Part III · 第 10 章相机与投影 · 10.2Part III · Ch. 10 Cameras & Projection · 10.2

射影几何与齐次坐标Projective Geometry & Homogeneous Coordinates

针孔相机的投影要除以深度，是个非线性运算；铁轨在远处会相交，而"过两点的直线""两条直线的交点"在平行线面前又要分情况讨论。给每个点多添一个坐标 w，只在"差一个比例"的意义下看待它，这些麻烦一并消失：投影变成矩阵乘法，平行线交在无穷远点，连线与求交都是同一个叉积。本节把 5.3 节里"补一个 1"的小技巧扩展成射影几何，它是 DLT、单应、对极几何、三角化等后面所有算法的共同语言。 A pinhole camera divides by depth, a non-linear operation; railway tracks meet in the distance, while "the line through two points" and "the intersection of two lines" need special cases for parallel lines. Give every point one extra coordinate w and only look at it up to scale, and these headaches vanish together: projection becomes a matrix product, parallel lines meet at points at infinity, and joining and intersecting are the same cross product. This lesson grows the "append a 1" trick of Lesson 5.3 into projective geometry, the common language of DLT, homographies, epipolar geometry, triangulation and everything else that follows.

约 35 分钟~35 min 前置：1.6 逆、秩与零空间 · 1.8 叉积 · 5.3 齐次变换 · 10.1 针孔相机Prereqs: 1.6 Inverse, Rank & Null Space · 1.8 Cross Product · 5.3 Homogeneous Transforms · 10.1 The Pinhole Camera 5 个交互5 interactives

1为什么机器人需要它Why robots need this

10.1 节的针孔模型是 $u = f_x X/Z + c_x,\ v = f_y Y/Z + c_y$。分母里的 $Z$ 使投影成为非线性映射：它不能写成一个矩阵乘法，不能把两个相机的变换连乘，也不能直接取逆。除此之外，你在图像里还会遇到这几件事：

平行线会相交。路面上的车道线、桌子的边、走廊的墙角线，在三维里互相平行，在图像里却汇聚到一个消失点。消失点的位置藏着相机的朝向，甚至焦距。
点和线的运算有特例。"过两点的直线""两条直线的交点"，在普通坐标里要解方程组，还要单独讨论竖直线与平行线。
后面的算法都是线性代数。由 4 个角点求单应（10.5）、由 6 个点标定相机（10.4）、由 8 对点求基础矩阵（11.1）、两视图三角化（11.3）：它们统统写成"$A\mathbf h = \mathbf 0$，取零空间"（1.6、1.12），前提是先把问题写成齐次向量上的线性方程。

5.3 节已经用过齐次坐标：补一个 1，让平移也变成矩阵乘法。这一节把末位的数当作一个真正的坐标，并且不再要求它等于 1。这个小小的放松，就是射影几何的全部入口。

Lesson 10.1's pinhole model is $u = f_x X/Z + c_x,\ v = f_y Y/Z + c_y$. The $Z$ in the denominator makes projection non-linear: it cannot be written as one matrix product, two camera transforms cannot be chained as matrices, and it cannot simply be inverted. On top of that, images confront you with a few more things:

Parallel lines meet. Lane markings, table edges and corridor corners are parallel in 3D, yet in the image they converge to a vanishing point. Where it lies encodes the camera's orientation, even its focal length.
Points and lines have special cases. "The line through two points" and "the intersection of two lines" mean solving equations in ordinary coordinates, with extra cases for vertical and parallel lines.
Everything later is linear algebra. A homography from 4 corners (10.5), a camera from 6 points (10.4), a fundamental matrix from 8 pairs (11.1), two-view triangulation (11.3): all of them become "$A\mathbf h = \mathbf 0$, take the null space" (1.6, 1.12), provided the problem is first written as linear equations on homogeneous vectors.

Lesson 5.3 already used homogeneous coordinates: append a 1 so that translation also becomes a matrix product. Here the last number becomes a genuine coordinate and is no longer required to equal 1. That small relaxation is the whole doorway into projective geometry.

2直觉：射影平面 = 过原点的射线Intuition: the projective plane = lines through the origin

让相机中心 $O$ 位于原点，成像平面取 $w = 1$（10.1 节的归一化像平面，这里把光轴记作第三个坐标 $w$）。从 $O$ 出发的每一条射线恰好穿过成像平面一次，所以一个像点就是一条过原点的射线，这条射线上任何一个非零向量 $\tilde{\mathbf x}=(x,y,w)$ 都可以代表它。把向量放大、缩小，甚至反号，射线没变，点就没变。要读出普通坐标，只需让射线与平面 $w=1$ 相交：$(x/w,\ y/w)$。

有意思的是平行于成像平面的射线（$w=0$）：它们永远打不到平面，却仍然是射线，是齐次向量。这就是无穷远点，它的坐标 $(x,y,0)$ 记录的是方向。而且一族平行线恰好对应这样一个点：每条直线与原点张成一个平面，这一族平面共享同一条射线。下图先拖滑块看一个点，再切到"平行线"看这件事。

Put the camera center $O$ at the origin and take the image plane to be $w = 1$ (Lesson 10.1's normalized image plane; here the optical axis is called the third coordinate $w$). Every ray from $O$ pierces the image plane exactly once, so an image point is a ray through the origin, and any nonzero vector $\tilde{\mathbf x}=(x,y,w)$ on that ray can stand for it. Scale the vector up, down, or flip its sign: the ray is unchanged, so the point is unchanged. To read off ordinary coordinates, intersect the ray with the plane $w=1$: $(x/w,\ y/w)$.

The interesting rays are the ones parallel to the image plane ($w=0$): they never hit the plane, yet they are still rays, still homogeneous vectors. These are the points at infinity; their coordinates $(x,y,0)$ record a direction. And a family of parallel lines gives exactly one such point: each line spans a plane together with the origin, and these planes all share one ray. Drag the sliders to inspect a point first, then switch to "parallel lines" to see it.

