One Point, Two Coordinate Descriptions
A point in a drawing can be described using coordinates measured from different origins and along different axes. The point itself does not move when its description changes; only its coordinates do. The main challenge is keeping track of which frame a coordinate column belongs to and applying transformations in the correct order. Homogeneous coordinates make this bookkeeping a matrix calculation.
Work in \(\mathbb{R}^2\). A planar coordinate frame \(A\) consists of an origin \(\mathbf{o}_A\) and two linearly independent axis vectors \(\mathbf{b}_{A,1}\) and \(\mathbf{b}_{A,2}\), all expressed in a common world coordinate system. A point with coordinates \((u,v)\) in frame \(A\) has world position \(\mathbf{o}_A+u\mathbf{b}_{A,1}+v\mathbf{b}_{A,2}\). Independence of the axes ensures that the frame gives unique coordinates to every point in the plane.
The upper-left \(2\times2\) block records the frame’s axis directions and scales; the third column records its origin. Since the two axes are linearly independent, that block is invertible, and therefore \(\mathbf{F}_A\) is invertible. In particular, knowing a point’s world position determines its coordinates in frame \(A\) by \(\mathbf{q}_A=\mathbf{F}_A^{-1}\mathbf{x}\).
The Conversion Formula and Its Direction
Suppose frames \(A\) and \(B\) are both expressed relative to the same world coordinates. Start with a point’s coordinates in \(A\). Multiplying by \(\mathbf{F}_A\) expresses the point in world coordinates; multiplying by \(\mathbf{F}_B^{-1}\) converts that world description into coordinates in \(B\). Thus the conversion matrix from \(A\)-coordinates to \(B\)-coordinates is \(\mathbf{F}_B^{-1}\mathbf{F}_A\), not \(\mathbf{F}_A^{-1}\mathbf{F}_B\).
Proof. Let \(\mathbf{x}\) be the point’s world homogeneous coordinate column. By the definition of a frame matrix, \(\mathbf{x}=\mathbf{F}_A\mathbf{q}_A\) and \(\mathbf{x}=\mathbf{F}_B\mathbf{q}_B\). Equating these expressions gives \(\mathbf{F}_B\mathbf{q}_B=\mathbf{F}_A\mathbf{q}_A\). Since \(\mathbf{F}_B\) is invertible, multiplying on the left by \(\mathbf{F}_B^{-1}\) gives \(\mathbf{q}_B=\mathbf{F}_B^{-1}\mathbf{F}_A\mathbf{q}_A\). For the reverse direction, start from \(\mathbf{F}_A\mathbf{q}_A=\mathbf{F}_B\mathbf{q}_B\) and multiply on the left by \(\mathbf{F}_A^{-1}\), obtaining \(\mathbf{q}_A=\mathbf{F}_A^{-1}\mathbf{F}_B\mathbf{q}_B\). \(\square\)
The order of the factors follows the route the point takes: first from frame \(A\) to the world description, then from the world description to frame \(B\). Matrix products act from right to left. The product \(\mathbf{F}_B^{-1}\mathbf{F}_A\) converts coordinates; neither factor alone does so. The expression also assumes that both frames are described relative to the same world system.
Finding and Checking a Frame Inverse
Worked Example: Inverting a Frame Matrix
Consider a frame \(B\) whose first axis is \((1,1)^\top\), second axis is \((-1,1)^\top\), and origin is \((1,0)^\top\). Its frame matrix is
The \(2\times2\) axis block has inverse \(\begin{bmatrix}\frac12&\frac12\\-\frac12&\frac12\end{bmatrix}\). Applying that inverse to the origin gives \((\frac12,-\frac12)^\top\); negating this result gives the translation column of the homogeneous inverse. Therefore
For example, the first row of \(\mathbf{F}_B\) multiplied by the third column of \(\mathbf{F}_B^{-1}\) gives \(-\frac12-\frac12+1=0\). Checking the whole product gives
Now convert the world point \(\mathbf{x}=(4,3,1)^\top\) into frame \(B\):
Check: Reconstruct the world point with frame \(B\): \(\mathbf{F}_B(3,0,1)^\top=(3+1,3,1)^\top=(4,3,1)^\top\), the original point.
Converting Between Two Nonstandard Frames
Worked Example: From Frame A to Frame B
Let frame \(A\) have origin \((1,2)^\top\) and axes \((1,1)^\top\), \((-1,1)^\top\). Let frame \(B\) have origin \((0,1)^\top\) and axes \((2,0)^\top\), \((1,1)^\top\). Their frame matrices are
The inverse for \(B\) is
Indeed, multiplying \(\mathbf{F}_B\mathbf{F}_B^{-1}\) gives first row \((1,0,1-1+0)=(1,0,0)\), second row \((0,1,-1+1)=(0,1,0)\), and bottom row \((0,0,1)\). The conversion matrix is
For a point with \(\mathbf{q}_A=(2,1,1)^\top\), multiply to obtain \(\mathbf{q}_B=(-1,4,1)^\top\). Check in world coordinates: \(\mathbf{F}_A\mathbf{q}_A=(2,5,1)^\top\), while \(\mathbf{F}_B\mathbf{q}_B=(2(-1)+4,\;4+1,\;1)^\top=(2,5,1)^\top\). Both coordinate descriptions identify the same point.
