11x_3^2 - 2x_1x_2 + 4x_1x_3 + 8x_2x_3.$$
By using Lagrange's Reduction, we have the canonical expression of $f,$
$$g(y_1,y_2,y_3) = y_1^2 - 2y_2^2 + 3y_3^2,$$ where
$$y_1 = x_1 - x_2 + 2x_3,\ y_2 = x_2 - 3x_3,\ y_3 = x_3.$$
My question is:
How to find the sets $f(mathbb R^3)$ and $f(mathbb R^3setminus {(0,0,0)})$?
Thank you for your help!
linear-algebra quadratic-forms
asked Nov 22 at 23:16
Success
112
Suppose the quadratic form $f: mathbb R^3to mathbb R$ with $$f(x_1,x_2,x_3) = x_1^2 - x_2^2 - 11x_3^2 - 2x_1x_2 + 4x_1x_3 + 8x_2x_3.$$
By using Lagrange's Reduction, we have the canonical expression of $f,$
$$g(y_1,y_2,y_3) = y_1^2 - 2y_2^2 + 3y_3^2,$$ where
$$y_1 = x_1 - x_2 + 2x_3,\ y_2 = x_2 - 3x_3,\ y_3 = x_3.$$
My question is:
How to find the sets $f(mathbb R^3)$ and $f(mathbb R^3setminus {(0,0,0)})$?
Thank you for your help!
linear-algebra quadratic-forms
linear-algebra quadratic-forms
asked Nov 22 at 23:16
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112
asked Nov 22 at 23:16
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112
edited Nov 22 at 23:31
asked Nov 22 at 23:16
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asked Nov 22 at 23:16
Success
112
asked Nov 22 at 23:16
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112
112
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Since $f(x_1,0,0)=x_1^2$ and $f(0,x_2,0)=-x_2^2,$ it's clear that $f(mathbb{R}^3)=mathbb{R},$ and since $f(0,0,0)=0,$ the only question is whether $f(mathbb{R}^3 setminus {(0,0,0)})=mathbb{R}$ or whether $f(mathbb{R}^3 setminus {(0,0,0)})=mathbb{R} setminus {0}.$ That is, does $f$ have any zeros other than $(0,0,0)?$
It should be easy to find another zero using your expression for $g$.
answered Nov 22 at 23:35
saulspatz
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Since $f(x_1,0,0)=x_1^2$ and $f(0,x_2,0)=-x_2^2,$ it's clear that $f(mathbb{R}^3)=mathbb{R},$ and since $f(0,0,0)=0,$ the only question is whether $f(mathbb{R}^3 setminus {(0,0,0)})=mathbb{R}$ or whether $f(mathbb{R}^3 setminus {(0,0,0)})=mathbb{R} setminus {0}.$ That is, does $f$ have any zeros other than $(0,0,0)?$
It should be easy to find another zero using your expression for $g$.
answered Nov 22 at 23:35
saulspatz
13.7k21328
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up vote
1
down vote
Since $f(x_1,0,0)=x_1^2$ and $f(0,x_2,0)=-x_2^2,$ it's clear that $f(mathbb{R}^3)=mathbb{R},$ and since $f(0,0,0)=0,$ the only question is whether $f(mathbb{R}^3 setminus {(0,0,0)})=mathbb{R}$ or whether $f(mathbb{R}^3 setminus {(0,0,0)})=mathbb{R} setminus {0}.$ That is, does $f$ have any zeros other than $(0,0,0)?$
It should be easy to find another zero using your expression for $g$.
answered Nov 22 at 23:35
saulspatz
13.7k21328
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up vote
1
down vote
up vote
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down vote
Since $f(x_1,0,0)=x_1^2$ and $f(0,x_2,0)=-x_2^2,$ it's clear that $f(mathbb{R}^3)=mathbb{R},$ and since $f(0,0,0)=0,$ the only question is whether $f(mathbb{R}^3 setminus {(0,0,0)})=mathbb{R}$ or whether $f(mathbb{R}^3 setminus {(0,0,0)})=mathbb{R} setminus {0}.$ That is, does $f$ have any zeros other than $(0,0,0)?$
It should be easy to find another zero using your expression for $g$.
answered Nov 22 at 23:35
saulspatz
13.7k21328
Since $f(x_1,0,0)=x_1^2$ and $f(0,x_2,0)=-x_2^2,$ it's clear that $f(mathbb{R}^3)=mathbb{R},$ and since $f(0,0,0)=0,$ the only question is whether $f(mathbb{R}^3 setminus {(0,0,0)})=mathbb{R}$ or whether $f(mathbb{R}^3 setminus {(0,0,0)})=mathbb{R} setminus {0}.$ That is, does $f$ have any zeros other than $(0,0,0)?$
It should be easy to find another zero using your expression for $g$.
answered Nov 22 at 23:35
saulspatz
13.7k21328
answered Nov 22 at 23:35
saulspatz
13.7k21328
answered Nov 22 at 23:35
saulspatz
13.7k21328
answered Nov 22 at 23:35
saulspatz
13.7k21328
13.7k21328
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