Equivalence of quadratic forms over arbitrary field (char(K)≠2)
$begingroup$
Let $mathbb{K}$ be an arbitrary field with $char(mathbb{K})neq 2$ and let $a_1, ldots ,a_n$ be Elements of $mathbb{K}$.
I want to proof the following equivalence of $n-$quadratic forms in diagonal form over $mathbb{K}$:
$ begin{align}
[a_1, ldots ,a_n] simeq [c_1^2a_1, ldots , c_n^2a_n],
end{align}
$
where $c_1, ldots , c_n in mathbb{K}^* forall n in mathbb{N}$.
According to Lam's Book $textit{Introduction to quadratic forms}$ (p. 1-2) it suffices to find an
"$textit{invertible matrix}$ $C in GL_n(mathbb{K})$ $textit{such that}$ $ g(Ccdot X)=f(X)$. (1) $textit{This means there exists a nonsingular, homogenuous linear substitution of the variables}$ $X_1, ldots ,X_n $ $ textit{that takes the form}$ $g$ $textit{to the form}$ $f$."
Now I was going to define the isometry
$C:=diag(sqrt{c_1},ldots ,sqrt{c_n})$.
Clearly $C in GL_n(mathbb{K})$ and (1) would be established. But I cannot really say if this substitution is "homogenuous" as understood in Lem's sense.
Can someone give me insight into this?
linear-algebra quadratic-forms
asked Jan 5 at 18:25
Jean PaulJean Paul
11
$endgroup$
add a comment |
$begingroup$
Let $mathbb{K}$ be an arbitrary field with $char(mathbb{K})neq 2$ and let $a_1, ldots ,a_n$ be Elements of $mathbb{K}$.
I want to proof the following equivalence of $n-$quadratic forms in diagonal form over $mathbb{K}$:
$ begin{align}
[a_1, ldots ,a_n] simeq [c_1^2a_1, ldots , c_n^2a_n],
end{align}
$
where $c_1, ldots , c_n in mathbb{K}^* forall n in mathbb{N}$.
According to Lam's Book $textit{Introduction to quadratic forms}$ (p. 1-2) it suffices to find an
"$textit{invertible matrix}$ $C in GL_n(mathbb{K})$ $textit{such that}$ $ g(Ccdot X)=f(X)$. (1) $textit{This means there exists a nonsingular, homogenuous linear substitution of the variables}$ $X_1, ldots ,X_n $ $ textit{that takes the form}$ $g$ $textit{to the form}$ $f$."
Now I was going to define the isometry
$C:=diag(sqrt{c_1},ldots ,sqrt{c_n})$.
Clearly $C in GL_n(mathbb{K})$ and (1) would be established. But I cannot really say if this substitution is "homogenuous" as understood in Lem's sense.
Can someone give me insight into this?
linear-algebra quadratic-forms
asked Jan 5 at 18:25
Jean PaulJean Paul
11
$endgroup$
$begingroup$
I think homogeneous linear means $x mapsto Cx$ not $x mapsto Cx+b$. Anyway a quadratic form on $K^n$ is a function of the form $q_A(x) = x^top A x, x in K^n$ for some matrix $A = A^top in M_n(K)$ and $q_A simeq q_B$ over $K$ iff $q_A(x) =q_B(Cx)$ for some $C in GL_n(K)$ ie. $A = C^top B C$ and $B = C^{-top} A C^{-1}$ and $q_B(x) = q_A(C^{-1}x)$
$endgroup$
– reuns
Jan 5 at 18:35
$begingroup$
@reuns So I'm done, plain and simple?
$endgroup$
– Jean Paul
Jan 5 at 18:41
add a comment |
$begingroup$
Let $mathbb{K}$ be an arbitrary field with $char(mathbb{K})neq 2$ and let $a_1, ldots ,a_n$ be Elements of $mathbb{K}$.
I want to proof the following equivalence of $n-$quadratic forms in diagonal form over $mathbb{K}$:
$ begin{align}
[a_1, ldots ,a_n] simeq [c_1^2a_1, ldots , c_n^2a_n],
end{align}
$
where $c_1, ldots , c_n in mathbb{K}^* forall n in mathbb{N}$.
According to Lam's Book $textit{Introduction to quadratic forms}$ (p. 1-2) it suffices to find an
"$textit{invertible matrix}$ $C in GL_n(mathbb{K})$ $textit{such that}$ $ g(Ccdot X)=f(X)$. (1) $textit{This means there exists a nonsingular, homogenuous linear substitution of the variables}$ $X_1, ldots ,X_n $ $ textit{that takes the form}$ $g$ $textit{to the form}$ $f$."
Now I was going to define the isometry
$C:=diag(sqrt{c_1},ldots ,sqrt{c_n})$.
Clearly $C in GL_n(mathbb{K})$ and (1) would be established. But I cannot really say if this substitution is "homogenuous" as understood in Lem's sense.
Can someone give me insight into this?
linear-algebra quadratic-forms
asked Jan 5 at 18:25
Jean PaulJean Paul
11
$endgroup$
Let $mathbb{K}$ be an arbitrary field with $char(mathbb{K})neq 2$ and let $a_1, ldots ,a_n$ be Elements of $mathbb{K}$.
