Sign of quadratic form with parameter
$begingroup$
Given $Q(x,y,z;alpha)=x^2+z^2+2alpha xy+2xz$, i have to study the sign of quadratic form. Obviously i can use eigenvalues or studying the sign of minors, but in this case i have a hard time to understand how to act. In fact, for $A=begin{bmatrix}
1
& alpha
& 1\
alpha
&
0&0\
1& 0&1
end{bmatrix}$ I have:
$A_1=1$,
$A_2=det begin{bmatrix}
1&alpha
\
alpha
& 0
end{bmatrix}=-alpha^2>0Rightarrow alpha^2<0$,
$A_3=det begin{bmatrix}
1& alpha
&1\
alpha
&0 &0\
1
&0 &1
end{bmatrix}=-alpha^2>0Rightarrow alpha^2<0$
So, $left{begin{matrix}
1>0
\
alpha^2<0
\
alpha^2<0
end{matrix}right.=left{begin{matrix}
1>0
\
alpha^2<0
end{matrix}right.$.
Now, since for $alpha^2<0$ no solutions exist, how can I conclude? Is the matrix positive-definite or what?
linear-algebra geometry inequality algebraic-geometry quadratic-forms
asked Jan 8 at 10:35
Marco PittellaMarco Pittella
1338
$endgroup$
add a comment |
$begingroup$
Given $Q(x,y,z;alpha)=x^2+z^2+2alpha xy+2xz$, i have to study the sign of quadratic form. Obviously i can use eigenvalues or studying the sign of minors, but in this case i have a hard time to understand how to act. In fact, for $A=begin{bmatrix}
1
& alpha
& 1\
alpha
&
0&0\
1& 0&1
end{bmatrix}$ I have:
$A_1=1$,
$A_2=det begin{bmatrix}
1&alpha
\
alpha
& 0
end{bmatrix}=-alpha^2>0Rightarrow alpha^2<0$,
$A_3=det begin{bmatrix}
1& alpha
&1\
alpha
&0 &0\
1
&0 &1
end{bmatrix}=-alpha^2>0Rightarrow alpha^2<0$
So, $left{begin{matrix}
1>0
\
alpha^2<0
\
alpha^2<0
end{matrix}right.=left{begin{matrix}
1>0
\
alpha^2<0
end{matrix}right.$.
Now, since for $alpha^2<0$ no solutions exist, how can I conclude? Is the matrix positive-definite or what?
linear-algebra geometry inequality algebraic-geometry quadratic-forms
asked Jan 8 at 10:35
Marco PittellaMarco Pittella
1338
$endgroup$
$begingroup$
Why do you impose $-alpha^2 > 0$? The computation of $A_2$ tells you exactly that the upper-left $2 times 2$ minor defines a quadratic form of signature $(1, 1)$ (or a degenerate, nonzero positive semidefinite quadratic form iff $alpha = 0$).
$endgroup$
– Travis
Jan 8 at 10:39
$begingroup$
@Travis Thanks for your answer. I thought that $>0$ was the default sign from which to start the analysis. But if not, what I have to impose for $-alpha^2$?
$endgroup$
– Marco Pittella
Jan 8 at 10:54
$begingroup$
There's no notion of "default sign", and there's nothing to impose. Rather, $-alpha^2$ controls the signature of the resulting form.
$endgroup$
– Travis
Jan 8 at 11:21
1
$begingroup$
Note, too, by the way, that when determining signature using eigenvalues, you don't actually need to determine the eigenvalues themselves (which in this case would involve solving a cubic), you only need their signs, and you can extract that information from Descartes' Rule of Signs. If it would be useful, I'm happy to write up a short solution explaining this method.
$endgroup$
– Travis
Jan 8 at 11:23
$begingroup$
@Travis Thanks again for your help. $-alpha^2$ controls the sign of the form, of course. So, how can I conclude in this case? Any value I put on place of $alpha$ I obtain an indefinite form. Could you show me the steps to follow? Instead, for the method that you mentioned well, ok, thank you so much for your time.
$endgroup$
– Marco Pittella
Jan 8 at 11:40
add a comment |
$begingroup$
Given $Q(x,y,z;alpha)=x^2+z^2+2alpha xy+2xz$, i have to study the sign of quadratic form. Obviously i can use eigenvalues or studying the sign of minors, but in this case i have a hard time to understand how to act. In fact, for $A=begin{bmatrix}
1
& alpha
& 1\
alpha
&
0&0\
1& 0&1
end{bmatrix}$ I have:
$A_1=1$,
$A_2=det begin{bmatrix}
1&alpha
\
alpha
& 0
end{bmatrix}=-alpha^2>0Rightarrow alpha^2<0$,
$A_3=det begin{bmatrix}
1& alpha
&1\
alpha
&0 &0\
1
&0 &1
end{bmatrix}=-alpha^2>0Rightarrow alpha^2<0$
So, $left{begin{matrix}
1>0
\
alpha^2<0
\
alpha^2<0
end{matrix}right.=left{begin{matrix}
1>0
\
alpha^2<0
end{matrix}right.$.
