2D rotation

In two dimensions, the standard rotation matrix has the following form: $${\displaystyle R(\theta )={\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \\\end{bmatrix}}.}$$

This rotates column vectors by means of the following matrix multiplication, $${\displaystyle {\begin{bmatrix}x'\\y'\\\end{bmatrix}}={\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \\\end{bmatrix}}{\begin{bmatrix}x\\y\\\end{bmatrix}}.}$$

Thus, the new coordinates (x′, y′) of a point (x, y) after rotation are $${\displaystyle {\begin{aligned}x'&=x\cos \theta -y\sin \theta \,\\y'&=x\sin \theta +y\cos \theta \,\end{aligned}}.}$$

For example, when the vector (initially aligned with the x-axis of the Cartesian coordinate system) $${\displaystyle \mathbf {\hat {x}} ={\begin{bmatrix}1\\0\\\end{bmatrix}}}$$ is rotated by an angle θ, its new coordinates are $${\displaystyle {\begin{bmatrix}\cos \theta \\\sin \theta \\\end{bmatrix}},}$$

and when the vector (initially aligned with the y-axis of the coordinate system) $${\displaystyle \mathbf {\hat {y}} ={\begin{bmatrix}0\\1\\\end{bmatrix}}}$$ is rotated by an angle θ, its new coordinates are $${\displaystyle {\begin{bmatrix}-\sin \theta \\\cos \theta \\\end{bmatrix}}.}$$

The sense or "direction" of vector rotation (not to be confused with the vector direction) is counterclockwise if θ is positive (e.g. 90°), and clockwise if θ is negative (e.g. −90°) for ${\displaystyle R(\theta )}$. Thus the clockwise rotation matrix is found as (by replacing θ with -θ and using the trigonometric symmetry of ${\textstyle \sin(-\theta )=-\sin(\theta )}$ and ${\textstyle \cos(-\theta )=\cos(\theta )}$) $${\displaystyle R(-\theta )={\begin{bmatrix}\cos \theta &\sin \theta \\-\sin \theta &\cos \theta \\\end{bmatrix}}.}$$

An alternative convention uses rotating axes (instead of rotating a vector),² and the above matrices also represent a rotation of the axes clockwise through an angle θ.

The two-dimensional case is the only non-trivial case where the rotation matrices group is commutative; it does not matter in which order rotations are multiply performed. For the 3-dimensional case, for example, a different order of multiple rotations gives a different result. (E.g., rotating a cell phone along z-axis then y-axis is not equal to rotations along the y-axis then z-axis.)

If a standard right-handed Cartesian coordinate system is used, with the x-axis to the right and the y-axis up, the rotation R(θ) is counterclockwise. If a left-handed Cartesian coordinate system is used, with x directed to the right but y directed down, R(θ) is clockwise. Such non-standard orientations are rarely used in mathematics but are common in 2D computer graphics, which often have the origin in the top left corner and the y-axis down the screen or page.³

See below for other alternative conventions which may change the sense of the rotation produced by a rotation matrix.

Matrices $${\displaystyle {\begin{bmatrix}1&0\\[3pt]0&1\\\end{bmatrix}},\quad {\begin{bmatrix}0&-1\\[3pt]1&0\\\end{bmatrix}},\quad {\begin{bmatrix}-1&0\\[3pt]0&-1\\\end{bmatrix}},\quad {\begin{bmatrix}0&1\\[3pt]-1&0\\\end{bmatrix}}}$$ are 2D rotation matrices corresponding to counter-clockwise rotations of respective angles of 0°, 90°, 180°, and 270°.

The matrices of the shape $${\displaystyle {\begin{bmatrix}x&-y\\y&x\end{bmatrix}}}$$ form a ring, since their set is closed under addition and multiplication. Since $${\displaystyle {\begin{bmatrix}0&-1\\1&0\end{bmatrix}}^{2}\ =\ {\begin{bmatrix}-1&0\\0&-1\end{bmatrix}}\ =-I}$$ (where ${\textstyle I}$ is the identity matrix), the map $${\displaystyle {\begin{bmatrix}x&-y\\y&x\end{bmatrix}}=x{\begin{bmatrix}1&0\\0&1\end{bmatrix}}+y{\begin{bmatrix}0&-1\\1&0\end{bmatrix}}\mapsto x+iy}$$ (where ${\displaystyle {\begin{bmatrix}0&-1\\1&0\end{bmatrix}}}$ corresponds to ${\displaystyle i}$) is a ring isomorphism from this ring to the field of the complex numbers ⁠${\displaystyle \mathbb {C} }$⁠ (incidentally, this shows that this ring is a field). Under this isomorphism, the rotation matrices ${\displaystyle {\begin{bmatrix}\cos t&-\sin t\\\sin t&\cos t\\\end{bmatrix}}=\cos t{\begin{bmatrix}1&0\\0&1\end{bmatrix}}+\sin t{\begin{bmatrix}0&-1\\1&0\end{bmatrix}}}$correspond to the circle of the unit complex numbers, the complex numbers of modulus 1, since ${\displaystyle \cos t^{2}+\sin t^{2}=1}$. As a result, the following equality holds,$${\displaystyle e^{it}=\cos t+i\sin t=\cos t{\begin{bmatrix}1&0\\0&1\end{bmatrix}}+\sin t{\begin{bmatrix}0&-1\\1&0\end{bmatrix}}={\begin{pmatrix}\cos t&-\sin t\\\sin t&\cos t\end{pmatrix}}}$$where the first equality is Euler's formula, the matrix ${\displaystyle I={\begin{bmatrix}1&0\\0&1\end{bmatrix}}}$corresponds to 1, and the matrix ${\displaystyle {\begin{bmatrix}0&-1\\1&0\end{bmatrix}}}$ corresponds to the imaginary unit ${\textstyle i}$.

If one identifies ${\displaystyle \mathbb {R} ^{2}}$ with ${\displaystyle \mathbb {C} }$ through the linear isomorphism ${\displaystyle (a,b)\mapsto a+ib}$, where ${\displaystyle (a,b)\in \mathbb {R} ^{2}}$ and ${\displaystyle a+ib\in \mathbb {C} }$, the action of a matrix ${\displaystyle {\begin{bmatrix}x&-y\\y&x\end{bmatrix}}}$ on a vector ${\displaystyle (a,b)}$ corresponds to multiplication on the complex number ${\displaystyle a+ib}$ by x + iy. In other words, a vector rotation corresponds to multiplication on a complex number (corresponding to the vector being rotated) by a complex number of modulus 1 (corresponding to the rotation matrix).