Simultaneously vanishing quadratic forms The Next CEO of Stack OverflowSimultaneous diagonalization of quadratic formsQuadratic forms…Feasibility of a given set of Quadratic FormsExistence of complex solutions satisfying two quadratic formsHow to show that two quadratic forms are equivalent?Simultaneously diagonalise two real quadratic formsPrimitive Integral Quadratic forms of fixed discriminantAre two positive defined quadractic forms simultaneously diagonalizable?Determine whether two quadratic forms are simultaneously diagonalizableQuadratic forms + and -
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Simultaneously vanishing quadratic forms
The Next CEO of Stack OverflowSimultaneous diagonalization of quadratic formsQuadratic forms…Feasibility of a given set of Quadratic FormsExistence of complex solutions satisfying two quadratic formsHow to show that two quadratic forms are equivalent?Simultaneously diagonalise two real quadratic formsPrimitive Integral Quadratic forms of fixed discriminantAre two positive defined quadractic forms simultaneously diagonalizable?Determine whether two quadratic forms are simultaneously diagonalizableQuadratic forms + and -
$begingroup$
Given a set of Hermitian matrices $A_i$, is there a simple way to check if there exists a vector $c$ such that for all $i$:
$$c^* A_i c = 0?$$
Namely, when can the quadratic forms defined by the $A_i$ simultaneously vanish?
linear-algebra matrices systems-of-equations quadratic-forms
edited Mar 18 at 19:21
Rodrigo de Azevedo
13.2k41960
asked Jul 31 '15 at 19:06
Eric MetodievEric Metodiev
663
$endgroup$
add a comment |
$begingroup$
Given a set of Hermitian matrices $A_i$, is there a simple way to check if there exists a vector $c$ such that for all $i$:
$$c^* A_i c = 0?$$
Namely, when can the quadratic forms defined by the $A_i$ simultaneously vanish?
linear-algebra matrices systems-of-equations quadratic-forms
edited Mar 18 at 19:21
Rodrigo de Azevedo
13.2k41960
asked Jul 31 '15 at 19:06
Eric MetodievEric Metodiev
663
$endgroup$
1
$begingroup$
Conceptually you're just checking the intersection of the radicals of the forms... but I guess that isn't concrete enough?
$endgroup$
– rschwieb
Jul 31 '15 at 19:11
1
$begingroup$
probably not. Also, if the number of $A_i$ is at least the dimension $n,$ then you do not expect any nonzero $c.$ Why do you want to do this?
$endgroup$
– Will Jagy
Jul 31 '15 at 19:13
1
$begingroup$
If such $c$ exists then it should necessarily satisfy $c^*sum_k=1^nx_kA_k c=0$ for all $x$. If there exists such $x$ that $sum_k=1^nx_kA_k$ is positive definite then $c=0$. Finding $x$ is a semidefinite programming.
$endgroup$
– A.Γ.
Jul 31 '15 at 19:37
$begingroup$
Do you mean $c^HA_ic=0$ by $c^*A_ic=0$? ($(cdot)^H$ is the Hermitian operator)
$endgroup$
– Mostafa Ayaz
Mar 19 at 23:18
add a comment |
$begingroup$
Given a set of Hermitian matrices $A_i$, is there a simple way to check if there exists a vector $c$ such that for all $i$:
$$c^* A_i c = 0?$$
Namely, when can the quadratic forms defined by the $A_i$ simultaneously vanish?
linear-algebra matrices systems-of-equations quadratic-forms
edited Mar 18 at 19:21
Rodrigo de Azevedo
13.2k41960
asked Jul 31 '15 at 19:06
Eric MetodievEric Metodiev
663
$endgroup$
Given a set of Hermitian matrices $A_i$, is there a simple way to check if there exists a vector $c$ such that for all $i$:
$$c^* A_i c = 0?$$
Namely, when can the quadratic forms defined by the $A_i$ simultaneously vanish?
linear-algebra matrices systems-of-equations quadratic-forms
linear-algebra matrices systems-of-equations quadratic-forms
edited Mar 18 at 19:21
Rodrigo de Azevedo
13.2k41960
asked Jul 31 '15 at 19:06
Eric MetodievEric Metodiev
663
edited Mar 18 at 19:21
Rodrigo de Azevedo
13.2k41960
asked Jul 31 '15 at 19:06
Eric MetodievEric Metodiev
663
edited Mar 18 at 19:21
Rodrigo de Azevedo
13.2k41960
edited Mar 18 at 19:21
Rodrigo de Azevedo
13.2k41960
edited Mar 18 at 19:21
Rodrigo de Azevedo
13.2k41960
13.2k41960
asked Jul 31 '15 at 19:06
Eric MetodievEric Metodiev
663
asked Jul 31 '15 at 19:06
Eric MetodievEric Metodiev
663
asked Jul 31 '15 at 19:06
Eric MetodievEric Metodiev
663
663
1
$begingroup$
Conceptually you're just checking the intersection of the radicals of the forms... but I guess that isn't concrete enough?
