# 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 $\mathbb{K}$ to denote one of { $\mathbb {R, C, H}$ where $\mathbb{H}$ is the set of all quaternions]

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

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

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

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

Definition 1.13

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

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

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

Proposition 1.14

${GL}_n$( $\mathbb{K}$) := {A $\in$ $M_n$( $\mathbb{K}$) | $R_A$ : $\mathbb{K}^n$ $\rightarrow$ $\mathbb{K}^n$ is a linear transformation}

Now the proof for this proposition starts as follows:

Proof:

If A $\in$ ${GL}_n$( $\mathbb{K}$) and B is such that BA = I then

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

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My questions are as follows:

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

How does this work since

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

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

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

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

[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 $\circ$ means composition of mappings. For all X in $\mathbb{K}^n$,

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

Therefore $R_A\circ R_B = R_{BA}.$
• Dec 26th 2011, 10:50 AM
Deveno
Re: Linear Transformations and the General Linear Group
$R_A(X)$ is "multiplication of the nxn matrix X on the right by the nxn matrix A". it is usually NOT the case that $R_{BA} = R_{AB}$, 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 $L_A$, which does not reverse the order of composition (there are parallel left- and right- constructions for any non-commutative operation).