A Comprehensive Guide To Projection Matrices In Computer Graphics

Introduction¶

We construct the perspective and orthographic projection matrices common to computer graphics in a very general way. The usual parametrizations are defined with respect to specific coordinate system and a specific canonical view volume, such that the view volume parameters define the view volume planes directly in terms of coordinates. This is how most computer graphics books construct the projection transformations.

Rendering is done in the manifold $\mathbb{RP}^{3}$ in homogeneous coordinates for multiple reasons:

1. Affine transformations become linear transformations one dimension higher, so translations can be treated like any other linear transformations.
2. Transformations in $\mathbb{RP}^{3}$ are well-defined under changes in scale, so we can handle projective transformations in a unified fashion with affine transformations.
3. Coordinate systems and scales become equivalent, allowing spatial simulation problems to be expressed directly in a coordinate independent way.
4. We can change coordinate systems to whatever coordinate system makes the problem at hand convenient to work with.
5. For very practical reasons, we can express the problem in a coordinate system that affords the best numerical precision possible on the hardware.

A platform such as OpenGL, Vulkan, DirectX, Metal, or WebGPU typically defines its normalized device coordinate system as one that is a numerically favorable coordinate system. Any spatial computing domain whose problems are formulated in Euclidean space can take advantage of the manifold $\mathbb{RP}^{3}$. Such domains include computer graphics, computer vision, geometric modeling, and robotics. In the context of computer graphics, the platform coordinate system is whichever one maps the view volume in the viewing space to the canonical view volume defined by the platform interface. In particular, the canonical view volume tends to be parametrized by either $[-1, 1] \times [-1, 1] \times [-1, 1]$ or $[-1, 1] \times \ [-1, 1] \times [0, 1]$. In either case, transforming the problem to a unit interval adds one free bit of extra precision when working with floating point numbers. This maximizes the accuracy of floating point computations on the GPU, including tasks such as intersection testing, depth testing, texture sampling, and clipping algorithms.

We parametrize the view space view volume in a slightly different way than the usual one to make the perspective view volume specification coordinate independent. This allows us to construct the matrices for any specific view space coordinate system, perspective view volume, orthographic view volume, normalized device coordinate system, and canonical view volume. We define a canonical set of transformations where the view space is a left-handed orthonormal frame where the horizontal axis points right, the vertical axis points up, and the depth axis points into the view volume, and a clip coordinate system with a left-handed orthonormal frame where the horizontal axis points right, the vertical axis points up, and the depth axis points into the view volume. We show how to construct a general perspective projection or orthographic projection from any source view coordinate system to any target clip coordinate system.

The Topological Manifold Structure Of Real Projective Space¶

This section establishes the topological structure and manifold structure of the real projective space $\mathbb{RP}^{3}$. It can be skipped if the reader so desires.

We define real projective space $\mathbb{RP}^{3}$ as follows. Define $\mathbf{w_{1}} \sim \mathbf{w_{2}}$ if an only if there exists a nonzero real number $\lambda \in \mathbb{R} - \{0\}$ such that $\mathbf{w_{1}} = \lambda \mathbf{w_{2}}$. The relation $\sim$ is an equivalence relation. To prove this, we need to show that $\sim$ is reflexive, symmetric, and transitive. For reflexivity, trivially $\mathbf{w} = \mathbf{w}$ so we can take $\lambda = 1$ which shows that $\mathbf{w} \sim \mathbf{w}$. To show symmetry, suppose that $\mathbf{w}_{1} \sim \mathbf{w}_{2}$. Then there exists $\lambda \in \mathbb{R} - \{ 0 \}$ such that $\mathbf{w}_{2} = \lambda \mathbf{w}_{1}$ implying that $\mathbf{w}_{1} = \frac{1}{\lambda} \mathbf{w}_{2}$ hence $\mathbf{w}_{2} \sim \mathbf{w}_{1}$. Now it remains to prove that $\sim$ is transitive. Suppose that $\mathbf{w}_{1} \sim \mathbf{w}_{2}$ and $\mathbf{w}_{2} \sim \mathbf{w}_{3}$. Then there exist real numbers $\mathbf{\lambda}, \mathbf{\mu} \in \mathbb{R} - \{ 0 \}$ such that $\mathbf{w}_{2} = \lambda \mathbf{w}_{1}$ and $\mathbf{w}_{3} = \mu \mathbf{w}_{2}$. This implies that

implying that $\mathbf{w}_{1} \sim \mathbf{w}_{3}$. This proves transitivity. Therefore, $\sim$ is and equivalence relation.

