congrats, you just won the game! :D

nope the final ring is the ring of fractions of said ring, but you only quotient by elements with a non zero constant term, again the localization at the maximal ideal generated by all x_q

this email could have been a battle to the death

why does the amount of linear algebra any graduate math course expects you to know greatly exceed the amount of linear algebra any math course will ever teach you

I’d really like to see the answer to this

don't you have a trivial counterexample where x is the zero vector?

how about for all nonzero n dimensional vectors x

yeah that's what i'm working through now - i think it only works for n = 1. otherwise you always can find some vector x that forms a linearly dependent set. having trouble formalizing it tho, several years since i've written a linalg proof.

this shouuuld just be asking what n dimensional subspaces of the space of nxn matrices intersect trivially the hypersurface det(X)=0, for complex matrices it should like probably basically always do that (except for boring dimensions) , so rip, for odd dimensions and for matrices over R, the matrices X and -X will both be in a given n dimensional subspace, and taking a continuous path from X to -X, not crossing 0, since det(X)=-det(-X), eventually you'll end up w a matrix in your subspace w 0 determinant so no odds. Some candidates could be the dimensions 2^n, by starting in the 2x2 case w kaiasky's example and iteratively taking a matrix M in our choise of 2^n matrices and making the bigger matrices [M 0] [0 M] and [0 M][-M 0]. I suspect the fact that this would create a map from R^n x R^n->R^n which restricts to an isomorphism in {x}xR^n and in R^n x{x} is enough through alg topology magic to say the dimension should be 2^n, but i'll check tmrw i desperately need sleep rn

ok like you can't expand to 2^n like i thought, unfortunately if you take the determinant you get a polynomial on n variables which is greater *or equal* to zero and it indeed does "or equal" 0 at times. The last thingy should work, given how you get a map from P^n-1 x P^n-1 to P^n-1, you look at the cohomology rings over Z/2Z and you get that n has to be a power of two so that's nice. for n=4 a 4x4 matrix representation of the quaternions should work instead of my construction (wikipedia has one which seems to work) if the 8 case can be solved somehow through octonions it'd be great, not sure how since non associative algebras can't be really like, represented w real matrices. I've seen ppl mention k-theory, which i do not know, but regardless it'd be nice if it was possible to somehow get a division algebra from the existence of those n-matrices, so that the result could be at least come from a more well known result. Still gotta think how to get a n=8 solution to work if there is one, or how to quotient stuff the right way to get another cool topological map.

why don't the octonions give you the solution? representing multiplication by an octonion as a real matrix the obvious way should just work since you don't need associativity, just distributivity and to be able to freely commute reals (which you have), right?

oh shee ye i think you're right, as long as it's a division algebra it works fine, in fact, maybe the map R^n xR^n into R^n can just make R^n into a non associative division algebra directly? It's non associative in general, it would be a division algebra by hypothesis tho i think maybe possibly idk

Oh ye i think this should work, checking the axioms for an algebra is obvious, to show divisibility we need to solve uniquely a=b x for any non null a and b, but this is like ezy because the map {b}xR^n->R^n is an isomorphism because b as a matrix has determinant 0, so there is exactly one element x which works, and the other way round works as well cuz R^n x{ b}->R^n is also an isomorphism, since no matrix sends b to 0. (To reiterate better, i'm identifying the span of the n matrices assumed to exist w the first copy of R^n, while the 2nd copy of R^n is just the set of vectors they act upon, so i get a map R^nxR^n->R^n by applying a matrix on the left to an element on the right) So this problem is exactly asking for what n there is an n-dimensional division algebra over R, so the answer do be 1,2,4,8 in virtue of a well known result (albeit hard to prove)

also for finite fields there's a similar set of matrices for all n, since for a field of cardinality p^k you can just see the field of cardinality p^(k*n) as a vector space over our original field, which gives us a division algebra so it gives us the solution

why does the amount of linear algebra any graduate math course expects you to know greatly exceed the amount of linear algebra any math course will ever teach you

I’d really like to see the answer to this

don't you have a trivial counterexample where x is the zero vector?

how about for all nonzero n dimensional vectors x

yeah that's what i'm working through now - i think it only works for n = 1. otherwise you always can find some vector x that forms a linearly dependent set. having trouble formalizing it tho, several years since i've written a linalg proof.

