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Thread: Linear Transformations and the General Linear Group

  1. #1
    Super Member Bernhard's Avatar
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    Linear Transformations and the General Linear Group

    I am reading Kristopher Tapp's book "Matix Groups for Undergraduates".

    [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)

    ==================================================

    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
    Last edited by Bernhard; Dec 25th 2011 at 07:53 PM.
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  2. #2
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    Opalg's Avatar
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    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 $\displaystyle R_A\circ R_B = R_{BA}.$
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  3. #3
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    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).
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