We define the real projective space by $\mathbb{RP}^{3} = ( \mathbb{R}^{4} - \{ \mathbf{0}\} )/\sim$. The real projective space identifies lines through the origin in $\mathbb{R}^{3}$ with points in $\mathbb{RP}^{3}$.

Define a map $\pi : \mathbb{R}^{4} - \{\mathbf{0}\} \rightarrow \mathbb{RP}^{3}$ by

where $[.]$ on the right-hand side indicates the equivalence class of $\begin{pmatrix} P^{T}, w \end{pmatrix}^{T}$. The function $\pi$ is surjective. To show this, suppose that $[P]$ is a homogeneous point in $\mathbb{RP}^{3}$. Since $[P]$ is an equivalence class, it is nonempty, so there is at least one element in $[P]$, namely $P$ itself. Since the map $\pi$ maps elements to its equivalence class, we obtain $\pi(P) = [P]$. Since the equivalence class $[P]$ was chosen arbitrarily, the function $\pi$ is surjective.

Now that we have established that $\pi$ is surjective, we can use $\pi$ to define a topology on $\mathbb{RP}^{3}$. We say that a set $U \subset \mathbb{RP}^{3}$ is open if and only if the inverse image $\pi^{-1}(U)$ is open in $\mathbb{R}^{4}$. In particular, we define the topology of $\mathbb{RP}^{3}$ to be the quotient topology induced by $\pi$. The real projective space $\mathbb{RP}^{3}$ in conjunction with the quotient toplogy induced by $\pi$ is a topological space. The quotient topology automatically makes the surjection $\pi$ a continuous function.

Now we want to create a topological manifold out of $\mathbb{RP}^{3}$. We need to define an atlas, then show that the charts in the atlas are homeomorphic to open subsets of $\mathbb{R}^{3} - \{ \mathbf{0} \}$. Then we need to show that the resulting atlas makes $\mathbb{RP}^{3}$ locally Euclidean. That is, every point in $\mathbb{RP}^{3}$ has an open neighborhood homeomorphic to $\mathbb{R}^{3} - \{ \mathbf{0} \}$. After that, we show that $\mathbb{RP}^{3}$ is Hausdorff and second countable. All of this together shows that $\mathbb{RP}^{3}$ is a topological 3-manifold.

For every $i \in \{ 0, 1, 2, 3 \}$, define the set $U_{i}$ by

The set $U_{i}$ is an open set in $\mathbb{R}^{4} - \{ \mathbf{0} \}$. To prove this, we will show that $U_{i}$ can be covered by open balls of elements in $U_{i}$. Suppose that $\mathbf{p} \in U_{i}$. Let $h = |\mathbf{p} \cdot \mathbf{\hat{e}}_{i}| = |p_{i}|$ be the distance of the point $\mathbf{p}$ from the hyperplane defined by $x_{i} = 0$. Let $d = \Vert \mathbf{p} \Vert = \Vert \mathbf{p} - \mathbf{0} \Vert$ be the Euclidean norm of the vector $\mathbf{p}$. Let $r = \min \{ d, h \}$ and consider the open ball $B_{r}(\mathbf{p})$ in $\mathbb{R}^{4}$. Choose $\mathbf{q} \in B_{r}(\mathbf{p})$. By the triangle inequality