Now, since for $alpha^2<0$ no solutions exist, how can I conclude? Is the matrix positive-definite or what?
linear-algebra geometry inequality algebraic-geometry quadratic-forms
asked Jan 8 at 10:35
Marco PittellaMarco Pittella
1338
$endgroup$
Given $Q(x,y,z;alpha)=x^2+z^2+2alpha xy+2xz$, i have to study the sign of quadratic form. Obviously i can use eigenvalues or studying the sign of minors, but in this case i have a hard time to understand how to act. In fact, for $A=begin{bmatrix}
1
& alpha
& 1\
alpha
&
0&0\
1& 0&1
end{bmatrix}$ I have:
$A_1=1$,
$A_2=det begin{bmatrix}
1&alpha
\
alpha
& 0
end{bmatrix}=-alpha^2>0Rightarrow alpha^2<0$,
$A_3=det begin{bmatrix}
1& alpha
&1\
alpha
&0 &0\
1
&0 &1
end{bmatrix}=-alpha^2>0Rightarrow alpha^2<0$
So, $left{begin{matrix}
1>0
\
alpha^2<0
\
alpha^2<0
end{matrix}right.=left{begin{matrix}
1>0
\
alpha^2<0
end{matrix}right.$.
Now, since for $alpha^2<0$ no solutions exist, how can I conclude? Is the matrix positive-definite or what?
linear-algebra geometry inequality algebraic-geometry quadratic-forms
linear-algebra geometry inequality algebraic-geometry quadratic-forms
asked Jan 8 at 10:35
Marco PittellaMarco Pittella
1338
asked Jan 8 at 10:35
Marco PittellaMarco Pittella
1338
edited Jan 8 at 11:05
Marco Pittella
asked Jan 8 at 10:35
Marco PittellaMarco Pittella
1338
asked Jan 8 at 10:35
Marco PittellaMarco Pittella
1338
asked Jan 8 at 10:35
Marco PittellaMarco Pittella
1338
1338
$begingroup$
Why do you impose $-alpha^2 > 0$? The computation of $A_2$ tells you exactly that the upper-left $2 times 2$ minor defines a quadratic form of signature $(1, 1)$ (or a degenerate, nonzero positive semidefinite quadratic form iff $alpha = 0$).
$endgroup$
– Travis
Jan 8 at 10:39
$begingroup$
@Travis Thanks for your answer. I thought that $>0$ was the default sign from which to start the analysis. But if not, what I have to impose for $-alpha^2$?
$endgroup$
– Marco Pittella
Jan 8 at 10:54
$begingroup$
There's no notion of "default sign", and there's nothing to impose. Rather, $-alpha^2$ controls the signature of the resulting form.
$endgroup$
– Travis
Jan 8 at 11:21
1
$begingroup$
Note, too, by the way, that when determining signature using eigenvalues, you don't actually need to determine the eigenvalues themselves (which in this case would involve solving a cubic), you only need their signs, and you can extract that information from Descartes' Rule of Signs. If it would be useful, I'm happy to write up a short solution explaining this method.
$endgroup$
– Travis
Jan 8 at 11:23
$begingroup$
@Travis Thanks again for your help. $-alpha^2$ controls the sign of the form, of course. So, how can I conclude in this case? Any value I put on place of $alpha$ I obtain an indefinite form. Could you show me the steps to follow? Instead, for the method that you mentioned well, ok, thank you so much for your time.
$endgroup$
– Marco Pittella
Jan 8 at 11:40
add a comment |
$begingroup$
Why do you impose $-alpha^2 > 0$? The computation of $A_2$ tells you exactly that the upper-left $2 times 2$ minor defines a quadratic form of signature $(1, 1)$ (or a degenerate, nonzero positive semidefinite quadratic form iff $alpha = 0$).
$endgroup$
– Travis
Jan 8 at 10:39
$begingroup$
@Travis Thanks for your answer. I thought that $>0$ was the default sign from which to start the analysis. But if not, what I have to impose for $-alpha^2$?
$endgroup$
– Marco Pittella
Jan 8 at 10:54
$begingroup$
There's no notion of "default sign", and there's nothing to impose. Rather, $-alpha^2$ controls the signature of the resulting form.