FIG 1射影平面 = 过原点的射线The projective plane = rays through the origin

青色半透明方块是像平面 $w=1$，底部的网格是平面 $w=0$（理想点所在），紫色圆是所有方向。琥珀色箭头是齐次向量 $\tilde{\mathbf x}=\lambda\mathbf d$，青色虚线是它所在的射线。"点"模式：$\alpha$ 是射线与光轴（$w$ 轴）的夹角，$\varphi$ 是方位角，$\lambda$ 是向量的缩放（可以为负）。"平行线"模式：三条平行线（红、绿、蓝）各与原点张成一个平面，三个平面交于琥珀色的射线。可以用鼠标拖动画布旋转视角。 The translucent cyan square is the image plane $w=1$, the grid underneath is the plane $w=0$ (home of the ideal points), and the purple circle is the set of all directions. The amber arrow is the homogeneous vector $\tilde{\mathbf x}=\lambda\mathbf d$ and the cyan dashed line is its ray. "Point" mode: $\alpha$ is the angle between the ray and the optical ($w$) axis, $\varphi$ the azimuth, $\lambda$ a scale on the vector (it may be negative). "Parallel lines" mode: three parallel lines (red, green, blue) each span a plane with the origin, and the three planes meet in the amber ray. Drag the canvas to orbit the view.

试一试Try this

在"点"模式下把 $\lambda$ 拖到负数：箭头反向了，可射线上的交点有没有动？这说明 $\tilde{\mathbf x}$ 和 $-\tilde{\mathbf x}$ 是什么关系？In "Point" mode drag $\lambda$ to a negative value: the arrow flips, but did the point where the ray meets the plane move? What does that say about $\tilde{\mathbf x}$ and $-\tilde{\mathbf x}$?
把 $\alpha$ 一路拖到 $90^\circ$：交点飞向哪里？$w$ 变成多少？此时读数里的"点"变成了什么？Drag $\alpha$ all the way to $90^\circ$: where does the intersection fly to? What does $w$ become, and what does the readout call the point now?

核心洞见Key insight

射影平面 $\mathbb P^2$ 就是 $\R^3$ 中过原点的直线（射线）的集合。普通点是穿过 $w=1$ 平面的射线，无穷远点是平行于它的射线，二者在代数上没有区别，只是 $w$ 恰好为 0。"平行线交于无穷远"不是一个约定，而是一个事实：平行线族对应的平面共享一条射线。

The projective plane $\mathbb P^2$ is the set of lines through the origin of $\R^3$. Ordinary points are rays that pierce the plane $w=1$; points at infinity are rays parallel to it; algebraically there is no difference, only that $w$ happens to be 0. "Parallel lines meet at infinity" is not a convention but a fact: the planes of a parallel family share one ray.

3动手玩：连线、求交与对偶Play: join, meet and duality

在齐次坐标里，直线 $ax+by+c=0$ 就是向量 $\mathbf l=(a,b,c)$，点 $\tilde{\mathbf x}$ 在直线上当且仅当 $\mathbf l^{\top}\tilde{\mathbf x}=0$。过两点 $\mathbf a,\mathbf b$ 的直线必须与两者都"垂直"（内积为 0），而叉积恰好给出同时垂直于两个向量的向量（1.8 节），所以 $\mathbf l=\mathbf a\times\mathbf b$。两条直线的交点同理：$\tilde{\mathbf x}=\mathbf l\times\mathbf m$。连线与求交是同一个运算，只是把点和线的角色对调，这叫对偶。

下图拖动四个点：绿线过 $\mathbf a,\mathbf b$，紫线过 $\mathbf c,\mathbf d$，琥珀色点是它们的交点。读数里逐项验证了关联关系 $\mathbf l^\top\mathbf a=0$ 等。点击"使 m ∥ l"让两线平行，再轻轻拖动一个点。

In homogeneous coordinates the line $ax+by+c=0$ is just the vector $\mathbf l=(a,b,c)$, and a point $\tilde{\mathbf x}$ lies on it exactly when $\mathbf l^{\top}\tilde{\mathbf x}=0$. The line through two points $\mathbf a,\mathbf b$ must be "perpendicular" (zero inner product) to both, and the cross product is precisely a vector perpendicular to two given vectors (Lesson 1.8), so $\mathbf l=\mathbf a\times\mathbf b$. The intersection of two lines works the same way: $\tilde{\mathbf x}=\mathbf l\times\mathbf m$. Joining and intersecting are one operation with the roles of points and lines swapped, which is called duality.

Drag the four points below: the green line passes through $\mathbf a,\mathbf b$, the purple line through $\mathbf c,\mathbf d$, and the amber point is their intersection. The readout checks incidences such as $\mathbf l^\top\mathbf a=0$ numerically. Press "Make m ∥ l" to make the lines parallel, then nudge one point.

FIG 2连线与求交：两个叉积Join and meet: two cross products

点吸附到 0.25 的网格上，所以能做到精确平行。交点跑出画面时，边缘的琥珀色箭头指向它所在的方向；两线平行时交点是无穷远点 $(d_x,d_y,0)$，两个方向的箭头同时出现，因为 $(d_x,d_y,0)$ 与 $(-d_x,-d_y,0)$ 是同一个点。读数里的"相对残差"是 $|\mathbf l^\top\mathbf p|/(\|\mathbf l\|\|\mathbf p\|)$，应为 0（舍入误差在 $10^{-16}$ 量级，对网格点常常恰好为 0）。$\det[\mathbf a\ \mathbf b\ \mathbf c]$ 等于三角形 $abc$ 面积的两倍（带符号），为 0 当且仅当三点共线。 Points snap to a 0.25 grid so exactly parallel lines are reachable. When the intersection leaves the view, an amber arrow on the border points toward it; for parallel lines it is the point at infinity $(d_x,d_y,0)$ and arrows appear in both directions, since $(d_x,d_y,0)$ and $(-d_x,-d_y,0)$ are the same point. The "relative residual" is $|\mathbf l^\top\mathbf p|/(\|\mathbf l\|\|\mathbf p\|)$, zero up to rounding (around $10^{-16}$, often exactly 0 for grid points). $\det[\mathbf a\ \mathbf b\ \mathbf c]$ is twice the signed area of triangle $abc$, and zero exactly when the three points are collinear.

