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

#### Peter

##### Well-known member
MHB Site Helper
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

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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

#### Opalg

##### MHB Oldtimer
Staff member
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
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