MHB Understanding Browder's Remarks on Linear Transformations

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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
 
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