Matrix for transforming vector components under rotation

[saadhusayn]

Say we have a matrix $L$ that maps vector components from an unprimed basis to a rotated primed basis according to the rule $x'_{i} = L_{ij} x_{j}$. $x'_i$ is the $i$th component in the primed basis and $x_{j}$ the $j$ th component in the original unprimed basis. Now $x'_{i} = \overline{e}'_i. \overline{x} = \overline{e}'_i. \overline{e}_j x_{j}$. Hence $L_{ij} = \overline{e}'_i. \overline{e}_j$. Thus the matrix equation relating the primed co-ordinate system to the unprimed one in $\mathbb{R}^3$ 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 $\overline{e}'_i$s and $\overline{e}_j$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 $\mathbf{A} = \overline{e}_1$ and $\mathbf{B} = \overline{e}_2$. It turns out that the matrix that transforms $\mathbf{A} \rightarrow \mathbf{A}'$ and $\mathbf{B} \rightarrow \mathbf{B}'$ 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.

 

[WWGD]

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?

 

[saadhusayn]

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,
$$\mathbf{A} = \begin{pmatrix} 1 \\<br /> 0 \\<br /> 0<br /> \end{pmatrix}$$

and in the rotated frame, the components of $\mathbf{A}$ are given by $$\mathbf{A'} = \begin{pmatrix} \frac{\sqrt{3}}{2} \\<br /> 0 \\<br /> \frac{1}{2}<br /> \end{pmatrix}$$

and

in the original frame, $$\mathbf{B} = \begin{pmatrix} 0 \\<br /> 1 \\<br /> 0<br /> \end{pmatrix}$$

and in the rotated frame, the components of $\mathbf{B}$ are given by $$\mathbf{B'} = \begin{pmatrix} -\frac{{1}}{2} \\<br /> 0 \\<br /> \frac{\sqrt{3}}{2}<br /> \end{pmatrix}$$

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 $\mathbf{L}$ such that $\mathbf{x'} = \mathbf{L} \mathbf{x}$. Now from tensor analysis, we know that the $L_{ij} = e'_i.e_j$ where $e'_i$ = ith basis vector in rotated frame and $e_j$ = jth basis vector in original frame. I let $e_1 = \mathbf{A}$, $e'_1 = \mathbf{A'}$, $e_2 = \mathbf{B}$ and $e'_2 = \mathbf{B'}$
And in the original frame, $$\mathbf{e_3} = \begin{pmatrix}<br /> 0 \\<br /> 0 \\<br /> 1<br /> \end{pmatrix}$$

And in the rotated frame,

$$\mathbf{e_3}' = \begin{pmatrix}<br /> a \\<br /> b \\<br /> c<br /> \end{pmatrix}$$I used the fact that $\mathbf{L}$ has a determinant of 1 and that it has orthonormal rows to find $a,b$ and $c$.

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

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