Hence $\mathbf{q} \in \mathbb{R}^{4} - \{ \mathbf{0} \}$. Since $\mathbf{q}$ was chosen arbitrarily, this implies that $B_{r}(\mathbf{p}) \subset \mathbb{R}^{4} - \{ \mathbf{0} \}$. Applying the triangle inequality again,

Thus $\mathbf{q}$ does not lie on the hyperplane defined by $x_{i} = 0$. Therefore $\mathbf{q} \in U_{i}$ implying that $B_{r}(\mathbf{p}) \subset U_{i}$. Since the vector $\mathbf{p} \in U_{i}$ was arbitrary, every point in $U_{i}$ lies in an open neighborhood that contains only elements of $U_{i}$, we see that

The open balls constructed above cover $U_{i}$ in open sets. Since $\mathbb{R}^{4}$ is a topological space, and $U_{i}$ is the union of open sets in the topology of $\mathbb{R}^{4}$, then $U_{i}$ is an open set in $\mathbb{R}^{4}$. Moreover, since $\mathbb{R}^{4} - \{ \mathbf{0} \}$ is an open set in $\mathbb{R}^{4}$, the set $U_{i}$ is an open set of $\mathbb{R}^{4} - \{ \mathbf{0} \}$ in the subspace topology that $\mathbb{R}^{4} - \{ \mathbf{0} \}$ inherits from $\mathbb{R}^{4}$.

Define the set $V_{i} \subset \mathbb{RP}^{3}$ by $V_{i} = \pi\left(U_{i}\right)$. The set $V_{i}$ is open in $\mathbb{RP}^{3}$. To show this, observe the following equalities

where the last equality follows because the preimage of any element in $V_{i}$ contains all nonzero multiples of an element of $U_{i}$, which are still an elements of $U_{i}$. Therefore, the set $\pi^{-1}(V_{i})$ is an open set in $\mathbb{R}^{4} - \{ \mathbf{0} \}$, which makes $V_{i}$ an open set in $\mathbb{RP}^{3}$ under the quotient topology.

For each $i \in \{ 0, 1, 2, 3 \}$ define the map $\pi_{i} : U_{i} \rightarrow V_{i}$ by $\pi_{i} = \pi|_{U_{i}}$. Since the map $\pi_{i}$ is the restriction of a continuous map, it is also a continuous map. It is surjective by definition, since the codomain is the image $\pi(U_{i})$. To show that $\pi_{i}$ is a quotient map, we must prove that a subset $W \subset V_{i}$ is open in $V_{i}$ if and only if $\pi^{-1}_{i}(W)$ is open in $U_{i}$. Suppose that $W$ is open in $V_{i}$. Then the inverse image of $W$ under $\pi$ is open in $\mathbb{R}^{4} - \{ \mathbf{0} \}$. Since $U_{i}$ is open in $\mathbb{R}^{4} - \{ \mathbf{0} \}$ we see that $\pi^{-1}_{i}(W) = \pi^{-1}(W) \cap U_{i}$. Since $W$ is open in $V_{i}$, there exists an open set $X \subset \mathbb{RP}^{3}$ such that $W = X \cap V_{i}$. Since $\pi$ is continuous, $\pi^{-1}(X)$ is open in $\mathbb{R}^{4} - \{ \mathbf{0} \}$, which implies that $\pi^{-1}(X) \cap U_{i}$ is open in $U_{i}$. But

