Linear Transformations & Matrices: Armstrong, Tapp Chs. 9 & 1 - Explained

[Math Amateur]

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:View attachment 9568
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:View attachment 9569Note 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]

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 $f_{AB} = f_Bf_A$.

 

[Math Amateur]

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