Matrices of Linear Transformations .... Poole, Example 6.76 ....

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I am reading David Poole's book: "Linear Algebra: A Modern Introduction" (Third Edition) and am currently focused on Section 6.6: The Matrix of a Linear Transformation ... ...

I need some help in order to fully understand Example 6.76 ... ...

Example 6.76 reads as follows:View attachment 8770
View attachment 8771
My question or issue of concern is as follows:When we calculate transformation outputs from inputs using $$T \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} x - 2y \\ x + y - 3z \end{bmatrix}$$

... it appears, as if by default, that we are using $$\{ e_1, e_2 , e_3 \}$$ and $$\{ e_1, e_2 \}$$ as bases for $$\mathbb{R}^3$$ and $$\mathbb{R}^2$$ respectively ...

... BUT ...Poole states that the bases are $$\mathcal{B} = \{ e_1, e_2 , e_3 \}$$ and $$\mathcal{C} = \{ e_2, e_1 \}$$... ?So, in the example, it seems that $$T$$ is defined in terms of $$\{ e_1, e_2 , e_3 \}$$ and $$\{ e_1, e_2 \}$$ ... and then we recalculate to find the matrix of $$T$$ with respect to $$\mathcal{B}$$ and $$\mathcal{C}$$ ... ...Can someone explain what is going on here ... shouldn't $$T$$ be defined in terms of the declared bases $$\mathcal{B}$$ and $$\mathcal{C}$$ ... so that $$T$$ takes an input in terms of $$\mathcal{B}$$, and then give an output in terms of $$\mathcal{C}$$ ... ...Hope someone can clarify the above issue ...

Help will be appreciated ..

Peter
 

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  • Poole - 1 -  Example 6.76 ... ...  PART 1 ... .png
    Poole - 1 - Example 6.76 ... ... PART 1 ... .png
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  • Poole - 2 -  Example 6.76 ... ...  PART 2 ... .png
    Poole - 2 - Example 6.76 ... ... PART 2 ... .png
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Note that $\mathcal C$ has the standard unit vectors swapped.
So it is non-standard.
And that's what we see in the matrix of the end result as well. That is, the rows are swapped.
 
Klaas van Aarsen said:
Note that $\mathcal C$ has the standard unit vectors swapped.
So it is non-standard.
And that's what we see in the matrix of the end result as well. That is, the rows are swapped.
Thanks Klaas ...

Peter
 

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