让两条线精确平行（点"使 m ∥ l"，或自己拖到网格点上）。交点 $\tilde{\mathbf x}$ 的第三个分量变成了什么？程序有没有除以它？Make the two lines exactly parallel (press "Make m ∥ l" or place the points on the grid yourself). What does the third component of $\tilde{\mathbf x}$ become? Did the program divide by it?
让交点先在画面很远的一侧，再慢慢转到另一侧（经过平行的位置）。交点是从无穷远"绕回来"的：射影平面是闭合的。Park the intersection far away on one side, then rotate a line until it reappears far away on the other side, passing through the parallel position. The point wraps around through infinity: the projective plane is closed.

对偶：同一个公式，两种读法Duality: one formula, two readings

$\mathbf l^\top\tilde{\mathbf x}=0$ 对 $\mathbf l$ 和 $\tilde{\mathbf x}$ 是对称的。读成"点在线上"，叉积给出连线；读成"线过点"，叉积给出交点。凡是一个关于点和线的射影定理，把"点"与"线"、"共线"与"共点"互换，得到的仍是定理。这也是为什么本课之后会反复出现"一个 3×3 矩阵既可以看作点到点的映射，又可以看作点到线的映射"（11.1 节的基础矩阵就是后者）。

$\mathbf l^\top\tilde{\mathbf x}=0$ is symmetric in $\mathbf l$ and $\tilde{\mathbf x}$. Read it as "the point lies on the line" and the cross product gives the join; read it as "the line passes through the point" and it gives the meet. Any projective theorem about points and lines stays true if you swap "point" with "line" and "collinear" with "concurrent". This is why a 3×3 matrix will later be seen both as a point-to-point map and as a point-to-line map (the fundamental matrix of Lesson 11.1 is the latter).

4第二、第三视角：消失点与变换层级Second and third views: vanishing points and the hierarchy of maps

消失点与地平线Vanishing points and the horizon

三维里方向为 $\mathbf d$ 的所有平行线，在相机坐标里的方向都是 $R_{cw}\mathbf d$。把"方向"当作 $w=0$ 的齐次点 $(\mathbf d,0)$，投影矩阵作用上去，得到它们共同的像：消失点 $\tilde{\mathbf v}=K R_{cw}\mathbf d$（FIG 5 会验证）。几何上，过相机中心、平行于这族直线的那条射线，与像平面的交点就是消失点。地面上所有方向的消失点落在同一条线上，即地平线（地面的消失线）；地平线是地面的无穷远线在图像里的像。

下图左边是三维场景：地面、一座"房子"、相机和三条过相机中心的射线（沿 $x,y,z$ 方向），射线与像平面的交点就是消失点。右边是相机拍到的图像。地平线画的是 $K^{-\top}\mathbf n_c$（$\mathbf n_c$ 为地面法向在相机系中的方向）；读数里再用两个消失点的叉积算一遍，两者一致，即使其中一个消失点跑到无穷远，叉积公式依然成立。

All parallel lines of 3D direction $\mathbf d$ have direction $R_{cw}\mathbf d$ in camera coordinates. Treat the direction as the homogeneous point $(\mathbf d,0)$ with $w=0$ and let the projection act on it: the common image of the family is the vanishing point $\tilde{\mathbf v}=K R_{cw}\mathbf d$ (FIG 5 verifies this). Geometrically, the ray through the camera center parallel to the family pierces the image plane at the vanishing point. The vanishing points of all directions in the ground plane lie on one line, the horizon (the ground's vanishing line), which is the image of the ground plane's line at infinity.

On the left is the 3D scene: the ground, a "house", the camera, and three rays through the camera center (along $x,y,z$) whose intersections with the image plane are the vanishing points. On the right is what the camera sees. The horizon is drawn as $K^{-\top}\mathbf n_c$ ($\mathbf n_c$ the ground normal in the camera frame); the readout recomputes it as the cross product of two vanishing points and the two agree, and the cross product keeps working even when one vanishing point goes to infinity.

FIG 3消失点与地平线Vanishing points and the horizon

三维场景（可旋转）3D SCENE (drag to orbit)

相机图像（u 向右，v 向下）CAMERA IMAGE (u right, v down)

相机在以原点为中心、半径 7.5 m 的圆周上，朝向原点，再向下俯仰。图像为 640×480，主点在中心。红、绿、蓝是沿 $x,y,z$ 的平行线族；明亮部分在图像内，暗淡部分在图像外；琥珀色虚线是地平线。消失点在画面外时，边缘的箭头给出方向。图中的虚线是房子的边向各自消失点的延长。 The camera sits on a circle of radius 7.5 m around the origin, faces the origin, and pitches down. The image is 640×480 with the principal point at the center. Red, green, blue are the parallel families along $x,y,z$; bright inside the image, faint outside; the amber dashed line is the horizon. A vanishing point off-screen is shown as an arrow on the border. The dashed extensions are the house's edges continued to their vanishing points.

把方位角调到 0°，让相机正对房子的一面。红色 $x$ 线在图像里变成什么？$\tilde{\mathbf v}_x$ 的 $w$ 是多少？地平线还能算出来吗？Set the azimuth to 0° so the camera faces one wall squarely. What do the red $x$-lines look like in the image? What is $w$ of $\tilde{\mathbf v}_x$? Can the horizon still be computed?
把俯仰调到 0°：地平线落在图像的哪一行？再把相机升到 4 m 以上，地平线动了吗，地面网格呢？Set pitch to 0°: on which image row does the horizon sit? Then raise the camera above 4 m: did the horizon move, and the ground grid?

