Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.


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.

share|improve this question
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. –  Guess who it is. 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

2 Answers 2

up vote 3 down vote accepted

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"! ;-)

share|improve this answer
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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.