Mixing
bilinear form only positive or negative
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Let $f$ be a definite, symmetric bilinear form.
Show that $f$ is positive or negative.
Consider the quadratic form $q(x,y,z)=x^2+y^2-z^2$, then the polar form assocaited to $q$ is symmetric, definite but negative and positive.
Is there a problem with the statement of the problem ?
EDIT:
I attempt to prove this result:
Let $x,tilde y$ such that $q(x)>0$ and $q(tilde y)<0$ where $q$ is the quadratic form associated to $f$.
Let $lambda$ such that $y=lambda tilde y$ is such that $f(x,y-x)=0$.
Note that $q(y)<0$.
We have for $tin mathbb{R}$,
$$
q(x+t(y-x))=q(x)+t^2q(y-x)
$$
and we obtain:
$q(y-x)<-q(x)<0$
By continuity of $f$, there exists $tin (0,1)$ such that $q(x+t(y-x))=0$.
(This $t$ could be such that $x+t(y-x)=0$).
But, when $tto infty$, $q(x+t(y-x))to -infty$, so it means that $q$ has to for two different value of $t$ which is absurd.
Finally $q$ is of constant sign.
Any toughts about this proof?
proof-verification quadratic-forms bilinear-form positive-definite
asked Nov 12 at 22:54
Smilia
576516
add a comment |
up vote
0
down vote
favorite
Let $f$ be a definite, symmetric bilinear form.
Show that $f$ is positive or negative.
Consider the quadratic form $q(x,y,z)=x^2+y^2-z^2$, then the polar form assocaited to $q$ is symmetric, definite but negative and positive.
Is there a problem with the statement of the problem ?
EDIT:
I attempt to prove this result:
Let $x,tilde y$ such that $q(x)>0$ and $q(tilde y)<0$ where $q$ is the quadratic form associated to $f$.
Let $lambda$ such that $y=lambda tilde y$ is such that $f(x,y-x)=0$.
Note that $q(y)<0$.
We have for $tin mathbb{R}$,
$$
q(x+t(y-x))=q(x)+t^2q(y-x)
$$
and we obtain:
$q(y-x)<-q(x)<0$
By continuity of $f$, there exists $tin (0,1)$ such that $q(x+t(y-x))=0$.
(This $t$ could be such that $x+t(y-x)=0$).
But, when $tto infty$, $q(x+t(y-x))to -infty$, so it means that $q$ has to for two different value of $t$ which is absurd.
Finally $q$ is of constant sign.
Any toughts about this proof?
proof-verification quadratic-forms bilinear-form positive-definite
asked Nov 12 at 22:54
Smilia
576516
Why would your $f$ be definite?
– user10354138
Nov 12 at 22:58
I mean your assertion "the polar form associated to $q$ is [...] definite"
– user10354138
Nov 12 at 23:00
1
I don't understand. The form $x^2 + y^2 - z^2$ is called "indefinite"
– Will Jagy
Nov 12 at 23:00
ok yeah the quadratic form is not definite
– Smilia
Nov 12 at 23:12
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f$ be a definite, symmetric bilinear form.
Show that $f$ is positive or negative.
Consider the quadratic form $q(x,y,z)=x^2+y^2-z^2$, then the polar form assocaited to $q$ is symmetric, definite but negative and positive.
Is there a problem with the statement of the problem ?
EDIT:
I attempt to prove this result:
Let $x,tilde y$ such that $q(x)>0$ and $q(tilde y)<0$ where $q$ is the quadratic form associated to $f$.
Let $lambda$ such that $y=lambda tilde y$ is such that $f(x,y-x)=0$.
Note that $q(y)<0$.
We have for $tin mathbb{R}$,
$$
q(x+t(y-x))=q(x)+t^2q(y-x)
$$
and we obtain:
$q(y-x)<-q(x)<0$
By continuity of $f$, there exists $tin (0,1)$ such that $q(x+t(y-x))=0$.
(This $t$ could be such that $x+t(y-x)=0$).
But, when $tto infty$, $q(x+t(y-x))to -infty$, so it means that $q$ has to for two different value of $t$ which is absurd.
Finally $q$ is of constant sign.
Any toughts about this proof?
proof-verification quadratic-forms bilinear-form positive-definite
asked Nov 12 at 22:54
Smilia
576516
Let $f$ be a definite, symmetric bilinear form.
Show that $f$ is positive or negative.
Consider the quadratic form $q(x,y,z)=x^2+y^2-z^2$, then the polar form assocaited to $q$ is symmetric, definite but negative and positive.
Is there a problem with the statement of the problem ?
EDIT:
I attempt to prove this result:
Let $x,tilde y$ such that $q(x)>0$ and $q(tilde y)<0$ where $q$ is the quadratic form associated to $f$.
Let $lambda$ such that $y=lambda tilde y$ is such that $f(x,y-x)=0$.
Note that $q(y)<0$.
We have for $tin mathbb{R}$,
$$
q(x+t(y-x))=q(x)+t^2q(y-x)
$$
and we obtain:
$q(y-x)<-q(x)<0$
By continuity of $f$, there exists $tin (0,1)$ such that $q(x+t(y-x))=0$.
(This $t$ could be such that $x+t(y-x)=0$).
But, when $tto infty$, $q(x+t(y-x))to -infty$, so it means that $q$ has to for two different value of $t$ which is absurd.
