[math-fun] Symbolic null space solutions
Hi all, I'm not sure if Google's and my linear algebra education are failing me, but I've got a matrix problem that I have no idea how to solve. The basics go like this. I've got an n by (n-1) matrix Q = [ q[1,1] q[1,2] .... q[1,n] ] [ . ] [ . ] [ . ] [ q[n-1,1] .... q[n-1,n] ] I can guarantee that all (n-1)x(n-1) matrix minors are non-singular. I'm looking for the general solution of this matrix for a proof I'm working on. Any ideas how to go about doing this? Thanks, Scott
Work in n-dimensional space with basis e1, ... , en. You are given n-1 vectors, the rows of your matrix, by hypothesis linearly independent. The 1-dimensional null space is generated by a vector orthogonal to these. Use the n-dimensional generalization of the cross product. Append to your matrix the row [e1, ... , en]. Now you have an nxn matrix. Calculate its determinant, and express it as A1 e1 + A2 e2 + ... + An en. Then [A1, ... , An] is a basis for the null space. -- Gene
________________________________ From: Scott Fenton <sctfen@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Saturday, July 13, 2013 4:48 PM Subject: [math-fun] Symbolic null space solutions
Hi all,
I'm not sure if Google's and my linear algebra education are failing me, but I've got a matrix problem that I have no idea how to solve. The basics go like this. I've got an n by (n-1) matrix
Q = [ q[1,1] q[1,2] .... q[1,n] ] [ . ] [ . ] [ . ] [ q[n-1,1] .... q[n-1,n] ]
I can guarantee that all (n-1)x(n-1) matrix minors are non-singular. I'm looking for the general solution of this matrix for a proof I'm working on. Any ideas how to go about doing this?
Thanks, Scott _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
If you have access to a CAS, use the nullspace() facility provided; but beware of confusion between the matrix and its transpose! It's worthwhile understanding what nullspace() does, since it's much easier to use in a progam than fancy "equation solvers". WFL On 7/14/13, Eugene Salamin <gene_salamin@yahoo.com> wrote:
Work in n-dimensional space with basis e1, ... , en. You are given n-1 vectors, the rows of your matrix, by hypothesis linearly independent. The 1-dimensional null space is generated by a vector orthogonal to these. Use the n-dimensional generalization of the cross product. Append to your matrix the row [e1, ... , en]. Now you have an nxn matrix. Calculate its determinant, and express it as A1 e1 + A2 e2 + ... + An en. Then [A1, ... , An] is a basis for the null space.
-- Gene
________________________________ From: Scott Fenton <sctfen@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Saturday, July 13, 2013 4:48 PM Subject: [math-fun] Symbolic null space solutions
Hi all,
I'm not sure if Google's and my linear algebra education are failing me, but I've got a matrix problem that I have no idea how to solve. The basics go like this. I've got an n by (n-1) matrix
Q = [ q[1,1] q[1,2] .... q[1,n] ] [ . ] [ . ] [ . ] [ q[n-1,1] .... q[n-1,n] ]
I can guarantee that all (n-1)x(n-1) matrix minors are non-singular. I'm looking for the general solution of this matrix for a proof I'm working on. Any ideas how to go about doing this?
Thanks, Scott _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I mean, for my purposes, I'm trying to use Metamath (http://us.metamath.org) to prove the general form of Tarski's upper dimensional axiom for geometry. There's no CAS to work with here, since I'm doing it in the NxN case. Thank you all for the solution, but now I really need the proof that it's necessary. That is, given a vector in that null space, I've got to show that it's a scalar multiple of the vector from the whole determinant process. Any ideas? Again, my basic linear algebra has failed me. -Scott On Sat, Jul 13, 2013 at 8:41 PM, Fred lunnon <fred.lunnon@gmail.com> wrote:
If you have access to a CAS, use the nullspace() facility provided; but beware of confusion between the matrix and its transpose!
It's worthwhile understanding what nullspace() does, since it's much easier to use in a progam than fancy "equation solvers".
WFL
On 7/14/13, Eugene Salamin <gene_salamin@yahoo.com> wrote:
Work in n-dimensional space with basis e1, ... , en. You are given n-1 vectors, the rows of your matrix, by hypothesis linearly independent. The 1-dimensional null space is generated by a vector orthogonal to these. Use the n-dimensional generalization of the cross product. Append to your matrix the row [e1, ... , en]. Now you have an nxn matrix. Calculate its determinant, and express it as A1 e1 + A2 e2 + ... + An en. Then [A1, ... , An] is a basis for the null space.
-- Gene
________________________________ From: Scott Fenton <sctfen@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Saturday, July 13, 2013 4:48 PM Subject: [math-fun] Symbolic null space solutions
Hi all,
I'm not sure if Google's and my linear algebra education are failing me, but I've got a matrix problem that I have no idea how to solve. The basics go like this. I've got an n by (n-1) matrix
Q = [ q[1,1] q[1,2] .... q[1,n] ] [ . ] [ . ] [ . ] [ q[n-1,1] .... q[n-1,n] ]
I can guarantee that all (n-1)x(n-1) matrix minors are non-singular. I'm looking for the general solution of this matrix for a proof I'm working on. Any ideas how to go about doing this?