两个相互垂直的水平方向（$x$ 和 $y$）的消失点还满足 $\tilde{\mathbf v}_x^\top\omega\,\tilde{\mathbf v}_y=0$，$\omega=K^{-\top}K^{-1}$；主点已知、像素为方形时，这就是一行公式：$f^2=-(\mathbf v_x-\mathbf c)\cdot(\mathbf v_y-\mathbf c)$。读数里的 $\hat f$ 正是这样算出来的，10.4 节会把它推广成标定方法。

The vanishing points of two orthogonal horizontal directions ($x$ and $y$) also satisfy $\tilde{\mathbf v}_x^\top\omega\,\tilde{\mathbf v}_y=0$ with $\omega=K^{-\top}K^{-1}$; with a known principal point and square pixels it becomes one line: $f^2=-(\mathbf v_x-\mathbf c)\cdot(\mathbf v_y-\mathbf c)$. The readout's $\hat f$ is computed exactly this way, and Lesson 10.4 generalizes it into a calibration method.

变换的层级：保持什么，丢掉什么The hierarchy of transformations: what is kept, what is lost

平面上的 3×3 可逆矩阵 $H$ 作用在齐次点上，叫射影变换（单应）。限制 $H$ 的形状，得到越来越"刚"的子类：欧氏（刚体）→ 相似 → 仿射 → 射影。每放松一级，多丢掉一些几何性质，也多出几个自由度。下图让你亲手操作这四级，并实时测量哪些量还保持不变。最右的"四角点"模式直接拖动正方形的四个角，由 DLT 反求 $H$，那就是 10.5 节的单应。

A 3×3 invertible matrix $H$ acting on homogeneous points of the plane is a projective transformation (homography). Restricting the shape of $H$ gives ever more "rigid" subclasses: Euclidean (rigid) → similarity → affine → projective. Each relaxation gives up some geometric property and gains some degrees of freedom. Operate all four classes below and watch which quantities survive, measured live. The rightmost "4 corners" mode drags the four corners of the square directly and recovers $H$ by DLT: that is the homography of Lesson 10.5.

FIG 42D 变换的层级Hierarchy of 2D transformations

虚线是原图，实线是 $H$ 作用后的图像。两个直角标记 L₁、L₂ 的臂长都是 0.4（红臂沿 $x$，绿臂沿 $y$）；圆变成了圆、椭圆或一般圆锥曲线；琥珀色的四个点在 $x$ 轴上。紫色虚线是原图左右两条竖边的像，两线交点是消失点；琥珀色长虚线是 $H^{-\top}\mathbf l_\infty$，即无穷远线的像。表中 ✓ 表示该量保持不变（相对误差 < 0.1%）。"四角点"模式下拖动四个圆圈。 Dashed is the source, solid is the image under $H$. The two right-angle marks L₁, L₂ have arms of length 0.4 (red along $x$, green along $y$); the circle becomes a circle, an ellipse or a general conic; the four amber dots lie on the $x$-axis. The purple dashed lines are the images of the square's left and right sides and their intersection is a vanishing point; the long amber dashed line is $H^{-\top}\mathbf l_\infty$, the image of the line at infinity. A ✓ in the table means the quantity is preserved (relative error < 0.1%). In "4 corners" mode drag the four circles.

选"仿射"，把"剪切 k"拉到 0.5 以上：哪几行变成 ✗？平行性呢？Choose "Affine" and pull the shear k above 0.5: which rows turn ✗? What about parallelism?
选"射影"，调 $v_1$ 或 $v_2$ 使左右两边在图像里交于有限远的一点（距离 < 8）。这个点是 $H$ 作用下哪两条线的交点？地平线在哪？Choose "Projective" and adjust $v_1$ or $v_2$ so the left and right sides meet at a finite point (distance < 8). Which two lines is that the intersection of? Where is the horizon?
选"四角点"，把正方形拖成梯形。哪几行永远不会变成 ✗？Choose "4 corners" and drag the square into a trapezoid. Which rows can never turn ✗?

一个 3-向量，四种身份One 3-vector, four identities

同一个三元组可以是：图像点（$w\neq0$），方向/消失点（$w=0$），直线（用 $\mathbf l^\top\tilde{\mathbf x}=0$ 的方式读），或者相机坐标系里的射线（$K^{-1}\tilde{\mathbf x}$，10.1 节的反投影）。区别在于怎么变换：点用 $H$，线用 $H^{-\top}$，射线用 $K^{-1}$。写代码时给它们分别起名字（甚至用不同的类型），能避免一大类 bug。

The same triple can be an image point ($w\neq0$), a direction/vanishing point ($w=0$), a line (read through $\mathbf l^\top\tilde{\mathbf x}=0$), or a ray in the camera frame ($K^{-1}\tilde{\mathbf x}$, the back-projection of Lesson 10.1). What differs is how it transforms: points by $H$, lines by $H^{-\top}$, rays by $K^{-1}$. Giving them different names (even different types) in code prevents a whole class of bugs.

5再看数学The math

提示：悬停在公式中带底色的符号上，FIG 1 和 FIG 2 里对应的元素会被高亮。Tip: hover the highlighted symbols in the formulas; the matching elements in FIG 1 and FIG 2 light up.