this shouuuld just be asking what n dimensional subspaces of the space of nxn matrices intersect trivially the hypersurface det(X)=0, for complex matrices it should like probably basically always do that (except for boring dimensions) , so rip, for odd dimensions and for matrices over R, the matrices X and -X will both be in a given n dimensional subspace, and taking a continuous path from X to -X, not crossing 0, since det(X)=-det(-X), eventually you'll end up w a matrix in your subspace w 0 determinant so no odds. Some candidates could be the dimensions 2^n, by starting in the 2x2 case w kaiasky's example and iteratively taking a matrix M in our choise of 2^n matrices and making the bigger matrices [M 0] [0 M] and [0 M][-M 0]. I suspect the fact that this would create a map from R^n x R^n->R^n which restricts to an isomorphism in {x}xR^n and in R^n x{x} is enough through alg topology magic to say the dimension should be 2^n, but i'll check tmrw i desperately need sleep rn

ok like you can't expand to 2^n like i thought, unfortunately if you take the determinant you get a polynomial on n variables which is greater *or equal* to zero and it indeed does "or equal" 0 at times. The last thingy should work, given how you get a map from P^n-1 x P^n-1 to P^n-1, you look at the cohomology rings over Z/2Z and you get that n has to be a power of two so that's nice. for n=4 a 4x4 matrix representation of the quaternions should work instead of my construction (wikipedia has one which seems to work) if the 8 case can be solved somehow through octonions it'd be great, not sure how since non associative algebras can't be really like, represented w real matrices. I've seen ppl mention k-theory, which i do not know, but regardless it'd be nice if it was possible to somehow get a division algebra from the existence of those n-matrices, so that the result could be at least come from a more well known result. Still gotta think how to get a n=8 solution to work if there is one, or how to quotient stuff the right way to get another cool topological map.

why don't the octonions give you the solution? representing multiplication by an octonion as a real matrix the obvious way should just work since you don't need associativity, just distributivity and to be able to freely commute reals (which you have), right?

oh shee ye i think you're right, as long as it's a division algebra it works fine, in fact, maybe the map R^n xR^n into R^n can just make R^n into a non associative division algebra directly? It's non associative in general, it would be a division algebra by hypothesis tho i think maybe possibly idk

Oh ye i think this should work, checking the axioms for an algebra is obvious, to show divisibility we need to solve uniquely a=b x for any non null a and b, but this is like ezy because the map {b}xR^n->R^n is an isomorphism because b as a matrix has determinant 0, so there is exactly one element x which works, and the other way round works as well cuz R^n x{ b}->R^n is also an isomorphism, since no matrix sends b to 0. (To reiterate better, i'm identifying the span of the n matrices assumed to exist w the first copy of R^n, while the 2nd copy of R^n is just the set of vectors they act upon, so i get a map R^nxR^n->R^n by applying a matrix on the left to an element on the right) So this problem is exactly asking for what n there is an n-dimensional division algebra over R, so the answer do be 1,2,4,8 in virtue of a well known result (albeit hard to prove)

why does the amount of linear algebra any graduate math course expects you to know greatly exceed the amount of linear algebra any math course will ever teach you

I’d really like to see the answer to this

don't you have a trivial counterexample where x is the zero vector?

how about for all nonzero n dimensional vectors x

yeah that's what i'm working through now - i think it only works for n = 1. otherwise you always can find some vector x that forms a linearly dependent set. having trouble formalizing it tho, several years since i've written a linalg proof.

this shouuuld just be asking what n dimensional subspaces of the space of nxn matrices intersect trivially the hypersurface det(X)=0, for complex matrices it should like probably basically always do that (except for boring dimensions) , so rip, for odd dimensions and for matrices over R, the matrices X and -X will both be in a given n dimensional subspace, and taking a continuous path from X to -X, not crossing 0, since det(X)=-det(-X), eventually you'll end up w a matrix in your subspace w 0 determinant so no odds. Some candidates could be the dimensions 2^n, by starting in the 2x2 case w kaiasky's example and iteratively taking a matrix M in our choise of 2^n matrices and making the bigger matrices [M 0] [0 M] and [0 M][-M 0]. I suspect the fact that this would create a map from R^n x R^n->R^n which restricts to an isomorphism in {x}xR^n and in R^n x{x} is enough through alg topology magic to say the dimension should be 2^n, but i'll check tmrw i desperately need sleep rn

ok like you can't expand to 2^n like i thought, unfortunately if you take the determinant you get a polynomial on n variables which is greater *or equal* to zero and it indeed does "or equal" 0 at times. The last thingy should work, given how you get a map from P^n-1 x P^n-1 to P^n-1, you look at the cohomology rings over Z/2Z and you get that n has to be a power of two so that's nice. for n=4 a 4x4 matrix representation of the quaternions should work instead of my construction (wikipedia has one which seems to work) if the 8 case can be solved somehow through octonions it'd be great, not sure how since non associative algebras can't be really like, represented w real matrices. I've seen ppl mention k-theory, which i do not know, but regardless it'd be nice if it was possible to somehow get a division algebra from the existence of those n-matrices, so that the result could be at least come from a more well known result. Still gotta think how to get a n=8 solution to work if there is one, or how to quotient stuff the right way to get another cool topological map.