Finally $q$ is of constant sign.
Any toughts about this proof?
proof-verification quadratic-forms bilinear-form positive-definite
proof-verification quadratic-forms bilinear-form positive-definite
asked Nov 12 at 22:54
Smilia
576516
asked Nov 12 at 22:54
Smilia
576516
edited 2 days ago
asked Nov 12 at 22:54
Smilia
576516
asked Nov 12 at 22:54
Smilia
576516
asked Nov 12 at 22:54
Smilia
576516
576516
Why would your $f$ be definite?
– user10354138
Nov 12 at 22:58
I mean your assertion "the polar form associated to $q$ is [...] definite"
– user10354138
Nov 12 at 23:00
1
I don't understand. The form $x^2 + y^2 - z^2$ is called "indefinite"
– Will Jagy
Nov 12 at 23:00
ok yeah the quadratic form is not definite
– Smilia
Nov 12 at 23:12
add a comment |
Why would your $f$ be definite?
– user10354138
Nov 12 at 22:58
I mean your assertion "the polar form associated to $q$ is [...] definite"
– user10354138
Nov 12 at 23:00
1
I don't understand. The form $x^2 + y^2 - z^2$ is called "indefinite"
– Will Jagy
Nov 12 at 23:00
ok yeah the quadratic form is not definite
– Smilia
Nov 12 at 23:12
Why would your $f$ be definite?
– user10354138
Nov 12 at 22:58
Why would your $f$ be definite?
– user10354138
Nov 12 at 22:58
I mean your assertion "the polar form associated to $q$ is [...] definite"
– user10354138
Nov 12 at 23:00
I mean your assertion "the polar form associated to $q$ is [...] definite"
– user10354138
Nov 12 at 23:00
1
1
I don't understand. The form $x^2 + y^2 - z^2$ is called "indefinite"
– Will Jagy
Nov 12 at 23:00
I don't understand. The form $x^2 + y^2 - z^2$ is called "indefinite"
– Will Jagy
Nov 12 at 23:00
ok yeah the quadratic form is not definite
– Smilia
Nov 12 at 23:12
ok yeah the quadratic form is not definite
– Smilia
Nov 12 at 23:12
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
I found what seems to be the definition at Associated Bilinear Form to Q (Quadratic Form)
The short version: given a quadratic form, create the matrix $H$ which is the Hessian matrix of second partial derivatives.
Given column vectors $x,y:$ the original quadratic form applied to $x$ is given by
$$ (1/2) x^T H x $$
The bilinear form applied to $x,y$ is
$$ (1/2) x^T H y = (1/2) y^T H x $$
answered Nov 12 at 23:11
Will Jagy
100k597198
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I found what seems to be the definition at Associated Bilinear Form to Q (Quadratic Form)
The short version: given a quadratic form, create the matrix $H$ which is the Hessian matrix of second partial derivatives.
Given column vectors $x,y:$ the original quadratic form applied to $x$ is given by
$$ (1/2) x^T H x $$
The bilinear form applied to $x,y$ is
$$ (1/2) x^T H y = (1/2) y^T H x $$
answered Nov 12 at 23:11
Will Jagy
100k597198
add a comment |
up vote
0
down vote
I found what seems to be the definition at Associated Bilinear Form to Q (Quadratic Form)
The short version: given a quadratic form, create the matrix $H$ which is the Hessian matrix of second partial derivatives.
Given column vectors $x,y:$ the original quadratic form applied to $x$ is given by
$$ (1/2) x^T H x $$
The bilinear form applied to $x,y$ is
$$ (1/2) x^T H y = (1/2) y^T H x $$
answered Nov 12 at 23:11
Will Jagy
100k597198
add a comment |
up vote
0
down vote
up vote
0
down vote
I found what seems to be the definition at Associated Bilinear Form to Q (Quadratic Form)
The short version: given a quadratic form, create the matrix $H$ which is the Hessian matrix of second partial derivatives.
Given column vectors $x,y:$ the original quadratic form applied to $x$ is given by
$$ (1/2) x^T H x $$
The bilinear form applied to $x,y$ is
$$ (1/2) x^T H y = (1/2) y^T H x $$
answered Nov 12 at 23:11
Will Jagy
100k597198
I found what seems to be the definition at Associated Bilinear Form to Q (Quadratic Form)
The short version: given a quadratic form, create the matrix $H$ which is the Hessian matrix of second partial derivatives.
Given column vectors $x,y:$ the original quadratic form applied to $x$ is given by
$$ (1/2) x^T H x $$
The bilinear form applied to $x,y$ is
$$ (1/2) x^T H y = (1/2) y^T H x $$
answered Nov 12 at 23:11
Will Jagy
100k597198
answered Nov 12 at 23:11
Will Jagy
100k597198
answered Nov 12 at 23:11
Will Jagy
100k597198
answered Nov 12 at 23:11
Will Jagy
100k597198
100k597198
add a comment |
add a comment |
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Why would your $f$ be definite?
– user10354138
Nov 12 at 22:58
I mean your assertion "the polar form associated to $q$ is [...] definite"
– user10354138
Nov 12 at 23:00
1
I don't understand. The form $x^2 + y^2 - z^2$ is called "indefinite"
– Will Jagy
Nov 12 at 23:00
ok yeah the quadratic form is not definite
– Smilia
Nov 12 at 23:12