I want to proof the following equivalence of $n-$quadratic forms in diagonal form over $mathbb{K}$:
$ begin{align}
[a_1, ldots ,a_n] simeq [c_1^2a_1, ldots , c_n^2a_n],
end{align}
$
where $c_1, ldots , c_n in mathbb{K}^* forall n in mathbb{N}$.
According to Lam's Book $textit{Introduction to quadratic forms}$ (p. 1-2) it suffices to find an
"$textit{invertible matrix}$ $C in GL_n(mathbb{K})$ $textit{such that}$ $ g(Ccdot X)=f(X)$. (1) $textit{This means there exists a nonsingular, homogenuous linear substitution of the variables}$ $X_1, ldots ,X_n $ $ textit{that takes the form}$ $g$ $textit{to the form}$ $f$."
Now I was going to define the isometry
$C:=diag(sqrt{c_1},ldots ,sqrt{c_n})$.
Clearly $C in GL_n(mathbb{K})$ and (1) would be established. But I cannot really say if this substitution is "homogenuous" as understood in Lem's sense.
Can someone give me insight into this?
linear-algebra quadratic-forms
linear-algebra quadratic-forms
asked Jan 5 at 18:25
Jean PaulJean Paul
11
asked Jan 5 at 18:25
Jean PaulJean Paul
11
edited Jan 5 at 18:36
Jean Paul
asked Jan 5 at 18:25
Jean PaulJean Paul
11
asked Jan 5 at 18:25
Jean PaulJean Paul
11
asked Jan 5 at 18:25
Jean PaulJean Paul
11
11
$begingroup$
I think homogeneous linear means $x mapsto Cx$ not $x mapsto Cx+b$. Anyway a quadratic form on $K^n$ is a function of the form $q_A(x) = x^top A x, x in K^n$ for some matrix $A = A^top in M_n(K)$ and $q_A simeq q_B$ over $K$ iff $q_A(x) =q_B(Cx)$ for some $C in GL_n(K)$ ie. $A = C^top B C$ and $B = C^{-top} A C^{-1}$ and $q_B(x) = q_A(C^{-1}x)$
$endgroup$
– reuns
Jan 5 at 18:35
$begingroup$
@reuns So I'm done, plain and simple?
$endgroup$
– Jean Paul
Jan 5 at 18:41
add a comment |
$begingroup$
I think homogeneous linear means $x mapsto Cx$ not $x mapsto Cx+b$. Anyway a quadratic form on $K^n$ is a function of the form $q_A(x) = x^top A x, x in K^n$ for some matrix $A = A^top in M_n(K)$ and $q_A simeq q_B$ over $K$ iff $q_A(x) =q_B(Cx)$ for some $C in GL_n(K)$ ie. $A = C^top B C$ and $B = C^{-top} A C^{-1}$ and $q_B(x) = q_A(C^{-1}x)$
$endgroup$
– reuns
Jan 5 at 18:35
$begingroup$
@reuns So I'm done, plain and simple?
$endgroup$
– Jean Paul
Jan 5 at 18:41
$begingroup$
I think homogeneous linear means $x mapsto Cx$ not $x mapsto Cx+b$. Anyway a quadratic form on $K^n$ is a function of the form $q_A(x) = x^top A x, x in K^n$ for some matrix $A = A^top in M_n(K)$ and $q_A simeq q_B$ over $K$ iff $q_A(x) =q_B(Cx)$ for some $C in GL_n(K)$ ie. $A = C^top B C$ and $B = C^{-top} A C^{-1}$ and $q_B(x) = q_A(C^{-1}x)$
$endgroup$
– reuns
Jan 5 at 18:35
$begingroup$
I think homogeneous linear means $x mapsto Cx$ not $x mapsto Cx+b$. Anyway a quadratic form on $K^n$ is a function of the form $q_A(x) = x^top A x, x in K^n$ for some matrix $A = A^top in M_n(K)$ and $q_A simeq q_B$ over $K$ iff $q_A(x) =q_B(Cx)$ for some $C in GL_n(K)$ ie. $A = C^top B C$ and $B = C^{-top} A C^{-1}$ and $q_B(x) = q_A(C^{-1}x)$
$endgroup$
– reuns
Jan 5 at 18:35
$begingroup$
@reuns So I'm done, plain and simple?
$endgroup$
– Jean Paul
Jan 5 at 18:41
$begingroup$
@reuns So I'm done, plain and simple?
$endgroup$
– Jean Paul
Jan 5 at 18:41
add a comment |
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$begingroup$
I think homogeneous linear means $x mapsto Cx$ not $x mapsto Cx+b$. Anyway a quadratic form on $K^n$ is a function of the form $q_A(x) = x^top A x, x in K^n$ for some matrix $A = A^top in M_n(K)$ and $q_A simeq q_B$ over $K$ iff $q_A(x) =q_B(Cx)$ for some $C in GL_n(K)$ ie. $A = C^top B C$ and $B = C^{-top} A C^{-1}$ and $q_B(x) = q_A(C^{-1}x)$
$endgroup$
– reuns
Jan 5 at 18:35
$begingroup$
@reuns So I'm done, plain and simple?
$endgroup$
– Jean Paul
Jan 5 at 18:41