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 an 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 operations can be represented as

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

Separate $v$ in 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}