# Confusion on different representations of 2d rotation matrices

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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Two things: 1. $\begin{bmatrix}c&-s\\s&c\end{bmatrix}$ rotates anticlockwise 2. The inverse (the operation needed to undo the rotation) is identical to the transpose of the rotation matrix. – J. M. May 2 '11 at 8:29
People sometimes call the two phenomena 'active' and 'passive' transformations. If it helps, Wikipedia has an entry on this: en.wikipedia.org/wiki/Active_and_passive_transformation – Gerben May 2 '11 at 16:10
@Gerben: I don't think this is really the issue here (even though it's closely related). When talking about active transformation, one takes the transformation matrix times a column vector of coordinates (each of which is a scalar, of course). Here we instead have another type of construction, we're one (symbolically) multiplies a matrix by a "vector of vectors". – Hans Lundmark May 2 '11 at 18:03
@Hans: you might be right; it's just what I deduced from the excerpt. Anyway, you provided a good and complete answer. – Gerben May 2 '11 at 21:48

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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I'm confused still, in the first equation you have $\begin{bmatrix}i^\prime&&j^\prime\end{bmatrix} = \begin{bmatrix}i&&j\end{bmatrix}\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta‌​&\cos\theta\end{bmatrix}$ . This seems to only work if i and j are column vectors. i and j if left multiplied would have to be row vectors but in your example they are transposed. Why is this the case? – coderdave May 2 '11 at 17:48
When reading these equations you should think of the vectors $\mathbf{i}$ and $\mathbf{j}$ only as symbols. Don't try to substitute row or column vectors in their place, that doesn't make sense. This is not really the standard matrix product, but rather a convenient notational shorthand, where one writes a "row vector" in which each coordinate is actually a geometric vector instead of a scalar. (This notation by definition simply means what you get if you just multiply it out like you would do if all entries were scalars.) – Hans Lundmark May 2 '11 at 17:57
@Hans: you say that '(Note: No prime on u in the second formula, since it's the same geometrical vector as in the first formula.)' I don't see how u would be the same geometrical vector. x transforms to x' because the basis i,j are rotated counterclockwise to i'j'. I don't see how the two u are equivalent. – coderdave May 3 '11 at 15:53
Let's see if this explanation helps: First just draw an arrow on a blank piece of white paper. That represents your geometrical vector $\mathbf{u}$. Since it just sits there in the middle of nowhere on a white paper (no grid or anything), it's not meaningful to speak of "the coordinates of $\mathbf{u}$." In order to do that we have to introduce a coordinate system. So draw a coordinate system (two orthogonal axes, a square grid, etc.) on a transparent paper and put that on top of the white paper so that the tail of the vector $\mathbf{u}$ is at the origin of coordinates. (Cont.) – Hans Lundmark May 3 '11 at 16:29
Yes, it is arbitrary. I simply followed the convention that COordinates should be written as COlumns (this agrees with how one usually represents linear transformations: $m \times n$ matrix times $n \times 1$ column matrix). That requires me to write $\begin{bmatrix} \mathbf{i} & \mathbf{j} \end{bmatrix}$ as a row matrix. But the final relation between $(x,y)$ and $(x',y')$ can't depend on the book-keeping of the computations, so if you get the wrong sign, it must be because you made a mistake somewhere. – Hans Lundmark May 9 '11 at 6:43

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.

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