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  1. MHB Master
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    #1
    At the start of Chapter 9, M. A. Armstrong in his book, "Groups and Symmetry" (see text below) writes the following:

    " ... ... Each matrix $ \displaystyle A$ in this group determines an invertible linear transformation $ \displaystyle f_A: \mathbb{R} \to \mathbb{R}$ defined by $ \displaystyle f_A(x) = x A^t$ ... ... "


    I know that one may define entities how one wishes ... but why does Armstrong define $ \displaystyle f$ in terms of the transpose of $ \displaystyle A$ rather than just simply $ \displaystyle A$ ... there must be some reason or advantage to this ... but what is it? Can someone help to explain ...

    I note in passing that Kristopher Tapp in his book, "Matrix Groups for Undergraduates" (Chapter 1, Section 5) ... see text below ... defines the action of a linear transformation ( multiplication by a matrix $ \displaystyle A$) as $ \displaystyle R_A = X \cdot A$ ... thus not using the transpose of $ \displaystyle A$ ...


    Hope that someone can help ...

    Peter


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

    The above post refers to the start of Ch. 9 of M. A. Armstrong's book, "Groups and Symmetry" ... so I am providing the relevant text ... as follows:






    The above post also refers to Chapter 1, Section 5 of Kristopher Tapp's book, "Matrix Groups for Undergraduates" ... so I am providing the relevant text ... as follows:





    Note that Tapp uses $ \displaystyle \mathbb{K}$ to refer to one of the real numbers, the complex numbers or the quaternions ...


    Hope that helps,

    Peter

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    #2
    Quote Originally Posted by Peter View Post
    At the start of Chapter 9, M. A. Armstrong in his book, "Groups and Symmetry" (see text below) writes the following:

    " ... ... Each matrix $ \displaystyle A$ in this group determines an invertible linear transformation $ \displaystyle f_A: \mathbb{R} \to \mathbb{R}$ defined by $ \displaystyle f_A(x) = x A^t$ ... ... "


    I know that one may define entities how one wishes ... but why does Armstrong define $ \displaystyle f$ in terms of the transpose of $ \displaystyle A$ rather than just simply $ \displaystyle A$ ... there must be some reason or advantage to this ... but what is it?
    The answer is given in the text from Armstrong's book that you posted. If $f_A(x)$ is defined to be $xA^T$ then the map $A\mapsto f_A$ preserves multiplication: $f_{AB} = f_Af_B$. If the transpose does not occur in the definition then the map would reverse the order and you would get $f_{AB} = f_Bf_A$.

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    Quote Originally Posted by Opalg View Post
    The answer is given in the text from Armstrong's book that you posted. If $f_A(x)$ is defined to be $xA^T$ then the map $A\mapsto f_A$ preserves multiplication: $f_{AB} = f_Af_B$. If the transpose does not occur in the definition then the map would reverse the order and you would get $f_{AB} = f_Bf_A$.


    Thanks for the help, Opalg ...

    Peter

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