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5.5: Homogeneous Coordinates

  • Page ID
    9979
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    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.


    This page titled 5.5: Homogeneous Coordinates is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Louis Scharf (OpenStax CNX) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.