which is an open set in $U_{i}$. Thus $\pi^{-1}_{i}(W)$ is an open set in $U_{i}$. This proves that if $W \subset V_{i}$ is open, then $\pi^{-1}(W) \subset U_{i}$ is open. Conversely, suppose that $\pi^{-1}_{i}(W)$ is open in $U_{i}$. Since $U_{i}$ is open in $\mathbb{R}^{4} - \{ \mathbf{0} \}$, there exists an open set $X$ of $\mathbb{R}^{4} - \{ \mathbf{0} \}$ such that $\pi^{-1}_{i}(W) = X \cap U_{i}$. Define the set $X^{\prime} = X \cup ((\mathbb{R}^{4} - \{ \mathbf{0} \}) - U_{i})$. We show that the set $X^{\prime}$ is open in $\mathbb{R}^{4} - \{ \mathbf{0} \}$. Suppose that $\mathbf{q} \in (\mathbb{R}^{4} - \{ \mathbf{0} \}) - U_{i}$. We can find a radius $r > 0$ such that $B_{r}(\mathbf{q})$ does not intersect $U_{i}$ such that $B_{r}(\mathbf{q}) \cap U_{i} = \emptyset$. Indeed, we can choose $r$ to be less than the distance to the nearest point where the $i^{th}$ coordinate is zero, i.e. $r < |q_{i}|$. Choose $r = |q_{i}| / 2$ for concreteness. With the chosen $r$, we have $B_{r}(\mathbf{q}) \subset \mathbb{R}^{4} - \{ \mathbf{0} \} - U_{i}$ for each $\mathbf{q} \in \mathbb{R}^{4} - \{ \mathbf{0} \} - U_{i}$. This shows that $\mathbb{R}^{4} - \{ \mathbf{0} \} - U_{i}$ is open in $\mathbb{R}^{4} - \{ \mathbf{0} \}$. Since $X^{\prime}$ is the union of two open sets, it is also open. The set $X^{\prime}$ is open and saturated with respect to $\pi$, so that image $\pi(X^{\prime})$ is open in $\mathbb{RP}^{3}$ because $\pi$ is a quotient map. We have

and it follows that $\pi^{-1}_{i}(W)$ is open in $\mathbb{R}^{4} - \{ \mathbf{0} \}$. This demonstrates that $\pi_{i}$ is a quotient map.

We show that each set $V_{i}$ is homeomorphic to $\mathbb{R}^{4}$. For each $i \in \{ 0, 1, 2, 3 \}$, define the map $H_{i} : V_{i} \rightarrow \mathbb{RP}^{3}$ by

That is, the $i^{th}$ coordinate is 1, and the rest are divided by $P_{i}$. This map sends a homogeneous point to the homogeneous point with the $i^{th}$ coordinate normalized. Let $W_{i}$ be the set of points in $\mathbb{RP}^{3}$ whose $i^{th}$ coordinate is 1. Define the map $\text{proj}_{i} : \mathbb{RP}^{3} \supset W_{i} \rightarrow \mathbb{R}^3$ by

which maps the homogeneous point to a Euclidean point with the $i^{th}$ component removed. The maps $\text{proj}_{i}$ and $H_{i}$ together allow us to define our coordinate maps. Define the map $\psi_{i} : V_{i} \rightarrow \mathbb{R}^{3}$ by $\psi_{i} = \text{proj}_{i} \circ H_{i}$, written out as

Let $\varphi_{i} : U_{i} \rightarrow \mathbb{R}^{3}$ be given by $\varphi_{i} = \psi_{i} \circ \pi$, written out as

The map $\varphi_{i}$ is continuous by the universal property of product spaces applied to $\mathbb{R}^{4}$. Since the map $\varphi_{i}$ is continuous, the universal property of quotient maps implies that the map $\psi_{i}$ is continuous. We must show that $\psi_{i}$ is bijective. Let’s prove surjectivity: let $\begin{pmatrix} P_{0} \ P_{1} \ \dots \ P_{i-1} \ P_{i+1} \ \dots & P_{3} \end{pmatrix}^{T} \in \mathbb{R}^{3}$, let $\begin{bmatrix} \begin{pmatrix} P_{0} \ P_{1} \ \dots \ P_{i-1} \ 1 \ P_{i+1} \dots \ P_{3} \end{pmatrix} \end{bmatrix}^{T} \in V_{i}$, From the definition of $\psi_{i}$, we see that

