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Matrices and Linear Transformations ... Armstrong, Ch. 9 and Tapp Ch. 1, Section 5 ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,881
Hobart, Tasmania
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:


Armstrong - Start of Chapter 9 ... .png



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:


Tapp - Chater 5, Section 1... .png


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
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,647
Leeds, UK
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 $f_{AB} = f_Bf_A$.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,881
Hobart, Tasmania
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