Thanks, Scott _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Given any vector in the null space, construct an orthogonal basis comprising that vector together with N-1 orthogonal to it (via Gram-Schmidt algorithm). Your matrix is now the unit matrix missing one row, so the result is immediate. This argument relies on your geometry being Euclidean: it would fail in a more general setting given nonzero null vectors, or algebraic incompleteness. WFL On 7/14/13, Scott Fenton <sctfen@gmail.com> wrote:
I mean, for my purposes, I'm trying to use Metamath (http://us.metamath.org) to prove the general form of Tarski's upper dimensional axiom for geometry. There's no CAS to work with here, since I'm doing it in the NxN case.
Thank you all for the solution, but now I really need the proof that it's necessary. That is, given a vector in that null space, I've got to show that it's a scalar multiple of the vector from the whole determinant process. Any ideas? Again, my basic linear algebra has failed me.
-Scott
On Sat, Jul 13, 2013 at 8:41 PM, Fred lunnon <fred.lunnon@gmail.com> wrote:
If you have access to a CAS, use the nullspace() facility provided; but beware of confusion between the matrix and its transpose!
It's worthwhile understanding what nullspace() does, since it's much easier to use in a progam than fancy "equation solvers".
WFL
On 7/14/13, Eugene Salamin <gene_salamin@yahoo.com> wrote:
Work in n-dimensional space with basis e1, ... , en. You are given n-1 vectors, the rows of your matrix, by hypothesis linearly independent. The 1-dimensional null space is generated by a vector orthogonal to these. Use the n-dimensional generalization of the cross product. Append to your matrix the row [e1, ... , en]. Now you have an nxn matrix. Calculate its determinant, and express it as A1 e1 + A2 e2 + ... + An en. Then [A1, ... , An] is a basis for the null space.
-- Gene
________________________________ From: Scott Fenton <sctfen@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Saturday, July 13, 2013 4:48 PM Subject: [math-fun] Symbolic null space solutions
Hi all,
I'm not sure if Google's and my linear algebra education are failing me, but I've got a matrix problem that I have no idea how to solve. The basics go like this. I've got an n by (n-1) matrix
Q = [ q[1,1] q[1,2] .... q[1,n] ] [ . ] [ . ] [ . ] [ q[n-1,1] .... q[n-1,n] ]
I can guarantee that all (n-1)x(n-1) matrix minors are non-singular. I'm looking for the general solution of this matrix for a proof I'm working on. Any ideas how to go about doing this?
Thanks, Scott _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Cool. Thank you for your help. On Mon, Jul 15, 2013 at 2:31 AM, Fred lunnon <fred.lunnon@gmail.com> wrote:
Given any vector in the null space, construct an orthogonal basis comprising that vector together with N-1 orthogonal to it (via Gram-Schmidt algorithm). Your matrix is now the unit matrix missing one row, so the result is immediate.
This argument relies on your geometry being Euclidean: it would fail in a more general setting given nonzero null vectors, or algebraic incompleteness.
WFL
On 7/14/13, Scott Fenton <sctfen@gmail.com> wrote:
I mean, for my purposes, I'm trying to use Metamath (http://us.metamath.org) to prove the general form of Tarski's upper dimensional axiom for geometry. There's no CAS to work with here, since I'm doing it in the NxN case.
Thank you all for the solution, but now I really need the proof that it's necessary. That is, given a vector in that null space, I've got to show that it's a scalar multiple of the vector from the whole determinant process. Any ideas? Again, my basic linear algebra has failed me.
-Scott
On Sat, Jul 13, 2013 at 8:41 PM, Fred lunnon <fred.lunnon@gmail.com> wrote:
If you have access to a CAS, use the nullspace() facility provided; but beware of confusion between the matrix and its transpose!
It's worthwhile understanding what nullspace() does, since it's much easier to use in a progam than fancy "equation solvers".
WFL
On 7/14/13, Eugene Salamin <gene_salamin@yahoo.com> wrote:
Work in n-dimensional space with basis e1, ... , en. You are given n-1 vectors, the rows of your matrix, by hypothesis linearly independent. The 1-dimensional null space is generated by a vector orthogonal to these. Use the n-dimensional generalization of the cross product. Append to your matrix the row [e1, ... , en]. Now you have an nxn matrix. Calculate its determinant, and express it as A1 e1 + A2 e2 + ... + An en. Then [A1, ... , An] is a basis for the null space.
-- Gene
________________________________ From: Scott Fenton <sctfen@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Saturday, July 13, 2013 4:48 PM Subject: [math-fun] Symbolic null space solutions
Hi all,
I'm not sure if Google's and my linear algebra education are failing me, but I've got a matrix problem that I have no idea how to solve. The basics go like this. I've got an n by (n-1) matrix
Q = [ q[1,1] q[1,2] .... q[1,n] ] [ . ] [ . ] [ . ] [ q[n-1,1] .... q[n-1,n] ]
I can guarantee that all (n-1)x(n-1) matrix minors are non-singular. I'm looking for the general solution of this matrix for a proof I'm working on. Any ideas how to go about doing this?
Thanks, Scott _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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participants (3)
-
Eugene Salamin -
Fred lunnon -
Scott Fenton