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General Linear Group of a Vector Space

by Master J
Tags: linear, space, vector
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Master J
#1
Feb21-12, 10:46 AM
P: 225
The general linear group of a vector space GL(V) is the group who's set is the set of all linear maps from V to V that are invertible (automorphisms).


My question is, why is this a group? Surely the zero operator that sends all vectors in V to the zero vector is not invertible? But isn't it part of the definition of a group that an inverse exists for all elements?

My book tells me that it is not a vector space because of the above, but I can't see how it is even a group!

Can someone clear this up?
Thanks.
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micromass
#2
Feb21-12, 10:53 AM
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The zero operator doesn't have an inverse (except if the vector space is trivial), so it isn't an element of GL(V).

The only elements of GL(V) are the invertible linear maps. The zero operator isn't in the group. The identity of GL(V) is the map [itex]V\rightarrow V:x\rightarrow x[/itex].
HallsofIvy
#3
Feb21-12, 12:06 PM
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Also note that the operation involved here is composition of functions (or, if you write the automorphisms as matrices, multiplication of matrices) so the identity is, as micromass says, the identity function (or the identity matrix).

Master J
#4
Feb21-12, 03:57 PM
P: 225
General Linear Group of a Vector Space

I know that it cannot be, but I'm not seeing the reason clearly.

Why can we not consider the set of GL(V) as a vector space? The field of the vector space only requires an inverse for its non-zero elements (this is part of the definition of a field), so we could still include the zero operator even though it has no inverse, no?
morphism
#5
Feb21-12, 04:03 PM
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Quote Quote by Master J View Post
I know that it cannot be, but I'm not seeing the reason clearly.
Read HallsofIvy's post and make sure you understand it.
Why can we not consider the set of GL(V) as a vector space?
What would "vector addition" be in this case?
Master J
#6
Feb21-12, 04:10 PM
P: 225
I understand the post all right...but I think I'm confusing myself!

Well, could we define vector addition as the sum of two operations ie. for two linea transformations on V, say T and S, and u E V, we could have

(T + S)u = Tu + Su

Would this not make sense, adding two transformed vectors?
morphism
#7
Feb21-12, 04:15 PM
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But S+T doesn't have to be invertible, even if S and T are! So your addition doesn't really define an operation on GL(V).

The point here is that the group operation * on GL(V) isn't addition of operators: it's composition of operators. So the "zero" element of this group operation is the identity operator I.
Master J
#8
Feb21-12, 04:42 PM
P: 225
Could you perhaps demonstrate how S + T may not be invertible even if S and T are?
morphism
#9
Feb21-12, 04:46 PM
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Sure. Let S be your favorite invertible operator and take T=-S.

Another (perhaps less obtuse) example:
[tex]\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix}.[/tex]
Deveno
#10
Feb21-12, 04:49 PM
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Quote Quote by Master J View Post
Could you perhaps demonstrate how S + T may not be invertible even if S and T are?
let S = I, the nxn identity matrix. this is invertible, with inverse I (itself).

let T = -I = (-1)I, the nxn matrix with all -1's on the diagonal. this matrix, too is invertible, with inverse: (-I)-1 = -I:

(-I)(-Iv) = -I(-v) = -(-v) = v.

however, S+T = 0, the 0-matrix, which has no inverse.

EDIT: the point being, with matrices, we have TWO operations, an addition, and a multiplication. under addition, everything is hunky-dory, and we can make the set of nxn matrices (or even mxn matrices) into a vector space, no problem.

but Mat(n,F) is only a ring, not a field, it has "singular elements" (non-invertible matrices). in fact, it's not even a very GOOD ring, because it has "zero divisors", we can have AB = 0, even if neither A nor B is 0.

there is good reason for restricting our attention to just invertible linear transformations. non-invertible transformations "change the dimension of the space they transform", that is, "they lose information". invertible linear transformations, however, are "faithful" and can be thought of as "a suitable change of coordinates (i.e., a change of basis)". this is of practical importance in studying physical things, as objects don't change their dimensionality in most physical processes.
Norwegian
#11
Feb21-12, 06:20 PM
P: 144
The object that Master J is searching for here, is called the Endomorphism ring of V, End(V), consisting of all linear maps V-->V, with the operations indicated. Although the above poster doesnt consider this a "very good ring", it is interesting enough to have its own name, to be a useful objects in many contexts, as well as an object of study in its own right, particular in more general settings.
Deveno
#12
Feb21-12, 07:54 PM
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Quote Quote by Norwegian View Post
The object that Master J is searching for here, is called the Endomorphism ring of V, End(V), consisting of all linear maps V-->V, with the operations indicated. Although the above poster doesnt consider this a "very good ring", it is interesting enough to have its own name, to be a useful objects in many contexts, as well as an object of study in its own right, particular in more general settings.
as is often the case with algebraic structures and teen idols, it's interesting because it's "bad".

endomorphism rings are "natural" ways to get rings. if we have an abelian group A, and two endomorphisms φ,ζ on A, we can define φ+ζ "point-wise":

(φ+ζ)(x) = φ(x) + ζ(x)

which turns End(A) into an abelian group itself. then, since every element of End(A) is a group homomorphism, for η,φ,ζ in End(A), and all x in A:

(η(φ+ζ))(x) = η((φ+ζ)(x)) = η(φ(x) + ζ(x)) = η(φ(x)) + η(ζ(x))

= (ηφ)(x) + (ηζ)(x) = (ηφ + ηζ)(x), that is:

η(φ+ζ) = ηφ + ηζ <---distributive law

in other words, we get the ring structure for free.

in particular, we can identify End(Z) (for the free cyclic group (Z,+)) with Z itself:

n <=> φn, where φn in End(Z) takes k→k+k+...+k (n times).

that is: the ring of integers is isomorphic to the ring of (abelian group, i.e., additive) endomorphisms of an infinite cyclic group (free on one letter).

that said, one is often more interested in the automorphism group, and:

GL(V) = Aut(V).

or again: in a given ring, often the group of units is of special interest (being the "nice toys" to play with).

it turns out that Aut(M), where M is a (insert structure type here), is a useful concept in many settings, as well. i think of it like this:

in any (abstract) setting, if we imagine maps (morphisms) of a certain kind as being "allowable actions", we can impose various kinds of structure on this collection, which is often illuminating. but for solving problems, we often need to consider "reversible actions", so that we don't get stuck in "one-way traps" (an easy elementary example is "squaring"...if we are only dealing with positive numbers, squaring is reversible. but if we are dealing with arbitrary real numbers, after we square, we often don't know "which answer we want").

it would be nice if we could turn every ring into a field, just by "making it bigger". this is what makes domains nicer than arbitrary rings. alas, life seems to have dealt us a hand with many hidden "gotchas".


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