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Assume $\phi: \Bbb R^2 \rightarrow \Bbb R^2$ is an isometrie with $\det(\phi) = 1$. Then, $\phi$ is a rotation and its matrix can be denoted by

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So, I know that the unit circle is parameterized by the trigonometric functions $\sin$ and $\cos$, but for me, it's hard to imagine what this actually means when you try to interpret this in a geometrical way. Plus, why does this matrix has this specific form? It seems like it got out of nowhere. What are the actual results that I get by denoting a matrix in a way like this?

Julian
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  • Why a rotation can be given by that matrix of sines and cosines is a very different question from why an isometry with determinant $1$ is a rotation. Are you asking both these questions, or just one of them? And in the latter case, which one? – Arthur Oct 05 '16 at 09:31
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    The first one. Why an isometry with determinant 1 is a rotation is explained in the proof, but it doesn't tell me much about why the rotation matrix has this specific form. – Julian Oct 05 '16 at 09:33

3 Answers3

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There is a difficult, but direct way of proving this, and an easier, but slightly more opaque way. I will go for the more opaque.

A linear map is uniquely determined by where it sends your chosen basis vectors. In this case, your basis vectors are $(1,0)^T$ and $(0,1)^T$ (the standard ones). Given a linear map, its matrix representation is given the following way: the first column is where the transformation sends the first basis vector, the second column is where the second basis vector is sent, and so on.

In this case, the first basis vector, when rotated by $\theta$ counterclockwise, becomes $(\cos\theta, \sin\theta)^T$, so that's the first column. The second basis vector becomes $(-\sin\theta,\cos\theta)^T$, which is therefore the second column of the matrix.

The more direct route is taking a general vector $(a,b)^T$, figuring out where it ends up using trigonometry, and then reading the matrix from there. It's definitely doable, but in my opinion more work.

The link between the two methods is that $(a,b)^T=a(1,0)^T+b(0,1)^T$, and rotation is a linear transformation. So where $(a,b)^T$ ends up, and therefore the coefficients you read off are directly related to where the unit vectors end up. The $a$ in $(a,b)^T$ on the left-hand side only interacts with the first column of the transformation matrix, and that's the same as $(1,0)^T$ on the right-hand side. Same for the $b$ and the second column.

Arthur
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  • But why does it become the vectors that you mentioned when you rotate it counterclockwise? I guess my problem here is to understand how $\sin$ and $\cos$ do actually define a rotation. We never dove into that. – Julian Oct 05 '16 at 09:46
  • Draw a diagram! Take the first unit vector, rotate it by $\theta$, and draw where it ends up. Use trigonometry to divine the first and second coordinates of that new vector (this is the classic high school introduction application of trigonometry, it really is that basic). Do the same with the second basis vector. – Arthur Oct 05 '16 at 09:49
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Instead of saying that we are rotating the vector, let us view it as rotating the coordinate system.

Rotation of Coordinates

Consider the vector $\vec{r}$ in the figure. If we change coordinates from $x-y$ to $x'-y'$, the net effect is that the vector $\vec{r}$ is rotated anti-clockwise by $\theta$.

In the x-y coordinate, we can resolve $\vec{r}$ as $x\hat{i} + y \hat{j}$. Alternatively, we can say that $\vec{r}$ is sum of two vectors $x\hat{i}$ and $y \hat{j}$

Now, in the new-coordinate system $x'-y'$, we can resolve $x\hat{i}$ as -

$x\hat{i} = x\cos \theta \hat{i'} + x\sin \theta \hat{j'}$.

Similarly, we can resolve $y\hat{j}$ as

$y\hat{j} = -y\sin \theta \hat{i'} + y\cos \theta \hat{j'}$

$ \implies \vec{r} = x\hat{i} + y\hat{j} = (x\cos \theta - y\sin \theta) \hat{i'} + (x\sin \theta + y\cos \theta) \hat{j'}$

In matrix form, we can write it as $\begin {vmatrix} x'\\ y' \end{vmatrix} = \begin{vmatrix} \cos \theta && -\sin \theta \\ \sin \theta && \cos \theta \end{vmatrix} \cdot \begin {vmatrix} x\\ y \end{vmatrix}$

What is the advantage of using matrices in this way?

The advantage of this approach becomes clearer in 3-dimensions. In 2-d, if we just want to rotate a vector, complex numbers are easier than matrix multiplication. However, in 3-d, if we want to change co-ordinate systems, matrix multiplication is almost invariably our only option.

Pratyush Rathore
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  • "complex numbers are easier than matrix multiplication": can you substantiate that. I wouldn't see a big difference between $R_\theta v$ and $e^{i\theta}z$, and the expanded forms are identical. –  Oct 06 '16 at 09:51
  • Difficult to substantiate, it is more of an opinion. I will try to put in my line of thought. Experienced players find the two equal since, they do perform the exactly same task.

    But with young students, I have found that, ($\cos \theta + i\sin \theta$) is much easier to remember than the rotation matrix. A few days back, one of my students pointed out that it takes less effort to write it in the complex number form instead of viewing it as matrix.

     – Pratyush Rathore Oct 06 '16 at 10:57
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A rotation preserves the angles and preserves the lengths.

So the images of two perpendicular vectors, such as $(1,0)$ and $(0,1)$, which are transformed to the two columns of the array, have a null dot product.

So if a column is $(x,y)$, the other must be $(y,-x)$ to a constant factor. In addition, the distance remains $1$ if and only if $x^2+y^2=1$.

The vectors $(\cos\theta,\sin\theta)$ and $(-\sin\theta,\cos\theta)$ have precisely these properties, and an easy geometric interpretation: $\theta$ is just the rotation angle.