5.5: Homogeneous Coordinates

In the previous section we were able to combine rotation and scaling into a single composite operation by matrix multiplication. Unfortunately, translation cannot yet be included in the composite operator since we do it by addition rather than by multiplication.

Example \(\PageIndex{1}\)

Suppose we wish to rotate the image \(G\) by \(\frac{\pi}{3}\) about the point (−10,10). Our rotation matrix R(\(\theta\)) always rotates about the origin, so we must combine three transformations to accomplish this:

1. translate the point (−10,10) to the origin;
2. rotate \(\frac{\pi}{3}\) radians about the origin; and
3. translate the origin back to (−10,10).

For step (i), we have \(b_{0}=\left[\begin{array}{l}
10 \\
-10
\end{array}\right]\) and

\[\mathrm{G}_{1}=\mathrm{G}+\mathrm{b}_{0} 1^{T} \nonumber \]

For step (ii),

\[\begin{align}
\mathrm{G}_{2} &=\mathrm{R}\left(\frac{\pi}{3}\right) \mathrm{G}_{1} \nonumber \\
&=\mathrm{R}\left(\frac{\pi}{3}\right)\left[\mathrm{G}+\mathrm{b}_{0} 1^{T}\right] \nonumber \\
&=\mathrm{R}\left(\frac{\pi}{3}\right) \mathrm{G}+\mathrm{R}\left(\frac{\pi}{3}\right) \mathrm{b}_{0} 1^{T} .
\end{align} \nonumber \]

For step (iii), we can use −\(b_0\) from step (i):

\[\begin{align}
\mathrm{G}_{\text {new }} &=\mathrm{G}_{2}-\mathrm{b}_{0} 1^{T} \nonumber \\
&=\mathrm{R}\left(\frac{\pi}{3}\right) \mathrm{G}+\mathrm{R}\left(\frac{\pi}{3}\right) \mathrm{b}_{0} 1^{T}-\mathrm{b}_{0} 1^{T} \nonumber \\
&=\mathrm{R}\left(\frac{\pi}{3}\right) \mathrm{G}+\left[\left(\mathrm{R}\left(\frac{\pi}{3}\right)-\mathrm{I}\right) \mathrm{b}_{0}\right] 1^{T} .
\end{align} \nonumber \]

In this example we were unable to find a single matrix operator \(A\) to do the entire job. The total transformation took the form

\[\mathrm{G}_{\text {new }}=\mathrm{AG}+\mathrm{b} 1^{T} \nonumber \]

This is called an affine transformation because it involves both multiplication by \(A\) and addition of a constant matrix. This is in contrast to the more desirable linear transformation, which involves only multiplication by \(A\).

We will now move toward a modified representation of the image and the operators by rewriting the last equation as

\[\mathrm{G}_{\text {new }}=[\mathrm{Ab}]\left[\begin{array}{l}
\mathrm{G} \\
1^{T}
\end{array}\right] \nonumber \]

where in the example we had \(A = \mathrm{R}\left(\frac{\pi}{3}\right)\) and \(b=\left(\mathrm{R}\left(\frac{\pi}{3}\right)-\mathrm{I}\right) b_{0}\).

Exercise \(\PageIndex{1}\)

Show that, for any matrices A,B,C,D of compatible sizes,

\[\mathrm{AB}+\mathrm{CD}=\left[\begin{array}{ll}
A & \mathrm{C}
\end{array}\right]\left[\begin{array}{l}
\mathrm{B} \\
\mathrm{D}
\end{array}\right] \nonumber \]

The matrix \(\left[\begin{array}{l}
\mathrm{G} \\
1^{T}
\end{array}\right]\) looks like

\[\left[\begin{array}{lll}
x_{1} & x_{2} & x_{n} \\
y_{1} & y_{2} & y_{n} \\
1 & 1 & 1
\end{array}\right] \nonumber \]

and the points \((x_i,y_i,1)\) are called homogeneous coordinates. We can modify Equation 5 so that the new point matrix is also in homogeneous coordinates:

\[\left[\begin{array}{l}
\mathrm{G}_{\text {new }} \\
1^{T}
\end{array}\right]=\left[\begin{array}{ll}
\mathrm{A} & \mathrm{b} \\
\mathrm{O}^{T} & 1
\end{array}\right]\left[\begin{array}{l}
\mathrm{G} \\
1^{T}
\end{array}\right] \nonumber \]

In the new representation, each point in the image has a third coordinate, which is always a 1. The homogeneous transformation is a 3×3 matrix,

\[\mathrm{A}_{h}=\left[\begin{array}{ll}
\mathrm{A} & \mathrm{b} \\
0^{T} & 1
\end{array}\right] \text { , } \nonumber \]

which is capable of translation, rotation, and scaling all by matrix multiplication. Thus, using homogeneous coordinates, we can build composite transformations that include translation.

In homogeneous coordinates, we have

\[\mathrm{R}(\theta)=\left[\begin{array}{lll}
\cos \theta & -\sin \theta & 0 \\
\sin \theta & \cos \theta & 0 \\
0 & 0 & 1
\end{array}\right] \nonumber \]

\[\mathrm{S}\left(s_{x}, s_{y}\right)=\left[\begin{array}{lll}
s_{x} & 0 & 0 \\
0 & s_{y} & 0 \\
0 & 0 & 1
\end{array}\right] \nonumber \]

\[\mathrm{T}\left(t_{x}, t_{y}\right)=\left[\begin{array}{lll}
1 & 0 & t_{x} \\
0 & 1 & t_{y} \\
0 & 0 & 1
\end{array}\right] \nonumber \]

Example \(\PageIndex{2}\)

The composite transformation to triple the size of an image and then move it 2 units to the left is

\[\mathrm{A}=\left[\begin{array}{lll}
1 & 0 & -2 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\left[\begin{array}{lll}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{lll}
3 & 0 & -2 \\
0 & 3 & 0 \\
0 & 0 & 1
\end{array}\right] \nonumber \]

On the other hand, the composite transformation to move an image 2 units to the left and then triple its size is

\[\mathrm{B}=\left[\begin{array}{lll}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 1
\end{array}\right]\left[\begin{array}{lll}
1 & 0 & -2 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{lll}
3 & 0 & -6 \\
0 & 3 & 0 \\
0 & 0 & 1
\end{array}\right] \nonumber \]

In the latter case, the distance of the translation is also tripled.

Exercise \(\PageIndex{2}\)

Find a single composite transformation in homogeneous coordinates that rotates an image by angle \(\theta\) about point \((x_i,y_i)\) as in Example 1.