which establishes surjectivity. To show injectivity, we require the following fact: every element in $V_{i}$ has a unique representative whose $i^{th}$ coordinate is 1. To see this, suppose that $[Q]$ is a homogeneous point in $V_{i}$ with representatives $\begin{pmatrix} Q_{0} \ Q_{1} \ \dots \ Q_{i-1} \ 1 \ Q_{i+1} \dots \ Q_{3} \end{pmatrix}^{T} \in \mathbb{R}^{3}$ and $\begin{pmatrix} Q^{\prime}_{0} \ Q^{\prime}_{1} \ \dots \ Q^{\prime}_{i-1} \ 1 \ Q^{\prime}_{i+1} \dots \ Q^{\prime}_{3} \end{pmatrix}^{T} \in \mathbb{R}^{3}$ where the $i^{th}$ component is 1. Since both representatives represent the same point, they must be equivalent, i.e. there exists $\lambda \in \mathbb{R} - \{0\}$ such that

The fact that the $i^{th}$ component is 1 in both representatives implies that $Q^{\prime}_{j} = Q_{j}$ for each $j \in \{ 0, 1, 2, 3 \}$. Therefore

establishing uniqueness of the representative of $[Q]$ whose $i^{th}$ component is 1. Proving the injectivity of $\psi_{i}$ becomes easy. Recall the original definition of $\psi_{i}$ as $\psi_{i} = \text{proj}_{i} \circ H_{i}$. Since a given homogeneous point has a unique representative whose $i^{th}$ component is 1, the normalization map $H_{i}$ is injective. That is, given homogeneous points $[Q]$ and $[Q^{\prime}]$ with $H_{i}([Q]) = H_{i}([Q^{\prime}])$, we conclude $[Q] = [Q^{\prime}]$. Also, notice that $\text{proj}_{i}$ is bijective with inverse map $\text{proj}^{-1}_{i} : \mathbb{R}^{3} \rightarrow \mathbb{RP}^{3}$ defined by

Since $\text{proj}_{i}$ is bijective, it is injective, and since $H_{i}$ is injective, so is their composite $\psi_{i}$. This proves that $\psi_{i}$ is bijective.

Next, we show that $\psi^{-1}_{i}$ is continuous. Consider the map $\rho_{i} : \mathbb{R}^{3} \rightarrow \mathbb{R}^{4} - \{ \mathbf{0} \}$ given by

The map $\rho_{i}$ is continuous, its image is contained in $V_{i}$, and $\pi_{i} \circ \rho_{i} = \psi^{-1}_{i}$. This implies that $\psi^{-1}_{i}$ is continuous, since it is the composite of continuous functions. We have shown that $\psi_{i}$ is continuous, bijective, and has a continuous inverse. Therefore, it is a homeomorphism. Moreover, we can cover $\mathbb{RP}^{3}$ with the charts $\{ (V_{i}, \psi_{i}) \mid i \in \{ 0, 1, 2, 3 \} \}$, which means that $\mathbb{RP}^{3}$ is locally Euclidean.

We show that $\mathbb{RP}^{3}$ is Hausdorff. Suppose that $[P]$ and $[Q]$ are two distinct points in $\mathbb{RP}^{3}$. There are two possibilities: there is an $0 \leq i \leq 2$ such that both points lie in $V_{i}$, or no such $i$ exists such that both points lie in $V_{i}$. In the first case, $\psi_{i}([P])$ and $\psi_{i}([Q])$ are distinct points in $\mathbb{R}^{3}$. Since $\mathbb{R}^{3}$ is Hausdorff, there exist disjoint sets $A$ and $B$ such that $\psi_{i}([P]) \in A$ and $\psi_{i}([Q]) \in B$. Hence, $\psi^{-1}_{i}(A)$ and $\psi^{-1}_{i}$ are disjoint open subsets of $V_{i}$, and hence $\psi^{-1}_{i}(A)$ and $\psi^{-1}_{i}$ are open subsets of $\mathbb{RP}^{3}$. This yields the first case. Consider the second case. Suppose that no such $i$ exists such that $\psi_{i}([P])$ and $\psi_{i}([Q])$ are distinct points in $V_{i}$. Let $(x_{0}, x_{1}, x_{2}, x_{3})$ be a representative of $[P]$ and let $(y_{0}, y_{1}, y_{2}, y_{3})$ be a representative of $[Q]$. There exists $i \neq j$ or $0 \leq i, j \leq 3$ such that $x_{i} \neq 0, y_{i} = 0$ and $x_{j} = 0, y_{j} \neq 0$. Choose a representative such that $x_{i} = 1$ and $y_{j} = 1$. Assume without loss of generality that $i < j$, and choose $0 < \epsilon < 1$. The set

