Why isn't the set {M_{mxn}|m, n ℕ }of isomorphism class representative?
( M_{mxn}= set of matrices)
Thanks in advance
* set of class representatives, the real vector spaces , , etc.
* this material was posted here Linear Algebra/Dimension Characterizes Isomorphism - Wikibooks, open books for an open world
* appears as problem number 13
the trouble is, some "elements" (equivalence classes) are listed twice: for example the set of all 2x3 real matrices is isomorphic (as a vector space over R) to all 3x2 matrices (the isomorphism is the transpose map). for a collection of vector spaces to serve as a "class of representatives", we need one and only one from each isomorphism class. that is, the mapping:
NxN-->N
given by (m,n)-->mn is not 1-1 (it is onto, however).
that is (for example): M_{1x12}, M_{2x6},M_{3x4},M_{4x3},M_{6x2},M_{12x1} are all isomorphic to R^{12}, so we have "multiple representatives of R^{12}".
the isomorphism classes of FINITE-dimensional vector spaces are characterized by a single facet: their dimension. for an indexed family of vector spaces {V_{i}: i in I} to serve as a collection of representatives, we need two things:
1) a bijection I <--> N
2) V_{i} ≅ V_{j} iff i = j (so that we have a bijection V_{i}<-->N).
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the entire point of the wikibooks article is this: dimension is the single most important piece of information about a (finite-dimensional) vector space. if i tell you i have a finite-dimensional vector space V over a field F, there's no need to ask "which one" THEY ARE ALL ISOMORPHIC. pick one you like. most people use F^{n} (n-tuples in F), but you can use polynomials over F of degree n-1 or less if you prefer, or formal linear combinations of any n objects you like.
for example, with n = 3 (and say F = R, for definiteness), it doesn't matter if your basis is:
{red,green,blue}
{1,x,x^{2}}
{(1,0,0),(0,1,0),(0,0,1)}
{1, sin(x), cos(x)}
these all yield isomorphic vector spaces. that's why its important to understand that (a,b,c) does not define a vector. it means av_{1}+bv_{2}+cv_{3}, but first we have to have an agreement what the basis vectors are (so we know what the "coordinates" a,b and c MEAN). a lot of times linear algebra books "gloss over this", by using the standard basis for R^{3} almost exclusively (it becomes repetitious to say "(a,b,c) in the standard orthonormal basis for R^{3}").
in a sense ai+bj+ck is better, at least some dependence of the basis chosen is being indicated. triples of real numbers ARE NOT VECTORS, until we have decided which basis elements the coordinates "pair to". until then, they are just DATA. choosing a basis is the same thing as choosing "coordinate axes" (1-dimensional linearly independent subspaces of V). but: even a one-dimensional subspace U doesn't "uniquely" determine its basis, we have to decide on WHICH point u in U serves as "the unit distance".
in normal "euclidean space" we sort of cheat: we have geometric ideas of length and angle we use to define "unit lengths" and "perpendicular angles". these are encoded into an "inner product", which in euclidean spaces is the "dot-product". these ideas can be profitably generalized to other types of vector spaces...but vector spaces that are isomorphic as vector spaces may no longer be isomorphic as inner product spaces (an orthogonal basis in one space V, may no longer be orthogonal in that same space, under a different inner product). this is *additional* structure, and not part of the vector space definition.
matrix spaces are often treated as their corresponding "linear" vector space, it is common to "vectorize" matrices by identifying M_{mxn}(R) with R^{mn} (this can be thought of as lining up all the columns of a matrix A "head to toe"). surely you can see that both M_{2x3}(R) and M_{3x2}(R) both give us a 1x6 column this way, and so there's no clear choice for "which one" that should serve as the representative for a space of dimension 6.
for a collection of vector spaces to serve as a "class of representatives", we need one and only one from each isomorphism class. that is, the mapping:
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