How to interpret the listings of a rotation matrix?

Question

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

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?

Answer 1

Instead of saying that we are rotating the vector, let us view it as rotating the coordinate system.

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.

Answer 2

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.

Answer 3

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.