is an open set containing $[P]$ and the set

is an open set containing $[Q]$. The image $\psi_{i}(A)$ is an open rectangle in $\mathbb{R}^{3}$ centered on $\psi_{i}([P])$ with a side length of $2 \epsilon$. Similarly, the image $\psi_{i}(B)$ is an open rectangle in $\mathbb{R}^{3}$ centered on $\psi_{i}([Q])$ with a side length of $2 \epsilon$. The sets $A$ and $B$ are disjoint. To show this, suppose $A$ and $B$ are not disjoint. Then for $\begin{pmatrix} a_{0} \ a_{1} \ \dots a_{i-1} \ 1 \ a_{i + 1} \dots a_{3} \end{pmatrix}^{T} = \begin{pmatrix} b_{0} \ b_{1} \ \dots b_{j-1} \ 1 \ b_{j + 1} \dots b_{3} \end{pmatrix}^{T}$ we must have $a_{j} \neq 0$ and $b_{i} \neq 0$ which implies that $a_{j} b_{i} = 1$. But this is impossible, because $\lvert a_{j} \rvert < 1$ and $\lvert b_{i} \rvert < 1$. Therefore, the sets $A$ and $B$ must be disjoint. This proves that $\mathbb{RP}^{3}$ is Hausdorff.

Finally, second countability follows from the fact that $\mathbb{R}^{4} - \{ \mathbf{0} \}$ is second countable. More precisely, let $\mathcal{B}$ be a countable base for the topology on $\mathbb{R}^{4} - \{ \mathbf{0} \}$. The set

is a countable set of open sets in $\mathbb{RP}^{3}$ because $\pi$ is saturated. Since $\pi$ is surjective, $\cup_{B \in \mathcal{B}^{\prime}} B = \mathbb{RP}^{3}$, i.e. the set $\mathcal{B}^{\prime}$ covers $\mathbb{RP}^{3}$. Hence $\mathbb{RP}^{3}$ is second countable. Note that second countability implies that $\mathbb{RP}^{3}$ is compact. We sketch a proof here. It comes from the fact that the 3-sphere $S^{3}$ is a compact subset of $\mathbb{R}^{4}$, that the canonical surjection $\pi_{S^{3}}$ is continuous, and that the image of a compact set by a continuous map is compact. Since $\pi_{S^{3}}(S^{3})$ is compact, and $\pi_{S^{3}}(S^{3}) = \mathbb{RP}^{3}$ we immediately infer that $\mathbb{RP}^{3}$ is compact.

We have topologized real projective space, and proven that $\mathbb{RP}^{3}$ is second countable, Hausdorff and locally Euclidean. Therefore $\mathbb{RP}^{3}$ is a topological 3-manifold (or we just say that it is a 3-manifold, or a manifold). Finally, $\mathbb{RP}^{3}$ is compact, so it is a nice setting to work with topologically.