$endgroup$
– rschwieb
Jul 31 '15 at 19:11
1
$begingroup$
probably not. Also, if the number of $A_i$ is at least the dimension $n,$ then you do not expect any nonzero $c.$ Why do you want to do this?
$endgroup$
– Will Jagy
Jul 31 '15 at 19:13
1
$begingroup$
If such $c$ exists then it should necessarily satisfy $c^*sum_k=1^nx_kA_k c=0$ for all $x$. If there exists such $x$ that $sum_k=1^nx_kA_k$ is positive definite then $c=0$. Finding $x$ is a semidefinite programming.
$endgroup$
– A.Γ.
Jul 31 '15 at 19:37
$begingroup$
Do you mean $c^HA_ic=0$ by $c^*A_ic=0$? ($(cdot)^H$ is the Hermitian operator)
$endgroup$
– Mostafa Ayaz
Mar 19 at 23:18
add a comment |
1
$begingroup$
Conceptually you're just checking the intersection of the radicals of the forms... but I guess that isn't concrete enough?
$endgroup$
– rschwieb
Jul 31 '15 at 19:11
1
$begingroup$
probably not. Also, if the number of $A_i$ is at least the dimension $n,$ then you do not expect any nonzero $c.$ Why do you want to do this?
$endgroup$
– Will Jagy
Jul 31 '15 at 19:13
1
$begingroup$
If such $c$ exists then it should necessarily satisfy $c^*sum_k=1^nx_kA_k c=0$ for all $x$. If there exists such $x$ that $sum_k=1^nx_kA_k$ is positive definite then $c=0$. Finding $x$ is a semidefinite programming.
$endgroup$
– A.Γ.
Jul 31 '15 at 19:37
$begingroup$
Do you mean $c^HA_ic=0$ by $c^*A_ic=0$? ($(cdot)^H$ is the Hermitian operator)
$endgroup$
– Mostafa Ayaz
Mar 19 at 23:18
1
1
$begingroup$
Conceptually you're just checking the intersection of the radicals of the forms... but I guess that isn't concrete enough?
$endgroup$
– rschwieb
Jul 31 '15 at 19:11
$begingroup$
Conceptually you're just checking the intersection of the radicals of the forms... but I guess that isn't concrete enough?
$endgroup$
– rschwieb
Jul 31 '15 at 19:11
1
1
$begingroup$
probably not. Also, if the number of $A_i$ is at least the dimension $n,$ then you do not expect any nonzero $c.$ Why do you want to do this?
$endgroup$
– Will Jagy
Jul 31 '15 at 19:13
$begingroup$
probably not. Also, if the number of $A_i$ is at least the dimension $n,$ then you do not expect any nonzero $c.$ Why do you want to do this?
$endgroup$
– Will Jagy
Jul 31 '15 at 19:13
1
1
$begingroup$
If such $c$ exists then it should necessarily satisfy $c^*sum_k=1^nx_kA_k c=0$ for all $x$. If there exists such $x$ that $sum_k=1^nx_kA_k$ is positive definite then $c=0$. Finding $x$ is a semidefinite programming.
$endgroup$
– A.Γ.
Jul 31 '15 at 19:37
$begingroup$
If such $c$ exists then it should necessarily satisfy $c^*sum_k=1^nx_kA_k c=0$ for all $x$. If there exists such $x$ that $sum_k=1^nx_kA_k$ is positive definite then $c=0$. Finding $x$ is a semidefinite programming.
$endgroup$
– A.Γ.
Jul 31 '15 at 19:37
$begingroup$
Do you mean $c^HA_ic=0$ by $c^*A_ic=0$? ($(cdot)^H$ is the Hermitian operator)
$endgroup$
– Mostafa Ayaz
Mar 19 at 23:18
$begingroup$
Do you mean $c^HA_ic=0$ by $c^*A_ic=0$? ($(cdot)^H$ is the Hermitian operator)
$endgroup$
– Mostafa Ayaz
Mar 19 at 23:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The system of equations you present is a quadratic system of polynomial equations for the components of $c$. As discussed here, you can compute the Gröbner basis for this system of equations, which will tell you if it is inconsistent or not. In order to exclude the zero vector as a solution you could add a normalization condition $c^* c = 1$. I think this should work.