A conversion matrix has a useful structural feature: its bottom row is \((0,0,1)\). It takes a point column with last entry \(1\) to another point column with last entry \(1\). Its top two entries may change through a mixture of rotation, scaling, shear, and translation, because the frames may differ in their origins and axes.
Reversing and Composing Conversions
The reverse of a conversion from \(A\) to \(B\) must undo that conversion. The order of the inverse factors reverses as well. More generally, conversion through an intermediate frame gives the same result as converting directly, provided the intermediate coordinates are represented in the frame named by each factor.
Proof. Since each frame matrix is invertible, \(\left(\mathbf{F}_B^{-1}\mathbf{F}_A\right)^{-1}=\mathbf{F}_A^{-1}\mathbf{F}_B\), which is exactly \(\mathbf{C}_{A\leftarrow B}\). For composition, substitute the definitions and use associativity of matrix multiplication: \(\mathbf{C}_{C\leftarrow B}\mathbf{C}_{B\leftarrow A}=(\mathbf{F}_C^{-1}\mathbf{F}_B)(\mathbf{F}_B^{-1}\mathbf{F}_A)=\mathbf{F}_C^{-1}(\mathbf{F}_B\mathbf{F}_B^{-1})\mathbf{F}_A=\mathbf{F}_C^{-1}\mathbf{F}_A=\mathbf{C}_{C\leftarrow A}\). The cancellation is valid because \(\mathbf{F}_B\) is invertible. \(\square\)
Worked Example: A Rotated and Shifted Frame
Take frame \(A\) to be the world frame, so \(\mathbf{F}_A=\mathbf{I}_3\). Let frame \(B\) have origin \((2,1)^\top\), with axes \((0,1)^\top\) and \((-1,0)^\top\). Then
For a world point \(\mathbf{q}_A=(5,4,1)^\top\), the coordinates in \(B\) are \(\mathbf{q}_B=\mathbf{F}_B^{-1}\mathbf{q}_A=(3,-3,1)^\top\). Reconstructing gives \(\mathbf{F}_B\mathbf{q}_B=(3,-5+2+? ,1)^\top\) only if the row calculation is done incorrectly; compute each row directly: the first is \(0(3)-1(-3)+2=5\), and the second is \(1(3)+0(-3)+1=4\). Thus \(\mathbf{F}_B\mathbf{q}_B=(5,4,1)^\top\), as required.
For clarity, the inverse’s second coordinate is \(-5+2=-3\), and its first coordinate is \(4-1=3\). These values also follow geometrically: relative to the origin \((2,1)\), the point is displaced by \((3,3)\); the frame’s first axis accounts for the vertical displacement, while the second axis points left.
Common Mistake: Applying the Conversion in the Wrong Direction
In the nonstandard-frame example, the correct conversion from \(A\) to \(B\) is \(\mathbf{F}_B^{-1}\mathbf{F}_A\). A tempting but incorrect calculation is to apply \(\mathbf{F}_B^{-1}\) directly to \(\mathbf{q}_A=(2,1,1)^\top\), as if those entries were world coordinates. This produces \(\mathbf{F}_B^{-1}\mathbf{q}_A=(1,0,1)^\top\). But \((2,1,1)^\top\) was specified in frame \(A\), not in the world frame, so that multiplication uses the wrong input description.
The error is visible by reconstructing the supposed answer in frame \(B\): \(\mathbf{F}_B(1,0,1)^\top=(2,1,1)^\top\), which is not the original point’s world coordinate \((2,5,1)^\top\). The correct route is to compute \(\mathbf{F}_A\mathbf{q}_A=(2,5,1)^\top\), then apply \(\mathbf{F}_B^{-1}\), giving \((-1,4,1)^\top\). As a practical check, convert the result back to world coordinates; it must reproduce the same point.
Keep the labels attached to the matrices: \(\mathbf{F}_A\) maps \(A\)-coordinates to world coordinates, and \(\mathbf{F}_B^{-1}\) maps world coordinates to \(B\)-coordinates. The subscripted conversion \(\mathbf{C}_{B\leftarrow A}\) is a reminder that the input is in \(A\) and the output is in \(B\). This distinction matters in graphics pipelines, where a point’s numerical column may look identical even when its coordinate frame has changed.
Check Your Understanding
For each question, label the frame of every coordinate column and verify numerical conversions by reconstructing the world point.
- A frame has origin \((1,2)^\top\) and axes \((1,0)^\top\), \((0,2)^\top\). Write its homogeneous frame matrix and find the world coordinate of \((3,-1,1)^\top\) in that frame.
- If \(\mathbf{q}_A\) is a point’s coordinate column in frame \(A\), write the formula for its coordinates in frame \(B\). Explain why the factors occur in that order.
- For \(\mathbf{F}_B=\begin{bmatrix}1&0&2\\0&1&-1\\0&0&1\end{bmatrix}\), find \(\mathbf{F}_B^{-1}\) and check the product.
- Show from the conversion formulas that converting from \(A\) to \(B\) and then back to \(A\) recovers the original coordinate column.
- Explain what is wrong with applying \(\mathbf{F}_B^{-1}\) directly to a coordinate column known only to be expressed in frame \(A\).