Representing Transformations In Real Projective Space¶

Any orthonormal frame $(\tilde{O}_{frame}, \left( \mathbf{\hat{u}}_{h}, \mathbf{\hat{u}}_{v}, \mathbf{\hat{u}}_{d} \right))$ on $\mathbb{E}^{3}$ induces a coordinate chart as follows. A point $\tilde{P} \in \mathbb{E}^{3}$ is written as

and the orthonormal frame $(\tilde{O}_{frame}, \left( \mathbf{\hat{u}}_{h}, \mathbf{\hat{u}}_{v}, \mathbf{\hat{u}}_{d} \right))$ defines a coordinate chart $\varphi_{frame} : \mathbb{E}^{3} \rightarrow \mathbb{R}^{3}$ by $\varphi_{frame}( \tilde{P} ) = \tilde{P} - \tilde{O}$. Written out

is the representation of $\tilde{P}$ in $\mathbb{R}^{3}$. Also, the coordinate chart $\varphi_{frame}$ maps the origin $\tilde{O}_{frame}$ of the view coordinate system in $\mathbb{E}^{3}$ to $\mathbf{0}$: $O_{frame}$ = $\varphi_{frame}(\tilde{O}_{frame}) = \tilde{O}_{frame} - \tilde{O}_{frame} = \mathbf{0}$. This shows that the view space frame origin in $\mathbb{E}^{3}$ indeed maps to the vector space origin $\mathbf{0}$ in $\mathbb{R}^{3}$.

Recall that the definition of real projective space defines the manifold structure using the surjection $\pi : \mathbb{R}^{4} - \{\mathbf{0}\} \rightarrow \mathbb{RP}^{3}$ given by

where $[.]$ on the right-hand side indicates the equivalence class of $\begin{pmatrix} P^{T}, w \end{pmatrix}^{T}$.

Define a map $\rho : \mathbb{R}^{3} \rightarrow \mathbb{R}^{4} - \{\mathbf{0}\}$ by

The maps $\pi$ and $\rho$ together allow us to map from view space to projective view space with the camera orthonormal frame by

The map $\rho$ defines an embedding (injection) of $\mathbb{R}^{3}$ into $\mathbb{R}^{4} - \{ \mathbf{0} \}$, because each element in $\rho(\mathbb{R}^{3})$ has a unique elements $P \in \mathbb{R}^{3}$ such that $\rho(P) = \begin{pmatrix} P^{T} \ 1 \end{pmatrix}^{T}$. Since $\rho$ is surjective on its image, injective, and continuous with continuous inverse, it is a homeomorphism on its image, hence an embedding. Since every element of $\mathbb{R}^{3}$ has a unique homogeneous point $Q$ such that $Q = [\begin{pmatrix} P^{T} \ 1 \end{pmatrix}^{T}]$, the map $\pi \circ \rho$ is also an embedding. The set $\pi(\rho(\mathbb{R}^{3}))$ is sometimes called the affine patch or Euclidean patch of $\mathbb{RP}^{3}$.

We can lift points in $\mathbb{E}^{3}$ into $\mathbb{RP}^{3}$. Let’s do the same for transformations. Suppose that $A : \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ is an affine map. We define the lifted map of $A$ over to $\mathbb{RP}^{3}$ to be the map $\hat{A} : \mathbb{RP}^{3} \rightarrow \mathbb{RP}^{3}$ such that

Recall that an affine map has the form $A(P) = L(P) + \mathbf{t}$ where $L : \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ is a linear map, and $\mathbf{t}$ is a vector, i.e. a linear map plus a translation term. Expanding this out, the left-hand side of (27) becomes

and on the right-hand side we have

Equating both sides of (27) we have

or

so that $\hat{A}$ is unique. An affine map in $\mathbb{R}^{3}$ becomes a linear map in $\mathbb{RP}^{3}$. In the case of a linear map $L$, $\mathbf{t} = \mathbf{0}$ and

Now suppose that $T : \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ is a projective transformation. A projective transformation is a map of the form

where $A$ is an affine map and $g : \mathbb{R}^{3} \rightarrow \mathbb{R}$ is an affine scalar function. This means that $g$ has the form $g(P) = \mathbf{c} \cdot P + h$. We define the lifted map of $T$ over to $\mathbb{RP}^{3}$ to be the map $\hat{T} : \mathbb{RP}^{3} \rightarrow \mathbb{RP}^{3}$ such that

Expanding out the right-hand side of (34)

More interestingly, expanding out the left-hand side of (34) gives