edited Mar 19 at 21:53
Rodrigo de Azevedo
13.2k41960
answered Aug 30 '17 at 6:19
Teddy BakerTeddy Baker
601412
$endgroup$
$begingroup$
Few remarks. 1. We may assume that the $A_1,cdots, A_kin S_n$ constitute a free system. 2. See if there is $(j,k)$ s.t. the $(A_i)_j,k$ are zero. 3. Don't add the normalization condition $c^*c=1$ -that increases the calculation time of Grobner basis- 4. Indeed, it is better to calculate the Hilbert dimension $h$ of the Grobner basis; if $h>0$, then there is a non-zero solution. 5. If you want a calculation time <2', then you will have to settle for $n leq 9$.
$endgroup$
– loup blanc
Aug 30 '17 at 11:10
$begingroup$
NB. Calculations are carried out for real symmetric matrices. For complex hermitian matrices it is more complicated.
$endgroup$
– loup blanc
Aug 30 '17 at 11:16
add a comment |
Your Answer
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$begingroup$
The system of equations you present is a quadratic system of polynomial equations for the components of $c$. As discussed here, you can compute the Gröbner basis for this system of equations, which will tell you if it is inconsistent or not. In order to exclude the zero vector as a solution you could add a normalization condition $c^* c = 1$. I think this should work.
edited Mar 19 at 21:53
Rodrigo de Azevedo
13.2k41960
answered Aug 30 '17 at 6:19
Teddy BakerTeddy Baker
601412
$endgroup$
$begingroup$
Few remarks. 1. We may assume that the $A_1,cdots, A_kin S_n$ constitute a free system. 2. See if there is $(j,k)$ s.t. the $(A_i)_j,k$ are zero. 3. Don't add the normalization condition $c^*c=1$ -that increases the calculation time of Grobner basis- 4. Indeed, it is better to calculate the Hilbert dimension $h$ of the Grobner basis; if $h>0$, then there is a non-zero solution. 5. If you want a calculation time <2', then you will have to settle for $n leq 9$.
$endgroup$
– loup blanc
Aug 30 '17 at 11:10
$begingroup$
NB. Calculations are carried out for real symmetric matrices. For complex hermitian matrices it is more complicated.
$endgroup$
– loup blanc
Aug 30 '17 at 11:16
add a comment |
$begingroup$
The system of equations you present is a quadratic system of polynomial equations for the components of $c$. As discussed here, you can compute the Gröbner basis for this system of equations, which will tell you if it is inconsistent or not. In order to exclude the zero vector as a solution you could add a normalization condition $c^* c = 1$. I think this should work.
edited Mar 19 at 21:53
Rodrigo de Azevedo
13.2k41960
answered Aug 30 '17 at 6:19
Teddy BakerTeddy Baker
601412
$endgroup$
$begingroup$
Few remarks. 1. We may assume that the $A_1,cdots, A_kin S_n$ constitute a free system. 2. See if there is $(j,k)$ s.t. the $(A_i)_j,k$ are zero. 3. Don't add the normalization condition $c^*c=1$ -that increases the calculation time of Grobner basis- 4. Indeed, it is better to calculate the Hilbert dimension $h$ of the Grobner basis; if $h>0$, then there is a non-zero solution. 5. If you want a calculation time <2', then you will have to settle for $n leq 9$.
$endgroup$
– loup blanc
Aug 30 '17 at 11:10
$begingroup$
NB. Calculations are carried out for real symmetric matrices. For complex hermitian matrices it is more complicated.
$endgroup$
– loup blanc
Aug 30 '17 at 11:16
add a comment |
$begingroup$
The system of equations you present is a quadratic system of polynomial equations for the components of $c$. As discussed here, you can compute the Gröbner basis for this system of equations, which will tell you if it is inconsistent or not. In order to exclude the zero vector as a solution you could add a normalization condition $c^* c = 1$. I think this should work.
edited Mar 19 at 21:53
Rodrigo de Azevedo
13.2k41960
answered Aug 30 '17 at 6:19
Teddy BakerTeddy Baker
601412
$endgroup$
The system of equations you present is a quadratic system of polynomial equations for the components of $c$. As discussed here, you can compute the Gröbner basis for this system of equations, which will tell you if it is inconsistent or not. In order to exclude the zero vector as a solution you could add a normalization condition $c^* c = 1$. I think this should work.