$endgroup$
– Travis
Jan 8 at 11:21
1
$begingroup$
Note, too, by the way, that when determining signature using eigenvalues, you don't actually need to determine the eigenvalues themselves (which in this case would involve solving a cubic), you only need their signs, and you can extract that information from Descartes' Rule of Signs. If it would be useful, I'm happy to write up a short solution explaining this method.
$endgroup$
– Travis
Jan 8 at 11:23
$begingroup$
@Travis Thanks again for your help. $-alpha^2$ controls the sign of the form, of course. So, how can I conclude in this case? Any value I put on place of $alpha$ I obtain an indefinite form. Could you show me the steps to follow? Instead, for the method that you mentioned well, ok, thank you so much for your time.
$endgroup$
– Marco Pittella
Jan 8 at 11:40
$begingroup$
Why do you impose $-alpha^2 > 0$? The computation of $A_2$ tells you exactly that the upper-left $2 times 2$ minor defines a quadratic form of signature $(1, 1)$ (or a degenerate, nonzero positive semidefinite quadratic form iff $alpha = 0$).
$endgroup$
– Travis
Jan 8 at 10:39
$begingroup$
Why do you impose $-alpha^2 > 0$? The computation of $A_2$ tells you exactly that the upper-left $2 times 2$ minor defines a quadratic form of signature $(1, 1)$ (or a degenerate, nonzero positive semidefinite quadratic form iff $alpha = 0$).
$endgroup$
– Travis
Jan 8 at 10:39
$begingroup$
@Travis Thanks for your answer. I thought that $>0$ was the default sign from which to start the analysis. But if not, what I have to impose for $-alpha^2$?
$endgroup$
– Marco Pittella
Jan 8 at 10:54
$begingroup$
@Travis Thanks for your answer. I thought that $>0$ was the default sign from which to start the analysis. But if not, what I have to impose for $-alpha^2$?
$endgroup$
– Marco Pittella
Jan 8 at 10:54
$begingroup$
There's no notion of "default sign", and there's nothing to impose. Rather, $-alpha^2$ controls the signature of the resulting form.
$endgroup$
– Travis
Jan 8 at 11:21
$begingroup$
There's no notion of "default sign", and there's nothing to impose. Rather, $-alpha^2$ controls the signature of the resulting form.
$endgroup$
– Travis
Jan 8 at 11:21
1
1
$begingroup$
Note, too, by the way, that when determining signature using eigenvalues, you don't actually need to determine the eigenvalues themselves (which in this case would involve solving a cubic), you only need their signs, and you can extract that information from Descartes' Rule of Signs. If it would be useful, I'm happy to write up a short solution explaining this method.
$endgroup$
– Travis
Jan 8 at 11:23
$begingroup$
Note, too, by the way, that when determining signature using eigenvalues, you don't actually need to determine the eigenvalues themselves (which in this case would involve solving a cubic), you only need their signs, and you can extract that information from Descartes' Rule of Signs. If it would be useful, I'm happy to write up a short solution explaining this method.
$endgroup$
– Travis
Jan 8 at 11:23
$begingroup$
@Travis Thanks again for your help. $-alpha^2$ controls the sign of the form, of course. So, how can I conclude in this case? Any value I put on place of $alpha$ I obtain an indefinite form. Could you show me the steps to follow? Instead, for the method that you mentioned well, ok, thank you so much for your time.
$endgroup$
– Marco Pittella
Jan 8 at 11:40
$begingroup$
@Travis Thanks again for your help. $-alpha^2$ controls the sign of the form, of course. So, how can I conclude in this case? Any value I put on place of $alpha$ I obtain an indefinite form. Could you show me the steps to follow? Instead, for the method that you mentioned well, ok, thank you so much for your time.
$endgroup$
– Marco Pittella
Jan 8 at 11:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Following a remark in the comments, here's an efficient way to calculate the sign using the characteristic polynomial---recall that the signature is essentially a count of the positive, zero negative eigenvalues of the representative matrix---but without actually computing its roots.
The characteristic polynomial of the matrix representation $A$ of the quadratic form is
$$det left(t I_3 - pmatrix{1&alpha&1\ alpha&0&0\1&0&1}right) = t^3 - 2 t^2 - alpha^2 t + alpha^2 .$$
For $alpha neq 0$, we have $alpha^2 > 0$, in which case the number of sign changes of the coefficients is $2$, and thus by Descartes' Rule of Signs (and the fact that all of the eigenvalues of a real, symmetric matrix are real) the polynomial has (1) two positive roots, and, (2) since $0$ is not a root, one negative root. Thus, the signature is $(2, 1)$ (Lorentzian). NB we avoided computing explicitly the roots of the characteristic polynomial, which would have been difficult.