why don't the octonions give you the solution? representing multiplication by an octonion as a real matrix the obvious way should just work since you don't need associativity, just distributivity and to be able to freely commute reals (which you have), right?

oh shee ye i think you're right, as long as it's a division algebra it works fine, in fact, maybe the map R^n xR^n into R^n can just make R^n into a non associative division algebra directly? It's non associative in general, it would be a division algebra by hypothesis tho i think maybe possibly idk

why does the amount of linear algebra any graduate math course expects you to know greatly exceed the amount of linear algebra any math course will ever teach you

I’d really like to see the answer to this

don't you have a trivial counterexample where x is the zero vector?

how about for all nonzero n dimensional vectors x

yeah that's what i'm working through now - i think it only works for n = 1. otherwise you always can find some vector x that forms a linearly dependent set. having trouble formalizing it tho, several years since i've written a linalg proof.

this shouuuld just be asking what n dimensional subspaces of the space of nxn matrices intersect trivially the hypersurface det(X)=0, for complex matrices it should like probably basically always do that (except for boring dimensions) , so rip, for odd dimensions and for matrices over R, the matrices X and -X will both be in a given n dimensional subspace, and taking a continuous path from X to -X, not crossing 0, since det(X)=-det(-X), eventually you'll end up w a matrix in your subspace w 0 determinant so no odds. Some candidates could be the dimensions 2^n, by starting in the 2x2 case w kaiasky's example and iteratively taking a matrix M in our choise of 2^n matrices and making the bigger matrices [M 0] [0 M] and [0 M][-M 0]. I suspect the fact that this would create a map from R^n x R^n->R^n which restricts to an isomorphism in {x}xR^n and in R^n x{x} is enough through alg topology magic to say the dimension should be 2^n, but i'll check tmrw i desperately need sleep rn

ok like you can't expand to 2^n like i thought, unfortunately if you take the determinant you get a polynomial on n variables which is greater *or equal* to zero and it indeed does "or equal" 0 at times. The last thingy should work, given how you get a map from P^n-1 x P^n-1 to P^n-1, you look at the cohomology rings over Z/2Z and you get that n has to be a power of two so that's nice. for n=4 a 4x4 matrix representation of the quaternions should work instead of my construction (wikipedia has one which seems to work) if the 8 case can be solved somehow through octonions it'd be great, not sure how since non associative algebras can't be really like, represented w real matrices. I've seen ppl mention k-theory, which i do not know, but regardless it'd be nice if it was possible to somehow get a division algebra from the existence of those n-matrices, so that the result could be at least come from a more well known result. Still gotta think how to get a n=8 solution to work if there is one, or how to quotient stuff the right way to get another cool topological map.

why does the amount of linear algebra any graduate math course expects you to know greatly exceed the amount of linear algebra any math course will ever teach you

I’d really like to see the answer to this

don't you have a trivial counterexample where x is the zero vector?

how about for all nonzero n dimensional vectors x

yeah that's what i'm working through now - i think it only works for n = 1. otherwise you always can find some vector x that forms a linearly dependent set. having trouble formalizing it tho, several years since i've written a linalg proof.

this shouuuld just be asking what n dimensional subspaces of the space of nxn matrices intersect trivially the hypersurface det(X)=0, for complex matrices it should like probably basically always do that (except for boring dimensions) , so rip, for odd dimensions and for matrices over R, the matrices X and -X will both be in a given n dimensional subspace, and taking a continuous path from X to -X, not crossing 0, since det(X)=-det(-X), eventually you'll end up w a matrix in your subspace w 0 determinant so no odds. Some candidates could be the dimensions 2^n, by starting in the 2x2 case w kaiasky's example and iteratively taking a matrix M in our choise of 2^n matrices and making the bigger matrices [M 0] [0 M] and [0 M][-M 0]. I suspect the fact that this would create a map from R^n x R^n->R^n which restricts to an isomorphism in {x}xR^n and in R^n x{x} is enough through alg topology magic to say the dimension should be 2^n, but i'll check tmrw i desperately need sleep rn

glorious, wonderful, showstopping, breathtaking etc etc