edited Mar 19 at 21:53
Rodrigo de Azevedo
13.2k41960
answered Aug 30 '17 at 6:19
Teddy BakerTeddy Baker
601412
edited Mar 19 at 21:53
Rodrigo de Azevedo
13.2k41960
edited Mar 19 at 21:53
Rodrigo de Azevedo
13.2k41960
edited Mar 19 at 21:53
Rodrigo de Azevedo
13.2k41960
13.2k41960
answered Aug 30 '17 at 6:19
Teddy BakerTeddy Baker
601412
answered Aug 30 '17 at 6:19
Teddy BakerTeddy Baker
601412
answered Aug 30 '17 at 6:19
Teddy BakerTeddy Baker
601412
601412
$begingroup$
Few remarks. 1. We may assume that the $A_1,cdots, A_kin S_n$ constitute a free system. 2. See if there is $(j,k)$ s.t. the $(A_i)_j,k$ are zero. 3. Don't add the normalization condition $c^*c=1$ -that increases the calculation time of Grobner basis- 4. Indeed, it is better to calculate the Hilbert dimension $h$ of the Grobner basis; if $h>0$, then there is a non-zero solution. 5. If you want a calculation time <2', then you will have to settle for $n leq 9$.
$endgroup$
– loup blanc
Aug 30 '17 at 11:10
$begingroup$
NB. Calculations are carried out for real symmetric matrices. For complex hermitian matrices it is more complicated.
$endgroup$
– loup blanc
Aug 30 '17 at 11:16
add a comment |
$begingroup$
Few remarks. 1. We may assume that the $A_1,cdots, A_kin S_n$ constitute a free system. 2. See if there is $(j,k)$ s.t. the $(A_i)_j,k$ are zero. 3. Don't add the normalization condition $c^*c=1$ -that increases the calculation time of Grobner basis- 4. Indeed, it is better to calculate the Hilbert dimension $h$ of the Grobner basis; if $h>0$, then there is a non-zero solution. 5. If you want a calculation time <2', then you will have to settle for $n leq 9$.
$endgroup$
– loup blanc
Aug 30 '17 at 11:10
$begingroup$
NB. Calculations are carried out for real symmetric matrices. For complex hermitian matrices it is more complicated.
$endgroup$
– loup blanc
Aug 30 '17 at 11:16
$begingroup$
Few remarks. 1. We may assume that the $A_1,cdots, A_kin S_n$ constitute a free system. 2. See if there is $(j,k)$ s.t. the $(A_i)_j,k$ are zero. 3. Don't add the normalization condition $c^*c=1$ -that increases the calculation time of Grobner basis- 4. Indeed, it is better to calculate the Hilbert dimension $h$ of the Grobner basis; if $h>0$, then there is a non-zero solution. 5. If you want a calculation time <2', then you will have to settle for $n leq 9$.
$endgroup$
– loup blanc
Aug 30 '17 at 11:10
$begingroup$
Few remarks. 1. We may assume that the $A_1,cdots, A_kin S_n$ constitute a free system. 2. See if there is $(j,k)$ s.t. the $(A_i)_j,k$ are zero. 3. Don't add the normalization condition $c^*c=1$ -that increases the calculation time of Grobner basis- 4. Indeed, it is better to calculate the Hilbert dimension $h$ of the Grobner basis; if $h>0$, then there is a non-zero solution. 5. If you want a calculation time <2', then you will have to settle for $n leq 9$.
$endgroup$
– loup blanc
Aug 30 '17 at 11:10
$begingroup$
NB. Calculations are carried out for real symmetric matrices. For complex hermitian matrices it is more complicated.
$endgroup$
– loup blanc
Aug 30 '17 at 11:16
$begingroup$
NB. Calculations are carried out for real symmetric matrices. For complex hermitian matrices it is more complicated.
$endgroup$
– loup blanc
Aug 30 '17 at 11:16
add a comment |
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1
$begingroup$
Conceptually you're just checking the intersection of the radicals of the forms... but I guess that isn't concrete enough?
$endgroup$
– rschwieb
Jul 31 '15 at 19:11
1
$begingroup$
probably not. Also, if the number of $A_i$ is at least the dimension $n,$ then you do not expect any nonzero $c.$ Why do you want to do this?
$endgroup$
– Will Jagy
Jul 31 '15 at 19:13
1
$begingroup$
If such $c$ exists then it should necessarily satisfy $c^*sum_k=1^nx_kA_k c=0$ for all $x$. If there exists such $x$ that $sum_k=1^nx_kA_k$ is positive definite then $c=0$. Finding $x$ is a semidefinite programming.
$endgroup$
– A.Γ.
Jul 31 '15 at 19:37
$begingroup$
Do you mean $c^HA_ic=0$ by $c^*A_ic=0$? ($(cdot)^H$ is the Hermitian operator)
$endgroup$
– Mostafa Ayaz
Mar 19 at 23:18