Rotation and Uniform Scaling Matrices in the Plane

Two Basic Transformations of the Plane

A rotation about the origin changes the direction of a point while keeping its distance from the origin fixed. A uniform scaling about the origin changes that distance by the same factor for every point. These geometric descriptions become precise matrix rules once we determine where each transformation sends the standard basis vectors. That approach also makes the signs in the rotation matrix easier to track.

Throughout, points and vectors are columns in \(\mathbb{R}^2\). The standard basis vectors are \(\mathbf{e}_1=\begin{bmatrix}1\\0\end{bmatrix}\) and \(\mathbf{e}_2=\begin{bmatrix}0\\1\end{bmatrix}\). Each vector \(\mathbf{x}=\begin{bmatrix}x_1\\x_2\end{bmatrix}\) has the decomposition \(\mathbf{x}=x_1\mathbf{e}_1+x_2\mathbf{e}_2\). The standard matrix of a linear transformation \(T:\mathbb{R}^2\to\mathbb{R}^2\) has columns \(T(\mathbf{e}_1)\) and \(T(\mathbf{e}_2)\).

Angles may be specified in radians or degrees, provided the same unit is used when evaluating sine and cosine. We will use radians in the formulas. The scaling convention takes \(k\) to be nonnegative: for \(k>0\), every nonzero point keeps its direction; for \(k=0\), every point maps to \(\mathbf{0}\), so the image of a nonzero point has no direction.

Deriving the Rotation Matrix from Basis Images

To determine a rotation, first rotate the two standard basis vectors. The vector \(\mathbf{e}_1\) has length \(1\) and angle \(0\) from the positive \(x\)-axis. After a counterclockwise turn by \(\theta\), its coordinates are \((\cos\theta,\sin\theta)\). The vector \(\mathbf{e}_2\) has angle \(\frac{\pi}{2}\), so after the same turn its angle is \(\frac{\pi}{2}+\theta\). Its coordinates are \((\cos(\frac{\pi}{2}+\theta),\sin(\frac{\pi}{2}+\theta))=(-\sin\theta,\cos\theta)\).

Proof. The basis-vector images found from the geometry of the unit circle are \(R_\theta(\mathbf{e}_1)=\begin{bmatrix}\cos\theta\\\sin\theta\end{bmatrix}\) and \(R_\theta(\mathbf{e}_2)=\begin{bmatrix}-\sin\theta\\\cos\theta\end{bmatrix}\). Since the columns of a standard matrix are the images of \(\mathbf{e}_1\) and \(\mathbf{e}_2\), placing these images in order gives the displayed matrix. For any \(\mathbf{x}=\begin{bmatrix}x_1\\x_2\end{bmatrix}=x_1\mathbf{e}_1+x_2\mathbf{e}_2\), linearity gives \(R_\theta(\mathbf{x})=x_1R_\theta(\mathbf{e}_1)+x_2R_\theta(\mathbf{e}_2)=\begin{bmatrix}x_1\cos\theta-x_2\sin\theta\\x_1\sin\theta+x_2\cos\theta\end{bmatrix}=\mathbf{R}_\theta\mathbf{x}\). This verifies the matrix action for every input, not only for the basis vectors. \(\square\)

The first column records where the positive \(x\)-axis points after rotation; the second records where the positive \(y\)-axis points. This column interpretation is a useful sign check. In particular, for a positive quarter-turn, \(\mathbf{e}_1\) must point upward and \(\mathbf{e}_2\) must point left.

Deriving the Uniform Scaling Matrix

Scaling by \(k\) multiplies every vector by \(k\). In particular, it sends \(\mathbf{e}_1\) to \(\begin{bmatrix}k\\0\end{bmatrix}\) and \(\mathbf{e}_2\) to \(\begin{bmatrix}0\\k\end{bmatrix}\). These images determine the two columns of its standard matrix.

