Curl said:
Look at Exercise 8.3 on page 19. I got no idea how to do this, and actually I'm not even sure what it's asking
I scanned the notes. It's probably considered gauche to use linear algebra to talk about tensors, but by doing that, I formed an interpretation. I've changed his notation a little.
(This long post has many opportunities for errors. Check it!)
Three bases for the same vector space are [itex]{\bf e} = \{e_1,e_2,e_3 \}, {\bf f} = \{f_1,f_2,f_3\}[/itex] and [itex]{\bf g} = \{g_1, g_2, g_3\}[/itex].
Considering the elements of each basis to be a row vector, the change of coordinate formulae can be visualized as matrix multiplication.
Assume the following:
[tex]{\bf f} = {\bf e }{\bf S}[/tex]
[tex]{\bf e} = {\bf f }{\bf T }[/tex]
[tex]{\bf g} = {\bf f }{\bf P}[/tex]
[tex]{\bf g} = {\bf e }{\bf R}[/tex]
where each of [itex]{\bf S}, {\bf T}, {\bf P},{\bf R}[/itex] is a 3 by 3 matrix.
and [itex]{ \bf T } = { \bf S}^{-1}[/itex]
The convention for writing the matrix elements is illustrated by:
[itex]{\bf S } = \begin{pmatrix} S_1^1 & S_2^1 & S_3^1 \\ S_1^2 & S_2^2 & S_3^2 \\ S_1^3 & S_2^3 & S_3^3 \end{pmatrix}[/itex]
So my version of eq 5.7 is:
[tex]\begin{pmatrix} f_1 & f_2 & f_3 \end{pmatrix} = \begin{pmatrix} e_1 & e_2 & e_3 \end{pmatrix}\begin{pmatrix} S_1^1 & S_2^1 & S_3^1 \\ S_1^2 & S_2^2 & S_3^2 \\ S_1^3 & S_2^3 & S_3^3 \end{pmatrix}[/tex]
In order to be consistent the result of changing from basis [itex]{\bf e}[/itex] to basis [itex]{\bf g}[/itex] directly via the matrix [itex]{\bf R}[/itex] must produce the same result as changing from basis [itex]{\bf e }[/itex] to the basis [itex]{\bf f }[/itex] and then changing from the basis [itex]{\bf f }[/itex] to the basis [itex]{\bf g }[/itex]. This amounts to the equality:
[tex]{\bf e} {\bf R} = ( {\bf e}{\bf S} ){\bf P}[/tex]. For this to hold it is sufficient that [itex]{ \bf R } = {\bf S} {\bf P}[/itex] which we will assume.
The section that develops eq. 6.2 demonstrates the interesting fact that the coordinates [itex]{\bf x } = \{x_1,x_2,x_3\}[/itex] of a vector in basis [itex]{\bf e}[/itex] transform to its coordinates [itex]{\bf y}[/itex] in basis [itex]{\bf f}[/itex] by the rule:
[tex]{\bf y} = {\bf T} {\bf x}[/tex]
where the coordinates are written as column vectors. (This is contrary what we might naively expect - namely some equation involving the matrix [itex]{\bf S}[/itex] ).
The section with eq 8.1 hypothesizes that there is a quantity [itex]{\bf A }[/itex] that has a representation as 3 coordinates [itex]\{a_1, a_2, a_3\}[/itex] in basis [itex]{\bf e}[/itex]. Visualizing the coordinates as a row vector [itex]{\bf a }[/itex] , the rules for relating the coordinates for this type of quantity to its representation [itex]{\bf b}[/itex] in basis [itex]{\bf f}[/itex] are:
[tex]{\bf b } = {\bf a} {\bf S}[/tex]
[tex]{\bf a } = {\bf b} {\bf T}[/tex]
This representation is consistent with respect two ways of changing from basis [itex]{\bf e}[/itex] to basis [itex]g[/itex].
The first way is change from basis [itex]{\bf e }[/itex] to basis [itex]{\bf g}[/itex] directly by using matrix [itex]{\bf R}[/itex]. The second way is to change from basis [itex]{\bf e }[/itex] to basis [itex]{\bf f }[/itex] and then to change from basis [itex]{\bf f }[/itex] to basis [itex]{\bf g}[/itex]. This amounts to the equality:
[tex]{\bf a} {\bf R } = ({\bf a}{\bf S}){\bf P}[/tex]
Since we assumed above that [itex]{\bf R} = {\bf S}{\bf P}[/itex] , this equality holds.
My interpretation of exercise 8.3:
Let [itex]{\bf M }[/itex] be the matrix that transforms from basis [itex]{\bf v}[/itex] to basis [itex]{\bf w}[/itex]. Suppose there is a quantity [itex]{\bf A}[/itex] who coordinates ( as a row vector) [itex]{\bf b}[/itex] in basis [itex]{\bf w}[/itex] are given as a function of its coordinates [itex]{\bf a}[/itex] in basis [itex]{\bf v}[/itex] by:
[tex]{\bf b } = {\bf a} {\bf M}^{-1}[/tex]
Show the test of consistency may fail.
The test of consistency would be that this equality holds:
[tex]{\bf a}{\bf R}^{-1} = ( {\bf a {\bf S}^{-1}}) {\bf P}^{-1}[/tex]
Although we have assumed [itex]{\bf R} = {\bf S}{\bf P}[/itex], this does not imply what would be needed , which is [itex]{\bf R}^{-1} = {\bf S}^{-1}{\bf P}^{-1}[/itex], so we should be able to find a numerical example where the equality fails.