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When I first learned about 2d rotation matrices I read that you represented your point in your new coordinate system. That is you take the dot product of your vector in its current coordinate system and against the new i and j vector in the rotated coordinate system.

$\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}p=p^\prime$

The columns represent the new i and j vectors.

I'm reading a book and it says,

there can be confusion because their are two forms of the equation. One is through the transformation of the component of p (always with respect to x,y), x,y into x',y' and the other is through the transformation of the unit vectors i,j into i',j'.

He then shows the new equation

$\begin{bmatrix}i^\prime\\j^\prime\end{bmatrix} = \begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}\begin{bmatrix}i\\j\end{bmatrix}$

Anyway, my question is wouldn't you use the same matrix to change the vector i to i'? I don't see why the matrix would be the transpose. Additionally, if the vector i is on top of j then the vector is 4x1 rows by columns and you can't multiply 2x2 * 4x1. I'm quite confused.

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    @Hans: you might be $r$ight; it'$s$ just what I deduced f$r$om the excerpt. Anyway, you provided a good and complete answer.2011-05-02

2 Answers 2

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To begin with, in the second equation you shouldn't think of the vector $\mathbf{i}$ as $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$; as you say, that wouldn't make sense. That equation is just a symbolic way of writing the vector equations \mathbf{i}' = (\cos\theta) \mathbf{i} + (\sin\theta) \mathbf{j} and \mathbf{j}' = (-\sin\theta) \mathbf{i} + (\cos\theta) \mathbf{j}, which say that the new basis (\mathbf{i}',\mathbf{j}') is rotated by an angle $\theta$ (counterclockwise) relative to the old basis $(\mathbf{i},\mathbf{j})$.

As you'll see below, it's better to write it like this instead: \begin{bmatrix} \mathbf{i}' & \mathbf{j}' \end{bmatrix} = \begin{bmatrix} \mathbf{i} & \mathbf{j} \end{bmatrix} \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}.

Now for any vector $\mathbf{u}$, we can look at its coordinates with respect to the old basis, $\mathbf{u} = x \mathbf{i} + y \mathbf{j}$, or with respect to the new basis, \mathbf{u} = x' \mathbf{i}' + y' \mathbf{j}'. (Note: No prime on $\mathbf{u}$ in the second formula, since it's the same geometrical vector as in the first formula.) On matrix form, this can symbolically be written as \mathbf{u} = \begin{bmatrix} \mathbf{i} & \mathbf{j} \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \mathbf{i}' & \mathbf{j}' \end{bmatrix} \begin{bmatrix} x' \\ y' \end{bmatrix}. Here we replace the primed basis with the expression given by the formula above; this results in \begin{bmatrix} \mathbf{i} & \mathbf{j} \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \mathbf{i} & \mathbf{j} \end{bmatrix} \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} x' \\ y' \end{bmatrix}, hence (since coordinates with respect to a basis are unique) \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} x' \\ y' \end{bmatrix}, which is the relation between the coordinates of the vector $\mathbf{u}$ in the old and in the new basis, just as your textbook says.

As you yourself can testify, your book is also correct in its statement "there can be confusion"! ;-)

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    @Hans: I did make a mistake. I got [x,y] = [x',y']R and then wrote [x',y']=[x,y]R^T to get my final answer but that gives me the coordinates in the rotated frame rather then the coordinates in the original frame which is what I was looking for. Thank you! I'm glad I finally understand it. (phew!)2011-05-09
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A vector ($x$) is a combination of the magnitudes ($x_i, x_j$) and the base vectors of the coordinate system ($e_i$, $e_j$). When the coordinate system is transformed, the magnitude($x_i$, $x_j$) transform covariantly, while the base vector ($e_i$, $e_j$) transform contravariantly. This is needed in order to keep the vector invariant. The product of the covariant and the contravarient transformation matrices is an identity.