I 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 ith 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}'_is and \overline{e}_js 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&#039;} = \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&#039;} = \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&#039;} = \mathbf{L} \mathbf{x}. Now from tensor analysis, we know that the L_{ij} = e&#039;_i.e_j where e&#039;_i = ith basis vector in rotated frame and e_j = jth basis vector in original frame. I let e_1 = \mathbf{A}, e&#039;_1 = \mathbf{A&#039;}, e_2 = \mathbf{B} and e&#039;_2 = \mathbf{B&#039;}
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}&#039; = \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}&#039;_1. \overline{e}_1 &amp; \overline{e}&#039;_1. \overline{e}_2 &amp; \overline{e}&#039;_1. \overline{e}_3\\ \overline{e}&#039;_2. \overline{e}_1 &amp; \overline{e}&#039;_2. \overline{e}_2 &amp; \overline{e}&#039;_2. \overline{e}_1 \\ \overline{e}&#039;_3. \overline{e}_1 &amp; \overline{e}&#039;_3. \overline{e}_2 &amp; \overline{e}&#039;_3. \overline{e}_3 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} \sqrt{3} &amp; 0 &amp; 1 \\ -1 &amp; 0 &amp; \sqrt{3}\\ 0 &amp; -1 &amp; 0\end{pmatrix}

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