齐次坐标与无穷远点Homogeneous coordinates and points at infinity

射影平面是非零 3-向量模掉"相差一个非零倍数"的等价关系：

The projective plane is the set of nonzero 3-vectors modulo "differ by a nonzero multiple":

$$ \begin{gathered}\mathbb P^2=(\R^3\setminus\{\mathbf 0\})/\!\sim,\qquad \hl{x}{\tilde{\mathbf x}}\sim\hl{lam}{\lambda}\tilde{\mathbf x}\ \ (\lambda\neq0)\\[4pt] (x,y)\ \mapsto\ (x,y,1),\qquad (x,y,\hl{w}{w})\ \mapsto\ \Big(\tfrac{x}{w},\tfrac{y}{w}\Big)\ \ (w\neq0).\end{gathered} $$

三个数只有两个自由度（去掉一个整体比例）。$w=0$ 的向量 $(x,y,0)$ 没有对应的普通点，它们是无穷远点，编码方向 $(x,y)$，也就是 5.3 节里"方向用 $(\mathbf d,0)$"的那个 0。全体无穷远点构成一条线 $\mathbf l_\infty=(0,0,1)$：$\mathbf l_\infty^\top\tilde{\mathbf x}=w$。注意 $\lambda$ 可以是负数，$\tilde{\mathbf x}$ 与 $-\tilde{\mathbf x}$ 是同一个点。

Three numbers carry only two degrees of freedom (one overall scale is removed). Vectors $(x,y,0)$ with $w=0$ have no ordinary counterpart: they are the points at infinity, encoding a direction $(x,y)$, the very 0 of "directions are $(\mathbf d,0)$" in Lesson 5.3. All points at infinity form one line, $\mathbf l_\infty=(0,0,1)$, since $\mathbf l_\infty^\top\tilde{\mathbf x}=w$. Note that $\lambda$ may be negative: $\tilde{\mathbf x}$ and $-\tilde{\mathbf x}$ are the same point.

直线、连线与求交Lines, join and meet

直线 $ax+by+c=0$ 记作 $\hl{l}{\mathbf l}=(a,b,c)$，同样只差一个比例。$(a,b)$ 是法向，当 $\|(a,b)\|=1$ 时 $\mathbf l^\top\tilde{\mathbf x}$（$w=1$）就是点到直线的带符号距离。点与线的关联、连线、求交：

The line $ax+by+c=0$ is written $\hl{l}{\mathbf l}=(a,b,c)$, again only up to scale. $(a,b)$ is the normal, and when $\|(a,b)\|=1$, $\mathbf l^\top\tilde{\mathbf x}$ (with $w=1$) is the signed distance from the point to the line. Incidence, join and meet:

$$ \mathbf l^\top\tilde{\mathbf x}=0,\qquad \mathbf l=\tilde{\mathbf p}\times\tilde{\mathbf q}\ \ (\text{join}),\qquad \hl{x}{\tilde{\mathbf x}}=\hl{l}{\mathbf l}\times\hl{m}{\mathbf m}\ \ (\text{meet}). $$

验证：$\tilde{\mathbf p}^\top(\tilde{\mathbf p}\times\tilde{\mathbf q})=0$，$\tilde{\mathbf q}^\top(\tilde{\mathbf p}\times\tilde{\mathbf q})=0$（叉积垂直于两个因子，1.8 节）。当 $\tilde{\mathbf p}\sim\tilde{\mathbf q}$ 时叉积为 0，连线不唯一；当 $\mathbf l\sim\mathbf m$ 时同理。这是唯一的退化情形，不再需要"平行""竖直"之类的分类。两条直线 $ax+by+c_1=0$ 与 $ax+by+c_2=0$ 平行，叉积为 $(c_2-c_1)\,(b,\,-a,\,0)$，正是方向 $(b,-a)$ 的无穷远点。叉积也是线性的：用 1.8 节的反对称矩阵（第 5 章里的 hat 映射）写成

Check: $\tilde{\mathbf p}^\top(\tilde{\mathbf p}\times\tilde{\mathbf q})=0$ and $\tilde{\mathbf q}^\top(\tilde{\mathbf p}\times\tilde{\mathbf q})=0$ (the cross product is perpendicular to both factors, Lesson 1.8). If $\tilde{\mathbf p}\sim\tilde{\mathbf q}$ the cross product vanishes and the join is not unique; likewise for $\mathbf l\sim\mathbf m$. That is the only degenerate case: no more "parallel" or "vertical" cases. For $ax+by+c_1=0$ and $ax+by+c_2=0$ the cross product is $(c_2-c_1)\,(b,\,-a,\,0)$, exactly the point at infinity in direction $(b,-a)$. The cross product is also linear: with the skew-symmetric matrix of Lesson 1.8 (the hat map of Chapter 5)

$$ \mathbf a\times\mathbf b=[\mathbf a]_\times\mathbf b=\begin{bmatrix}0&-a_3&a_2\\ a_3&0&-a_1\\ -a_2&a_1&0\end{bmatrix}\mathbf b,\qquad \mathbf a\times\mathbf b=\mathbf 0\iff\mathbf a\sim\mathbf b. $$

于是"两个齐次向量表示同一点"可以写成线性方程 $[\mathbf a]_\times\mathbf b=\mathbf 0$。这是比较齐次向量的正确方式，也是 DLT 的出发点（见下）。

So "two homogeneous vectors represent the same point" becomes the linear equation $[\mathbf a]_\times\mathbf b=\mathbf 0$: the right way to compare homogeneous vectors, and the starting point of the DLT (below).

点怎么变换，线怎么变换How points transform, and how lines transform

可逆 3×3 矩阵 $H$ 把点映成点：$\tilde{\mathbf x}'=H\tilde{\mathbf x}$。线不能直接用 $H$，而要用 $H^{-\top}$。推导只要一行：点在线上的关系必须被保持，$\mathbf l'^\top\tilde{\mathbf x}'=0$。

An invertible 3×3 matrix $H$ maps points to points: $\tilde{\mathbf x}'=H\tilde{\mathbf x}$. Lines cannot use $H$ directly but need $H^{-\top}$. The derivation is one line: incidence must be preserved, $\mathbf l'^\top\tilde{\mathbf x}'=0$.

$$ \begin{gathered}0=\mathbf l^\top\tilde{\mathbf x}=\mathbf l^\top H^{-1}H\tilde{\mathbf x}=\big(H^{-\top}\mathbf l\big)^{\!\top}\tilde{\mathbf x}'\quad\Longrightarrow\quad \mathbf l'=H^{-\top}\mathbf l,\\[4pt] (H\mathbf a)\times(H\mathbf b)=\det(H)\,H^{-\top}(\mathbf a\times\mathbf b).\end{gathered} $$