On the other hand, if $alpha = 0$, the constant term of the characteristic polynomial is zero, so in that case the quadratic form is degenerate.
answered Jan 9 at 5:38
TravisTravis
64.4k769151
$endgroup$
add a comment |
$begingroup$
Here, it is an elementary solution.
After completing the square
$$
Q(x,y,z)=(x+alpha y +z)^2 -alpha^2(y+frac{1}{alpha}z)^2+z^2
$$
assuming $alphaneq 0$.
It means that $Q$ indefinite quadratic form if $alphaneq 0$, otherwise it is positive semi definite only.
answered Feb 18 at 15:34
MGyMGy
413
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
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$begingroup$
Following a remark in the comments, here's an efficient way to calculate the sign using the characteristic polynomial---recall that the signature is essentially a count of the positive, zero negative eigenvalues of the representative matrix---but without actually computing its roots.
The characteristic polynomial of the matrix representation $A$ of the quadratic form is
$$det left(t I_3 - pmatrix{1&alpha&1\ alpha&0&0\1&0&1}right) = t^3 - 2 t^2 - alpha^2 t + alpha^2 .$$
For $alpha neq 0$, we have $alpha^2 > 0$, in which case the number of sign changes of the coefficients is $2$, and thus by Descartes' Rule of Signs (and the fact that all of the eigenvalues of a real, symmetric matrix are real) the polynomial has (1) two positive roots, and, (2) since $0$ is not a root, one negative root. Thus, the signature is $(2, 1)$ (Lorentzian). NB we avoided computing explicitly the roots of the characteristic polynomial, which would have been difficult.
On the other hand, if $alpha = 0$, the constant term of the characteristic polynomial is zero, so in that case the quadratic form is degenerate.
answered Jan 9 at 5:38
TravisTravis
64.4k769151
$endgroup$
add a comment |
$begingroup$
Following a remark in the comments, here's an efficient way to calculate the sign using the characteristic polynomial---recall that the signature is essentially a count of the positive, zero negative eigenvalues of the representative matrix---but without actually computing its roots.
The characteristic polynomial of the matrix representation $A$ of the quadratic form is
$$det left(t I_3 - pmatrix{1&alpha&1\ alpha&0&0\1&0&1}right) = t^3 - 2 t^2 - alpha^2 t + alpha^2 .$$
For $alpha neq 0$, we have $alpha^2 > 0$, in which case the number of sign changes of the coefficients is $2$, and thus by Descartes' Rule of Signs (and the fact that all of the eigenvalues of a real, symmetric matrix are real) the polynomial has (1) two positive roots, and, (2) since $0$ is not a root, one negative root. Thus, the signature is $(2, 1)$ (Lorentzian). NB we avoided computing explicitly the roots of the characteristic polynomial, which would have been difficult.
On the other hand, if $alpha = 0$, the constant term of the characteristic polynomial is zero, so in that case the quadratic form is degenerate.
answered Jan 9 at 5:38
TravisTravis
64.4k769151
$endgroup$
add a comment |
$begingroup$
Following a remark in the comments, here's an efficient way to calculate the sign using the characteristic polynomial---recall that the signature is essentially a count of the positive, zero negative eigenvalues of the representative matrix---but without actually computing its roots.
The characteristic polynomial of the matrix representation $A$ of the quadratic form is
$$det left(t I_3 - pmatrix{1&alpha&1\ alpha&0&0\1&0&1}right) = t^3 - 2 t^2 - alpha^2 t + alpha^2 .$$
For $alpha neq 0$, we have $alpha^2 > 0$, in which case the number of sign changes of the coefficients is $2$, and thus by Descartes' Rule of Signs (and the fact that all of the eigenvalues of a real, symmetric matrix are real) the polynomial has (1) two positive roots, and, (2) since $0$ is not a root, one negative root. Thus, the signature is $(2, 1)$ (Lorentzian). NB we avoided computing explicitly the roots of the characteristic polynomial, which would have been difficult.
On the other hand, if $alpha = 0$, the constant term of the characteristic polynomial is zero, so in that case the quadratic form is degenerate.
answered Jan 9 at 5:38
TravisTravis
64.4k769151
$endgroup$
Following a remark in the comments, here's an efficient way to calculate the sign using the characteristic polynomial---recall that the signature is essentially a count of the positive, zero negative eigenvalues of the representative matrix---but without actually computing its roots.
