I Matrix for transforming vector components under rotation

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0
Say we have a matrix [itex]L[/itex] that maps vector components from an unprimed basis to a rotated primed basis according to the rule [itex]x'_{i} = L_{ij} x_{j}[/itex]. [itex]x'_i[/itex] is the [itex]i[/itex]th component in the primed basis and [itex]x_{j}[/itex] the [itex]j[/itex] th component in the original unprimed basis. Now [itex]x'_{i} = \overline{e}'_i. \overline{x} = \overline{e}'_i. \overline{e}_j x_{j} [/itex]. Hence [itex]L_{ij} = \overline{e}'_i. \overline{e}_j[/itex]. Thus the matrix equation relating the primed co-ordinate system to the unprimed one in [itex]\mathbb{R}^3[/itex] is

$$ \begin{pmatrix}x'_{1}\\ x'_{2}\\ x'_{3} \end{pmatrix} = \begin{pmatrix} \overline{e}'_1. \overline{e}_1 & \overline{e}'_1. \overline{e}_2 & \overline{e}'_1. \overline{e}_3\\ \overline{e}'_2. \overline{e}_1 & \overline{e}'_2. \overline{e}_2 & \overline{e}'_2. \overline{e}_1 \\ \overline{e}'_3. \overline{e}_1 & \overline{e}'_3. \overline{e}_2 & \overline{e}'_3. \overline{e}_3 \end{pmatrix}\begin{pmatrix}x_{1}\\ x_{2}\\ x_{3} \end{pmatrix} $$

Where the [itex]\overline{e}'_i[/itex]s and [itex]\overline{e}_j[/itex]s are unit basis vectors in the primed and unprimed co ordinate systems respectively.

Now I tried to apply the above idea to the following situation (Riley, Hobson and Bence Chapter 26, Problem 2).

I took [itex]\mathbf{A} = \overline{e}_1[/itex] and [itex]\mathbf{B} = \overline{e}_2[/itex]. It turns out that the matrix that transforms [itex]\mathbf{A} \rightarrow \mathbf{A}' [/itex] and [itex]\mathbf{B} \rightarrow \mathbf{B}' [/itex] is not the matrix that transforms the unprimed components to the primed components (that I used above) but the INVERSE (or transpose) of that matrix. I need to know where I am going wrong here. Thank you.
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WWGD

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I am not sure I understand your question, but if you want to go in the opposite direction, you use the inverse matrix. Was that the question?
 
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I am not sure I understand your question, but if you want to go in the opposite direction, you use the inverse matrix. Was that the question?
In the original frame,
[tex]\mathbf{A} = \begin{pmatrix} 1 \\
0 \\
0
\end{pmatrix}[/tex]

and in the rotated frame, the components of [itex] \mathbf{A}[/itex] are given by [tex] \mathbf{A'} = \begin{pmatrix} \frac{\sqrt{3}}{2} \\
0 \\
\frac{1}{2}
\end{pmatrix}[/tex]

and

in the original frame, [tex]\mathbf{B} = \begin{pmatrix} 0 \\
1 \\
0
\end{pmatrix}[/tex]

and in the rotated frame, the components of [itex] \mathbf{B}[/itex] are given by [tex] \mathbf{B'} = \begin{pmatrix} -\frac{{1}}{2} \\
0 \\
\frac{\sqrt{3}}{2}
\end{pmatrix}[/tex]

I want to find the matrix that gives you the components of any vector in the rotated frame if you have the components in the original frame. In other words, I want the matrix [itex] \mathbf{L}[/itex] such that [itex] \mathbf{x'} = \mathbf{L} \mathbf{x}[/itex]. Now from tensor analysis, we know that the [itex] L_{ij} = e'_i.e_j [/itex] where [itex]e'_i [/itex] = ith basis vector in rotated frame and [itex]e_j [/itex] = jth basis vector in original frame. I let [itex] e_1 = \mathbf{A} [/itex], [itex] e'_1 = \mathbf{A'}[/itex], [itex]e_2 = \mathbf{B}[/itex] and [itex]e'_2 = \mathbf{B'}[/itex]
And in the original frame, [tex]\mathbf{e_3} = \begin{pmatrix}
0 \\
0 \\
1
\end{pmatrix}[/tex]

And in the rotated frame,

[tex]\mathbf{e_3}' = \begin{pmatrix}
a \\
b \\
c
\end{pmatrix}[/tex]


I used the fact that [itex]\mathbf{L}[/itex] has a determinant of 1 and that it has orthonormal rows to find [itex]a,b [/itex] and [itex] c [/itex].

[tex] \mathbf{L} = \begin{pmatrix} \overline{e}'_1. \overline{e}_1 & \overline{e}'_1. \overline{e}_2 & \overline{e}'_1. \overline{e}_3\\ \overline{e}'_2. \overline{e}_1 & \overline{e}'_2. \overline{e}_2 & \overline{e}'_2. \overline{e}_1 \\ \overline{e}'_3. \overline{e}_1 & \overline{e}'_3. \overline{e}_2 & \overline{e}'_3. \overline{e}_3 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} \sqrt{3} & 0 & 1 \\ -1 & 0 & \sqrt{3}\\ 0 & -1 & 0\end{pmatrix} [/tex]

This, however, is not the correct matrix. It is the INVERSE of the correct matrix. Where am I going wrong?
 
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