第二个恒等式说：先变换两点再连线，等于先连线再用 $H^{-\top}$ 变换。FIG 3 里的地平线是它的一个实例：$K\mathbf r_1\times K\mathbf r_2=\det K\cdot K^{-\top}(\mathbf r_1\times\mathbf r_2)=\det K\cdot K^{-\top}\mathbf r_3$。对二次曲线 $\tilde{\mathbf x}^\top C\tilde{\mathbf x}=0$，同理 $C'=H^{-\top}CH^{-1}$（读数里不用，只作指针）。若 $H$ 是纯旋转（$R^{-\top}=R$），线和点用同一个矩阵变换，你不会发现差别；刚体运动下的法向量也只按 $R$ 旋转，同样看不出来。但齐次形式的平面在 $T\in\mathrm{SE}(3)$ 下仍要用 $T^{-\top}$（平移会改变平面的常数项）；一旦出现缩放、剪切或透视，差别立刻显现（见"误区 3"）。

The second identity says that transforming two points and then joining them equals joining first and transforming with $H^{-\top}$. The horizon in FIG 3 is an instance: $K\mathbf r_1\times K\mathbf r_2=\det K\cdot K^{-\top}(\mathbf r_1\times\mathbf r_2)=\det K\cdot K^{-\top}\mathbf r_3$. For a conic $\tilde{\mathbf x}^\top C\tilde{\mathbf x}=0$ the same argument gives $C'=H^{-\top}CH^{-1}$ (not used in the readouts, only a pointer). If $H$ is a pure rotation ($R^{-\top}=R$), lines and points transform by the same matrix and you never notice the difference; normal vectors under a rigid motion are also just rotated by $R$, which hides it too. A homogeneous plane under $T\in\mathrm{SE}(3)$ still needs $T^{-\top}$ (the translation changes the plane's constant term); scaling, shear or perspective make it appear at once (see Pitfall 3).

层级、自由度与不变量Hierarchy, degrees of freedom and invariants

类别class	H	自由度DoF	保持不变的量invariants
欧氏Euclidean	$\begin{bmatrix}R&\mathbf t\\ \mathbf 0^\top&1\end{bmatrix}$	3	长度、角度、面积length, angle, area
相似Similarity	$\begin{bmatrix}sR&\mathbf t\\ \mathbf 0^\top&1\end{bmatrix}$	4	角度、长度之比angle, ratio of lengths
仿射Affine	$\begin{bmatrix}A&\mathbf t\\ \mathbf 0^\top&1\end{bmatrix}$	6	平行性、平行线段的长度比、面积比；$\mathbf l_\infty$ 映到自身parallelism, ratio of lengths along parallel lines, area ratio; $\mathbf l_\infty$ maps to itself
射影Projective	$\begin{bmatrix}A&\mathbf t\\ \mathbf v^\top&1\end{bmatrix}$	8	关联性（共线、共点）、交比incidence (collinearity, concurrency), cross-ratio

射影类的 $H$ 有 9 个元素，去掉整体比例剩 8 个自由度。仿射类的最后一行是 $(0,0,1)$，$\mathbf l_\infty$ 被映到自己，所以平行线仍然平行；射影类的 $\mathbf v\neq\mathbf 0$，$\mathbf l_\infty$ 被映成有限远的一条线 $H^{-\top}\mathbf l_\infty$，这就是地平线（FIG 4 的琥珀色长虚线）。仿射部分 $A$ 可以分解为旋转、各向异性缩放、旋转（SVD，1.12 节）；FIG 4 中用 $R(\theta)\begin{bmatrix}s_x&k\\0&s_y\end{bmatrix}$ 参数化。

The projective class has 9 entries minus one overall scale: 8 degrees of freedom. In the affine class the last row is $(0,0,1)$, so $\mathbf l_\infty$ maps to itself and parallel lines stay parallel; in the projective class $\mathbf v\neq\mathbf 0$, and $\mathbf l_\infty$ maps to a finite line $H^{-\top}\mathbf l_\infty$, which is the horizon (the long amber dashed line in FIG 4). The affine part $A$ decomposes as rotation, anisotropic scaling, rotation (SVD, Lesson 1.12); FIG 4 parametrizes it as $R(\theta)\begin{bmatrix}s_x&k\\0&s_y\end{bmatrix}$.

交比：射影变换下仍然不变的"比"The cross-ratio: a ratio that survives projective maps

长度比在射影变换下不保持，但共线四点的交比保持。设四点在其所在直线上的一维坐标为 $t_1,\dots,t_4$：

Ratios of lengths are not preserved by projective maps, but the cross-ratio of four collinear points is. With the 1D coordinates $t_1,\dots,t_4$ of the points along their line:

$$ \operatorname{Cr}(t_1,t_2;t_3,t_4)=\frac{(t_1-t_2)(t_3-t_4)}{(t_1-t_3)(t_2-t_4)}. $$

一维射影变换是 $t\mapsto(\alpha t+\beta)/(\gamma t+\delta)$，代入后 $t_i'-t_j'=\det\cdot(t_i-t_j)/\big((\gamma t_i+\delta)(\gamma t_j+\delta)\big)$，分子分母里的 $\det$ 与 $(\gamma t_i+\delta)$ 恰好抵消。交比是"图像里的尺子"：已知三个点的真实间距，可以由图像上的四个点反推第四个点的位置（例如算出路面上某个标记的实际距离）。

A 1D projective map is $t\mapsto(\alpha t+\beta)/(\gamma t+\delta)$; substituting gives $t_i'-t_j'=\det\cdot(t_i-t_j)/\big((\gamma t_i+\delta)(\gamma t_j+\delta)\big)$, and the determinant and the factors $(\gamma t_i+\delta)$ cancel between numerator and denominator. The cross-ratio is a "ruler in the image": knowing the true spacing of three points, four image points give the position of the fourth (for example the real distance of a marking on the road).