The characteristic polynomial of the matrix representation $A$ of the quadratic form is
$$det left(t I_3 - pmatrix{1&alpha&1\ alpha&0&0\1&0&1}right) = t^3 - 2 t^2 - alpha^2 t + alpha^2 .$$
For $alpha neq 0$, we have $alpha^2 > 0$, in which case the number of sign changes of the coefficients is $2$, and thus by Descartes' Rule of Signs (and the fact that all of the eigenvalues of a real, symmetric matrix are real) the polynomial has (1) two positive roots, and, (2) since $0$ is not a root, one negative root. Thus, the signature is $(2, 1)$ (Lorentzian). NB we avoided computing explicitly the roots of the characteristic polynomial, which would have been difficult.
On the other hand, if $alpha = 0$, the constant term of the characteristic polynomial is zero, so in that case the quadratic form is degenerate.
answered Jan 9 at 5:38
TravisTravis
64.4k769151
answered Jan 9 at 5:38
TravisTravis
64.4k769151
answered Jan 9 at 5:38
TravisTravis
64.4k769151
answered Jan 9 at 5:38
TravisTravis
64.4k769151
64.4k769151
add a comment |
add a comment |
$begingroup$
Here, it is an elementary solution.
After completing the square
$$
Q(x,y,z)=(x+alpha y +z)^2 -alpha^2(y+frac{1}{alpha}z)^2+z^2
$$
assuming $alphaneq 0$.
It means that $Q$ indefinite quadratic form if $alphaneq 0$, otherwise it is positive semi definite only.
answered Feb 18 at 15:34
MGyMGy
413
$endgroup$
add a comment |
$begingroup$
Here, it is an elementary solution.
After completing the square
$$
Q(x,y,z)=(x+alpha y +z)^2 -alpha^2(y+frac{1}{alpha}z)^2+z^2
$$
assuming $alphaneq 0$.
It means that $Q$ indefinite quadratic form if $alphaneq 0$, otherwise it is positive semi definite only.
answered Feb 18 at 15:34
MGyMGy
413
$endgroup$
add a comment |
$begingroup$
Here, it is an elementary solution.
After completing the square
$$
Q(x,y,z)=(x+alpha y +z)^2 -alpha^2(y+frac{1}{alpha}z)^2+z^2
$$
assuming $alphaneq 0$.
It means that $Q$ indefinite quadratic form if $alphaneq 0$, otherwise it is positive semi definite only.
answered Feb 18 at 15:34
MGyMGy
413
$endgroup$
Here, it is an elementary solution.
After completing the square
$$
Q(x,y,z)=(x+alpha y +z)^2 -alpha^2(y+frac{1}{alpha}z)^2+z^2
$$
assuming $alphaneq 0$.
It means that $Q$ indefinite quadratic form if $alphaneq 0$, otherwise it is positive semi definite only.
answered Feb 18 at 15:34
MGyMGy
413
answered Feb 18 at 15:34
MGyMGy
413
answered Feb 18 at 15:34
MGyMGy
413
answered Feb 18 at 15:34
MGyMGy
413
413
add a comment |
add a comment |
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$begingroup$
Why do you impose $-alpha^2 > 0$? The computation of $A_2$ tells you exactly that the upper-left $2 times 2$ minor defines a quadratic form of signature $(1, 1)$ (or a degenerate, nonzero positive semidefinite quadratic form iff $alpha = 0$).
$endgroup$
– Travis
Jan 8 at 10:39
$begingroup$
@Travis Thanks for your answer. I thought that $>0$ was the default sign from which to start the analysis. But if not, what I have to impose for $-alpha^2$?
$endgroup$
– Marco Pittella
Jan 8 at 10:54
$begingroup$
There's no notion of "default sign", and there's nothing to impose. Rather, $-alpha^2$ controls the signature of the resulting form.
$endgroup$
– Travis
Jan 8 at 11:21
1
$begingroup$
Note, too, by the way, that when determining signature using eigenvalues, you don't actually need to determine the eigenvalues themselves (which in this case would involve solving a cubic), you only need their signs, and you can extract that information from Descartes' Rule of Signs. If it would be useful, I'm happy to write up a short solution explaining this method.
$endgroup$
– Travis
Jan 8 at 11:23
$begingroup$
@Travis Thanks again for your help. $-alpha^2$ controls the sign of the form, of course. So, how can I conclude in this case? Any value I put on place of $alpha$ I obtain an indefinite form. Could you show me the steps to follow? Instead, for the method that you mentioned well, ok, thank you so much for your time.
$endgroup$
– Marco Pittella
Jan 8 at 11:40