# Linear Transformations and the General Linear Group

• Dec 25th 2011, 07:39 PM
Bernhard
Linear Transformations and the General Linear Group

[note that Tapp uses the terminology $\displaystyle \mathbb{K}$ to denote one of {$\displaystyle \mathbb {R, C, H}$ where $\displaystyle \mathbb{H}$ is the set of all quaternions]

In section 5. Matrices as Linear Transformations we find the following definition:

Definition 1.10. If A $\displaystyle \in$ $\displaystyle M_n$($\displaystyle \mathbb{K}$), define $\displaystyle R_A$: $\displaystyle \mathbb{K}^n$$\displaystyle \rightarrow \displaystyle \mathbb{K}^n such that for X\displaystyle \in$$\displaystyle \mathbb{K}^n$,

$\displaystyle R_A$(X) := X$\displaystyle \circ$ A

Then, in section 6. The general Linear Groups we find

Definition 1.13

The general linear group over $\displaystyle \mathbb{K}$ is

$\displaystyle {GL}_n$($\displaystyle \mathbb{K}$) := {A$\displaystyle \in$$\displaystyle M_n(\displaystyle \mathbb{K}) | there exists B\displaystyle \in$$\displaystyle M_n$($\displaystyle \mathbb{K}$) with AB = BA = I}

The following more visual characterization of the general linear group is often useful:

Proposition 1.14

$\displaystyle {GL}_n$($\displaystyle \mathbb{K}$) := {A$\displaystyle \in$$\displaystyle M_n(\displaystyle \mathbb{K}) | \displaystyle R_A : \displaystyle \mathbb{K}^n$$\displaystyle \rightarrow$$\displaystyle \mathbb{K}^n is a linear transformation} Now the proof for this proposition starts as follows: Proof: If A \displaystyle \in$$\displaystyle {GL}_n$($\displaystyle \mathbb{K}$) and B is such that BA = I then

$\displaystyle R_A$ $\displaystyle \circ$ $\displaystyle R_B$ $\displaystyle R_{BA}$ = $\displaystyle R_I$ = id (the identity)

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My questions are as follows:

How is $\displaystyle R_A$ $\displaystyle \circ$ $\displaystyle R_B$ = $\displaystyle R_{BA}$???

How does this work since

$\displaystyle R_A$(X) is a 1 x n matrix
$\displaystyle R_B$(X) is a 1 x n matrix

so how does $\displaystyle R_A$ $\displaystyle \circ$ $\displaystyle R_B$ does work in matrix multplication.

Futher, why does the author express the product as $\displaystyle R_{BA}$ rather than $\displaystyle R_{AB}$ {even though in this case they are equal, I am assuming that there is some point in showing the product as $\displaystyle R_{BA}$}

A final question is - what is the group operation $\displaystyle \circ$ in $\displaystyle R_A$ $\displaystyle \circ$ $\displaystyle R_B$

[I am assuming it is matrix multiplication??]

Can someone please clarify these points?

Peter
• Dec 26th 2011, 12:36 AM
Opalg
Re: Linear Transformations and the General Linear Group
The group operation $\displaystyle \circ$ means composition of mappings. For all X in $\displaystyle \mathbb{K}^n$,

$\displaystyle (R_A\circ R_B)(X) = R_A(R_B(X)) = R_A(XB) = (XB)A = X(BA) = R_{BA}(X).$

Therefore $\displaystyle R_A\circ R_B = R_{BA}.$
• Dec 26th 2011, 10:50 AM
Deveno
Re: Linear Transformations and the General Linear Group
$\displaystyle R_A(X)$ is "multiplication of the nxn matrix X on the right by the nxn matrix A". it is usually NOT the case that $\displaystyle R_{BA} = R_{AB}$, because in general, and even for invertible matrices, A and B do not commute. however, if B is a (two-sided) inverse for A, then A and B DO commute.

usually, we are used to seeing the map $\displaystyle L_A$, which does not reverse the order of composition (there are parallel left- and right- constructions for any non-commutative operation).