Do two equivalent quadratic forms necessarily have the same solutions
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Do two equivalent quadratic forms necessarily have the same solutions? Suppose that I have $Q(x,y)= x^{2}- xy+ 8y^{2}$ and $R(x,y)= 2x^{2}+ 3xy+ 5y^{2}$ and the value of $Q(2,1)$ and $R(2,1$) are different. Does that mean they are not equivalent quadratic forms?
number-theory quadratic-forms
edited Dec 9 '18 at 17:34
Dr. Mathva
1,082316
asked Dec 9 '18 at 17:22
Jingting931015Jingting931015
828
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add a comment |
$begingroup$
Do two equivalent quadratic forms necessarily have the same solutions? Suppose that I have $Q(x,y)= x^{2}- xy+ 8y^{2}$ and $R(x,y)= 2x^{2}+ 3xy+ 5y^{2}$ and the value of $Q(2,1)$ and $R(2,1$) are different. Does that mean they are not equivalent quadratic forms?
number-theory quadratic-forms
edited Dec 9 '18 at 17:34
Dr. Mathva
1,082316
asked Dec 9 '18 at 17:22
Jingting931015Jingting931015
828
$endgroup$
$begingroup$
How do you define the equivalence between the quadratic forms?
$endgroup$
– Botond
Dec 9 '18 at 17:28
add a comment |
$begingroup$
Do two equivalent quadratic forms necessarily have the same solutions? Suppose that I have $Q(x,y)= x^{2}- xy+ 8y^{2}$ and $R(x,y)= 2x^{2}+ 3xy+ 5y^{2}$ and the value of $Q(2,1)$ and $R(2,1$) are different. Does that mean they are not equivalent quadratic forms?
number-theory quadratic-forms
edited Dec 9 '18 at 17:34
Dr. Mathva
1,082316
asked Dec 9 '18 at 17:22
Jingting931015Jingting931015
828
$endgroup$
Do two equivalent quadratic forms necessarily have the same solutions? Suppose that I have $Q(x,y)= x^{2}- xy+ 8y^{2}$ and $R(x,y)= 2x^{2}+ 3xy+ 5y^{2}$ and the value of $Q(2,1)$ and $R(2,1$) are different. Does that mean they are not equivalent quadratic forms?
number-theory quadratic-forms
number-theory quadratic-forms
edited Dec 9 '18 at 17:34
Dr. Mathva
1,082316
asked Dec 9 '18 at 17:22
Jingting931015Jingting931015
828
edited Dec 9 '18 at 17:34
Dr. Mathva
1,082316
asked Dec 9 '18 at 17:22
Jingting931015Jingting931015
828
edited Dec 9 '18 at 17:34
Dr. Mathva
1,082316
edited Dec 9 '18 at 17:34
Dr. Mathva
1,082316
edited Dec 9 '18 at 17:34
Dr. Mathva
1,082316
1,082316
asked Dec 9 '18 at 17:22
Jingting931015Jingting931015
828
asked Dec 9 '18 at 17:22
Jingting931015Jingting931015
828
asked Dec 9 '18 at 17:22
Jingting931015Jingting931015
828
828
$begingroup$
How do you define the equivalence between the quadratic forms?
$endgroup$
– Botond
Dec 9 '18 at 17:28
add a comment |
$begingroup$
How do you define the equivalence between the quadratic forms?
$endgroup$
– Botond
Dec 9 '18 at 17:28
$begingroup$
How do you define the equivalence between the quadratic forms?
$endgroup$
– Botond
Dec 9 '18 at 17:28
$begingroup$
How do you define the equivalence between the quadratic forms?
$endgroup$
– Botond
Dec 9 '18 at 17:28
add a comment |
2 Answers
2
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oldest
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Let us take the example of quadratic forms
$$Q_1(x,y)=x^2-y^2 text{and} Q_2(x,y)=2xy$$
They are equivalent, due to matrix identity : $A_1 = P^TA_2P$, for a certain matrix $P$, more precisely, with $a:=frac{1}{sqrt{2}}$ :
$$begin{pmatrix}1&0\0&-1end{pmatrix}=begin{pmatrix}a&a\a&-aend{pmatrix}
begin{pmatrix}0&1\1&0end{pmatrix}begin{pmatrix}a&a\a&-aend{pmatrix},$$
but the zero sets $x^2-y^2=0$ and $2xy=0$ are not at all the same (the two axes' bissectors in the first case and the two axes in the second one).
answered Dec 9 '18 at 17:37
Jean MarieJean Marie
29.1k42050
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add a comment |
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The quadratic forms $P(x,y)=2x^2+y^2$ and $Q(x,y)=6x^2+16xy+12y^2$ are equivalent. However, $P(2,1)neq Q(2,1)$.
answered Dec 9 '18 at 17:37
José Carlos SantosJosé Carlos Santos
156k22125227
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let us take the example of quadratic forms
$$Q_1(x,y)=x^2-y^2 text{and} Q_2(x,y)=2xy$$
They are equivalent, due to matrix identity : $A_1 = P^TA_2P$, for a certain matrix $P$, more precisely, with $a:=frac{1}{sqrt{2}}$ :
$$begin{pmatrix}1&0\0&-1end{pmatrix}=begin{pmatrix}a&a\a&-aend{pmatrix}
begin{pmatrix}0&1\1&0end{pmatrix}begin{pmatrix}a&a\a&-aend{pmatrix},$$
but the zero sets $x^2-y^2=0$ and $2xy=0$ are not at all the same (the two axes' bissectors in the first case and the two axes in the second one).