相机是 $\mathbb P^3\to\mathbb P^2$ 的线性映射The camera is a linear map $\mathbb P^3\to\mathbb P^2$

三维点也有齐次坐标 $\tilde{\mathbf X}=(\mathbf X,1)\in\mathbb P^3$，平面是 4-向量 $\boldsymbol\pi$（$\boldsymbol\pi^\top\tilde{\mathbf X}=0$），无穷远平面是 $\boldsymbol\pi_\infty=(0,0,0,1)$。10.3 节的相机矩阵 $P=K[R\,|\,\mathbf t]$ 是 3×4 矩阵：

3D points have homogeneous coordinates $\tilde{\mathbf X}=(\mathbf X,1)\in\mathbb P^3$, planes are 4-vectors $\boldsymbol\pi$ ($\boldsymbol\pi^\top\tilde{\mathbf X}=0$), and the plane at infinity is $\boldsymbol\pi_\infty=(0,0,0,1)$. The camera matrix $P=K[R\,|\,\mathbf t]$ of Lesson 10.3 is a 3×4 matrix:

$$ \begin{gathered}\lambda\tilde{\mathbf x}=P\tilde{\mathbf X}=K\big[R\,|\,\mathbf t\big]\begin{bmatrix}\mathbf X\\1\end{bmatrix}\ \ (\mathbf t={}^{c}\mathbf t_{w}),\qquad P\begin{bmatrix}\mathbf d\\0\end{bmatrix}=KR\mathbf d,\\[4pt] P\tilde{\mathbf C}=\mathbf 0\ \Longrightarrow\ \tilde{\mathbf C}=\begin{bmatrix}-R^\top\mathbf t\\1\end{bmatrix}.\end{gathered} $$

第二个式子：无穷远点 $(\mathbf d,0)$ 的像只与 $KR$ 有关，与平移无关，这就是消失点公式。第三个式子：$P$ 的秩为 3，所以有一维零空间；相机中心是唯一被映成 $(0,0,0)$（不是任何像点）的点，因此相机中心 = $P$ 的右零向量（FIG 5 用 SVD 数值验证，1.12 节）。这里的 $\lambda$ 是投影深度（带正负号：为负表示点在相机后面），所以在相机里符号是有意义的，下面"误区 2"会再谈。

The second formula: the image of a point at infinity $(\mathbf d,0)$ depends on $KR$ only, not on translation, which is the vanishing-point formula. The third: $P$ has rank 3, so its null space is one-dimensional; the camera center is the only point mapped to $(0,0,0)$ (not an image point), hence camera center = right null vector of $P$ (verified numerically with an SVD in FIG 5, Lesson 1.12). Here $\lambda$ is the projective depth (signed: negative means the point is behind the camera), so in a camera the sign does carry meaning; Pitfall 2 returns to this.

为什么 DLT 是零空间问题Why the DLT is a null-space problem

已知对应点 $\tilde{\mathbf x}\leftrightarrow\tilde{\mathbf x}'$ 求 $H$。不能写 $\tilde{\mathbf x}'=H\tilde{\mathbf x}$（只差比例，$\lambda$ 未知），而应写"平行"条件 $\tilde{\mathbf x}'\times H\tilde{\mathbf x}=\mathbf 0$，它对 $H$ 的 9 个元素 $\mathbf h=(\mathbf h_1;\mathbf h_2;\mathbf h_3)$ 是线性的。设 $\tilde{\mathbf x}'=(x',y',w')$：

Given correspondences $\tilde{\mathbf x}\leftrightarrow\tilde{\mathbf x}'$, find $H$. We cannot write $\tilde{\mathbf x}'=H\tilde{\mathbf x}$ (equal only up to scale, with $\lambda$ unknown); instead we demand that the two vectors be parallel, $\tilde{\mathbf x}'\times H\tilde{\mathbf x}=\mathbf 0$, which is linear in the 9 entries $\mathbf h=(\mathbf h_1;\mathbf h_2;\mathbf h_3)$ of $H$. With $\tilde{\mathbf x}'=(x',y',w')$:

$$ \begin{bmatrix}\mathbf 0^\top&-w'\tilde{\mathbf x}^\top&y'\tilde{\mathbf x}^\top\\ w'\tilde{\mathbf x}^\top&\mathbf 0^\top&-x'\tilde{\mathbf x}^\top\\ -y'\tilde{\mathbf x}^\top&x'\tilde{\mathbf x}^\top&\mathbf 0^\top\end{bmatrix}\begin{bmatrix}\mathbf h_1\\ \mathbf h_2\\ \mathbf h_3\end{bmatrix}=\mathbf 0. $$

三行里只有两行线性无关（叉积的三个分量满足一个线性关系），所以每对点提供 2 个方程；$H$ 有 8 个自由度，因此4 对点（任意三点不共线）刚好够。把 $n$ 对点的方程叠成 $A\mathbf h=\mathbf 0$，$\mathbf h$ 是 $A$ 的零空间，噪声下取最小奇异值对应的右奇异向量（1.12 节）。因为未知量只定义到比例，必须约束 $\|\mathbf h\|=1$，而不是 $h_{33}=1$（后者在 $h_{33}\approx0$ 时出问题）。同样的做法对相机矩阵 $P$（11 个自由度，每个 3D–2D 点对 2 个方程，至少 6 对点，10.4 节）、基础矩阵 $F$（$\tilde{\mathbf x}'^\top F\tilde{\mathbf x}=0$，一个线性方程，11.1 节）都成立。

Only two of the three rows are linearly independent (the three components of a cross product satisfy one linear relation), so each pair gives 2 equations; $H$ has 8 degrees of freedom, so 4 point pairs (no three collinear) are exactly enough. Stack the equations of $n$ pairs into $A\mathbf h=\mathbf 0$: $\mathbf h$ lies in the null space of $A$, and with noise one takes the right singular vector of the smallest singular value (Lesson 1.12). Because $\mathbf h$ is only defined up to scale, impose $\|\mathbf h\|=1$ rather than $h_{33}=1$ (the latter fails when $h_{33}\approx0$). The same recipe works for the camera matrix $P$ (11 degrees of freedom, 2 equations per 3D–2D pair, at least 6 pairs, Lesson 10.4) and for the fundamental matrix $F$ ($\tilde{\mathbf x}'^\top F\tilde{\mathbf x}=0$, one linear equation per pair, Lesson 11.1).

