MHB Understanding Browder's Remarks on Linear Transformations

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Browder's remarks in "Mathematical Analysis: An Introduction" highlight that each linear transformation in the set $$\mathscr{L} ( \mathbb{R^n, R^m} )$$ can be represented by an $m \times n$ matrix, which contains $nm$ elements. By arranging these matrix elements into a single row, they can be viewed as coordinates in the Euclidean space $$\mathbb{R^{nm}}$$. This correspondence allows for the discussion of open sets and continuous functions within the context of linear transformations. Understanding this relationship is crucial for grasping the broader implications of linear algebra in mathematical analysis. The assignment of matrices to linear transformations thus provides a foundational link between algebraic structures and geometric interpretations.
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I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am reading Chapter 8: Differentiable Maps ... ... and am currently focused on Section 8.1 Linear Algebra ... ...

I need some help in order to fully understand some remarks by Browder in Section 8.1, page 179 regarding the set of all linear transformations, $$\mathscr{L} ( \mathbb{R^n, R^m} )$$ ... ...


The relevant statements by Browder follow Definition 6.10 and read as follows:
View attachment 9363In the above text from Browder, we read the following:

" ... ... The assignment of a matrix to each linear transformation enables us to regard each element of $$\mathscr{L} ( \mathbb{R^n, R^m} )$$ as a point of Euclidean space $$\mathbb{R^{nm} }$$, and thus we can speak of open sets in $$\mathscr{L} ( \mathbb{R^n, R^m} )$$, of continuous functions of linear transformations, etc. ... ... "
My question is as follows:Can someone please explain, in some detail, how/why exactly the assignment of a matrix to each linear transformation enables us to regard each element of $$\mathscr{L} ( \mathbb{R^n, R^m} )$$ as a point of Euclidean space $$\mathbb{R^{nm} }$$ ... ...
Help will be much appreciated ...

Peter
 

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Peter said:
Can someone please explain, in some detail, how/why exactly the assignment of a matrix to each linear transformation enables us to regard each element of $$\mathscr{L} ( \mathbb{R^n, R^m} )$$ as a point of Euclidean space $$\mathbb{R^{nm} }$$ ... ...
A linear transformation in $$\mathscr{L} ( \mathbb{R^n, R^m} )$$ is specified by an $m\times n$ matrix, which consists of $nm$ elements. If you string out those elements into a single row, they form the coordinates of a point in $$\mathbb{R^{nm} }$$.
 
Opalg said:
A linear transformation in $$\mathscr{L} ( \mathbb{R^n, R^m} )$$ is specified by an $m\times n$ matrix, which consists of $nm$ elements. If you string out those elements into a single row, they form the coordinates of a point in $$\mathbb{R^{nm} }$$.

Thanks for the help, Opalg ...

Peter
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

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