Proof. The images of the standard basis vectors are \(S_k(\mathbf{e}_1)=\begin{bmatrix}k\\0\end{bmatrix}\) and \(S_k(\mathbf{e}_2)=\begin{bmatrix}0\\k\end{bmatrix}\). Hence the columns of the standard matrix are \(\begin{bmatrix}k\\0\end{bmatrix}\) and \(\begin{bmatrix}0\\k\end{bmatrix}\), giving \(\mathbf{S}_k=\begin{bmatrix}k&0\\0&k\end{bmatrix}\). For any \(\mathbf{x}=x_1\mathbf{e}_1+x_2\mathbf{e}_2\), linearity gives \(S_k(\mathbf{x})=x_1\begin{bmatrix}k\\0\end{bmatrix}+x_2\begin{bmatrix}0\\k\end{bmatrix}=\begin{bmatrix}kx_1\\kx_2\end{bmatrix}\). This is exactly \(k\mathbf{x}\), including when \(k=0\). \(\square\)

What Happens to Length and Direction?

For \(\mathbf{x}=\begin{bmatrix}x_1\\x_2\end{bmatrix}\), its Euclidean length is \(\|\mathbf{x}\|=\sqrt{x_1^2+x_2^2}\). The rotation matrix preserves this length, while scaling multiplies it by \(k\). These facts can be checked directly from the coordinate formulas.

Proof. Write \(c=\cos\theta\) and \(s=\sin\theta\). The rotation formula gives \(R_\theta(\mathbf{x})=\begin{bmatrix}cx_1-sx_2\\sx_1+cx_2\end{bmatrix}\). Its squared length is \((cx_1-sx_2)^2+(sx_1+cx_2)^2=(c^2+s^2)x_1^2+(c^2+s^2)x_2^2=x_1^2+x_2^2\), because the cross terms \(-2csx_1x_2\) and \(2csx_1x_2\) cancel and \(c^2+s^2=1\). Taking nonnegative square roots proves the rotation claim. For scaling, \(\|S_k(\mathbf{x})\|=\|k\mathbf{x}\|=\sqrt{(kx_1)^2+(kx_2)^2}=k\sqrt{x_1^2+x_2^2}=k\|\mathbf{x}\|\), since \(k\ge0\). If \(k>0\) and \(\mathbf{x}\ne\mathbf{0}\), the output \(k\mathbf{x}\) is a positive multiple of the input, so it points in the same direction. If \(k=0\), then \(S_0(\mathbf{x})=\mathbf{0}\) for every \(\mathbf{x}\); the zero vector has no direction. \(\square\)

The same length statements describe distances between any two points. For rotation, the difference between the images of \(\mathbf{x}\) and \(\mathbf{y}\) is \(R_\theta(\mathbf{x})-R_\theta(\mathbf{y})=R_\theta(\mathbf{x}-\mathbf{y})\), so its length equals \(\|\mathbf{x}-\mathbf{y}\|\). For scaling, the difference is \(k(\mathbf{x}-\mathbf{y})\), whose length is \(k\|\mathbf{x}-\mathbf{y}\|\). Thus rotation preserves all distances, and uniform scaling multiplies all distances by the same nonnegative factor.

Common Pitfalls

Reversing the rotation signs changes the direction. A common incorrect matrix for a counterclockwise rotation is \(\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}\). At \(\theta=\frac{\pi}{2}\), this becomes \(\begin{bmatrix}0&1\\-1&0\end{bmatrix}\), which sends \(\mathbf{e}_1=\begin{bmatrix}1\\0\end{bmatrix}\) to \(\begin{bmatrix}0\\-1\end{bmatrix}\). That points downward, so the matrix describes a clockwise quarter-turn, not the specified counterclockwise one. The correction is to use \(\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}\), whose first column sends \(\mathbf{e}_1\) upward at \(\frac{\pi}{2}\).

Do not claim that zero scaling preserves direction. For \(k=0\), \(\mathbf{S}_0\begin{bmatrix}1\\2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\). The input is nonzero, but the output is the zero vector, which has no direction. The accurate statement is that positive scaling preserves the direction of every nonzero point; zero scaling sends every point to the origin.

Both transformations are linear because their formulas are given by multiplication by a matrix. Their matrices also show why basis images are central: the columns record the transformed coordinate axes, and every other point follows by expressing it as a linear combination of those axes. Rotation keeps the two axis images at right angles and preserves length; uniform scaling multiplies each axis by the same factor.