
MHB Master
#1
February 26th, 2020,
23:37
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

February 26th, 2020 23:37
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MHB Oldtimer
#2
February 27th, 2020,
04:52
Originally Posted by
Peter
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 $f_{AB} = f_Bf_A$.

MHB Master
#3
March 1st, 2020,
00:19
Thread Author
Originally Posted by
Opalg
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