Scaling Objects with a Transformation Matrix

This article is part 2 in the series about transformation matrices:

Scaling Along the Cardinal Axes

Intuitively, the basis vectors should be multiplied by a scalar. Also, they are independently affected by the scale factors.

In 2D, the basis vectors become:

p^{'} = k_{x} p = k_{x} [\begin{matrix} 1 \\ 0 \end{matrix}] = [\begin{matrix} k_{x} \\ 0 \end{matrix}] q^{'} = k_{y} q = k_{y} [\begin{matrix} 0 \\ 1 \end{matrix}] = [\begin{matrix} 0 \\ k_{y} \end{matrix}]

Constructing the 2D scale matrix $S (k_{x}, k_{y})$ from these basis vectors:

S (k_{x}, k_{y}) = [\begin{matrix} k_{x} & 0 \\ 0 & k_{y} \end{matrix}]

Similarly, the 3D scale matrix is given by:

S (k_{x}, k_{y}, k_{z}) = [\begin{matrix} k_{x} & 0 & 0 \\ 0 & k_{y} & 0 \\ 0 & 0 & k_{z} \end{matrix}]

Scaling Along an Arbitrary Axis

Let $\hat{n}$ be the unit vector parallel to the direction of scale and $k$ to be the scale factor. A vector transformed by this scale operation can be represented as:

v^{'} = S (\hat{n}, k) v

Separate $v$ into two vectors: a vector parallel to $\hat{v}$ called $v_{∥}$ and a vector perpendicular to $\hat{v}$ called $v_{⊥}$ such that:

v = v_{∥} + v_{⊥}

Where:

\begin{aligned} v_{∥} & = (v \cdot \hat{n}) \hat{n} \\ v_{⊥} & = v - v_{∥} \end{aligned}

We can also represent $v^{'}$ as a sum of two vectors parallel and perpendicular to $\hat{n}$ :

v^{'} = v_{∥}^{'} + v_{⊥}^{'}

Note that any vector that lies in the 2D line or 3D plane perpendicular to $\hat{n}$ will not be affected by the scale operation, so $v^{'} = v_{∥}^{'} + v_{⊥}$ .

Since $v_{∥}$ is parallel to the direction of scale, then $v_{∥}^{'} = k v_{∥}$ .

Reconstructing the solution from the observations above:

\begin{aligned} v_{∥} & = (v \cdot \hat{n}) \hat{n} \\ v_{⊥}^{'} & = v_{⊥} \\ = v - v_{∥} \\ = v - (v \cdot \hat{n}) \hat{n} \\ v_{∥}^{'} & = k v_{∥} \\ = k (v \cdot \hat{n}) \hat{n} \\ v^{'} & = v_{⊥}^{'} + v_{∥}^{'} \\ = v - (v \cdot \hat{n}) \hat{n} + k (v \cdot \hat{n}) \hat{n} \\ = v + (k - 1) (v \cdot \hat{n}) \hat{n} \end{aligned}

We can construct a general scale matrix by computing the vectors resulting after transforming the basis vectors $p$ , $q$ , and $r$ . For example, let’s transform $p = {[\begin{matrix} 1 & 0 & 0 \end{matrix}]}^{T}$ :

\begin{aligned} p^{'} & = p + (k - 1) (p \cdot \hat{n}) \hat{n} \\ = [\begin{array}{c} 1 \\ 0 \\ 0 \end{array}] + (k - 1) ([\begin{array}{c} 1 \\ 0 \\ 0 \end{array}] {[\begin{array}{c} n_{x} \\ n_{y} \\ n_{z} \end{array}]}^{T}) [\begin{array}{c} n_{x} \\ n_{y} \\ n_{z} \end{array}] \\ = [\begin{array}{c} 1 \\ 0 \\ 0 \end{array}] + (k - 1) n_{x} [\begin{array}{c} n_{x} \\ n_{y} \\ n_{z} \end{array}] \\ = [\begin{array}{c} 1 + (k - 1) {n_{x}}^{2} \\ (k - 1) n_{x} n_{y} \\ (k - 1) n_{x} n_{z} \end{array}] \end{aligned}

Similarly, the values of $q^{'}$ and $r^{'}$ can be found, which make the general rotation matrix equal to:

\begin{aligned} S (\hat{n}, k) & = [\begin{array}{c} p^{'} & q^{'} & r^{'} \end{array}] \\ = [\begin{array}{c} 1 + (k - 1) {n_{x}}^{2} & (k - 1) n_{y} n_{x} & (k - 1) n_{z} n_{x} \\ (k - 1) n_{x} n_{y} & 1 + (k - 1) {n_{y}}^{2} & (k - 1) n_{z} n_{y} \\ (k - 1) n_{x} n_{z} & (k - 1) n_{y} n_{z} & 1 + (k - 1) {n_{z}}^{2} \end{array}] \end{aligned}