answered Dec 9 '18 at 17:37
Jean MarieJean Marie
29.1k42050
$endgroup$
add a comment |
$begingroup$
Let us take the example of quadratic forms
$$Q_1(x,y)=x^2-y^2 text{and} Q_2(x,y)=2xy$$
They are equivalent, due to matrix identity : $A_1 = P^TA_2P$, for a certain matrix $P$, more precisely, with $a:=frac{1}{sqrt{2}}$ :
$$begin{pmatrix}1&0\0&-1end{pmatrix}=begin{pmatrix}a&a\a&-aend{pmatrix}
begin{pmatrix}0&1\1&0end{pmatrix}begin{pmatrix}a&a\a&-aend{pmatrix},$$
but the zero sets $x^2-y^2=0$ and $2xy=0$ are not at all the same (the two axes' bissectors in the first case and the two axes in the second one).
answered Dec 9 '18 at 17:37
Jean MarieJean Marie
29.1k42050
$endgroup$
add a comment |
$begingroup$
Let us take the example of quadratic forms
$$Q_1(x,y)=x^2-y^2 text{and} Q_2(x,y)=2xy$$
They are equivalent, due to matrix identity : $A_1 = P^TA_2P$, for a certain matrix $P$, more precisely, with $a:=frac{1}{sqrt{2}}$ :
$$begin{pmatrix}1&0\0&-1end{pmatrix}=begin{pmatrix}a&a\a&-aend{pmatrix}
begin{pmatrix}0&1\1&0end{pmatrix}begin{pmatrix}a&a\a&-aend{pmatrix},$$
but the zero sets $x^2-y^2=0$ and $2xy=0$ are not at all the same (the two axes' bissectors in the first case and the two axes in the second one).
answered Dec 9 '18 at 17:37
Jean MarieJean Marie
29.1k42050
$endgroup$
Let us take the example of quadratic forms
$$Q_1(x,y)=x^2-y^2 text{and} Q_2(x,y)=2xy$$
They are equivalent, due to matrix identity : $A_1 = P^TA_2P$, for a certain matrix $P$, more precisely, with $a:=frac{1}{sqrt{2}}$ :
$$begin{pmatrix}1&0\0&-1end{pmatrix}=begin{pmatrix}a&a\a&-aend{pmatrix}
begin{pmatrix}0&1\1&0end{pmatrix}begin{pmatrix}a&a\a&-aend{pmatrix},$$
but the zero sets $x^2-y^2=0$ and $2xy=0$ are not at all the same (the two axes' bissectors in the first case and the two axes in the second one).
answered Dec 9 '18 at 17:37
Jean MarieJean Marie
29.1k42050
answered Dec 9 '18 at 17:37
Jean MarieJean Marie
29.1k42050
answered Dec 9 '18 at 17:37
Jean MarieJean Marie
29.1k42050
answered Dec 9 '18 at 17:37
Jean MarieJean Marie
29.1k42050
29.1k42050
add a comment |
add a comment |
$begingroup$
The quadratic forms $P(x,y)=2x^2+y^2$ and $Q(x,y)=6x^2+16xy+12y^2$ are equivalent. However, $P(2,1)neq Q(2,1)$.
answered Dec 9 '18 at 17:37
José Carlos SantosJosé Carlos Santos
156k22125227
$endgroup$
add a comment |
$begingroup$
The quadratic forms $P(x,y)=2x^2+y^2$ and $Q(x,y)=6x^2+16xy+12y^2$ are equivalent. However, $P(2,1)neq Q(2,1)$.
answered Dec 9 '18 at 17:37
José Carlos SantosJosé Carlos Santos
156k22125227
$endgroup$
add a comment |
$begingroup$
The quadratic forms $P(x,y)=2x^2+y^2$ and $Q(x,y)=6x^2+16xy+12y^2$ are equivalent. However, $P(2,1)neq Q(2,1)$.
answered Dec 9 '18 at 17:37
José Carlos SantosJosé Carlos Santos
156k22125227
$endgroup$
The quadratic forms $P(x,y)=2x^2+y^2$ and $Q(x,y)=6x^2+16xy+12y^2$ are equivalent. However, $P(2,1)neq Q(2,1)$.
answered Dec 9 '18 at 17:37
José Carlos SantosJosé Carlos Santos
156k22125227
answered Dec 9 '18 at 17:37
José Carlos SantosJosé Carlos Santos
156k22125227
answered Dec 9 '18 at 17:37
José Carlos SantosJosé Carlos Santos
156k22125227
answered Dec 9 '18 at 17:37
José Carlos SantosJosé Carlos Santos
156k22125227
156k22125227
add a comment |
add a comment |
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$begingroup$
How do you define the equivalence between the quadratic forms?
$endgroup$
– Botond
Dec 9 '18 at 17:28