6机器人中的它：地平线、车道线与"无穷远"In robotics: the horizon, lane lines and "infinity"

想象一台移动机器人（或一辆车）的前视相机，高度 $h$，向下俯仰 $\theta$。两条车道线平行，向前延伸到无穷远。把车道线上的点写成 $\tilde{\mathbf X}(w)=(\mathbf d+w\mathbf P_0,\ w)$，$w=1/s$ 是"离得多远"的倒数：

Imagine the front camera of a mobile robot (or a car), at height $h$, pitched down by $\theta$. Two lane lines run parallel into the distance. Write a point on a lane line as $\tilde{\mathbf X}(w)=(\mathbf d+w\mathbf P_0,\ w)$, where $w=1/s$ is the reciprocal of "how far away":

$$ \tilde{\mathbf x}(w)=P\tilde{\mathbf X}(w)=\cx{KR\mathbf d}+w\,P\tilde{\mathbf P}_0,\qquad \tilde{\mathbf P}_0=(\mathbf P_0,1). $$

$w$ 取 1 时是近处的点；$w\to0$ 时，点飞向无穷远，像素 $\tilde{\mathbf x}(w)$ 沿一条直线收敛到 $KR\mathbf d$，也就是消失点。它在 $w=0$ 处平滑地"到达"，因为齐次坐标把"走到无穷远"变成了 $w=0$ 这个普通数值。拖动距离滑块，再改变车道的坡度和相机的俯仰：平路的消失点落在地平线上，上坡的消失点跑到地平线之上。左边同时用 SVD 数值求出 $P$ 的零空间，核对相机中心。

At $w=1$ the point is near; as $w\to0$ it flies to infinity, and the pixel $\tilde{\mathbf x}(w)$ converges along a straight line to $KR\mathbf d$, the vanishing point. It "arrives" smoothly at $w=0$ because homogeneous coordinates turn "going to infinity" into the ordinary number $w=0$. Drag the distance slider, then change the lane's grade and the camera pitch: the vanishing point of a flat road sits on the horizon, that of an uphill road lies above it. An SVD of $P$ finds its null space numerically on the left, to check the camera center.

FIG 5相机矩阵 P：无穷远点的像是消失点The camera matrix P: the image of a point at infinity is the vanishing point

三维：机器人相机与车道（可旋转）3D · ROBOT CAMERA AND LANE (drag to orbit)

图像（v 向下）IMAGE (v down)

相机在原点正上方 $h$ 米处，朝 $+y$ 看，俯仰 $\theta$，焦距 500 px。两条车道 $x=\pm1.8$ m，方向 $\mathbf d=(0,1,g)$（$g$ 为坡度）；琥珀色点是右车道上 $\mathbf P_0+\mathbf d/w$ 处的点。右图中的细线是像素随 $w$ 从 1 到 0 走过的路径，它是一条直线。"零空间"一行是 $\operatorname{null}(P)$ 的 SVD 结果，应与相机中心一致。 The camera sits $h$ metres straight above the origin, looks along $+y$, pitched by $\theta$, with focal length 500 px. The two lanes are $x=\pm1.8$ m with direction $\mathbf d=(0,1,g)$ ($g$ the grade); the amber dot is the point $\mathbf P_0+\mathbf d/w$ on the right lane. The thin line in the image is the path the pixel takes as $w$ goes from 1 to 0; it is a straight line. The "null space" row is the SVD of $P$, which should agree with the camera center.

把距离拖到 $w\le0.01$（100 m 以外）：点距离消失点还有多少像素？它还在近处的车道线上吗？Drag the distance to $w\le0.01$ (beyond 100 m): how many pixels is the point from the vanishing point? Is the pixel still on the lane's image line?
把坡度 $g$ 调到 0.1 以上（约 6° 上坡）：消失点和地平线的关系变成了什么？这对"用消失点估计俯仰角"意味着什么？Raise the grade $g$ above 0.1 (about 6° uphill): how do the vanishing point and horizon relate now? What does that mean for "estimating pitch from the vanishing point"?

真实系统里In real systems

车道线与路面平面：车道检测常用"逆透视映射"（IPM）把前视图变成俯视图，它就是一个由相机高度、俯仰和内参决定的单应；地平线决定了哪里是路面的尽头。由消失点估计姿态：室内走廊和楼宇环境（Manhattan 世界）里，三个正交方向的消失点给出相机的旋转，也能给出焦距，用于无标定相机的初始化。基准标记：AprilTag、ArUco 的角点与标记平面之间是一个单应，OpenCV 的 findHomography、warpPerspective、solvePnP 都建立在这套齐次坐标的工具上。SLAM 前端：ORB-SLAM 一类系统的单目初始化会同时估计单应和基础矩阵，并用 DLT 三角化地图点，这些都是本课的"零空间问题"；OpenCV 提供 convertPointsToHomogeneous / convertPointsFromHomogeneous，Eigen 则是 .homogeneous() / .hnormalized()。图形学：GPU 里的裁剪坐标 $(x,y,z,w)$ 与"透视除法"正是同一件事（12.3 节）。tf / SE(3)：ROS tf 里的坐标变换（常被当作 4×4 矩阵处理）是 5.3 节的刚体变换，其中的 1 不能随意缩放；这里的齐次坐标则允许任意缩放，二者形式相似，含义不同。

Lane lines and the road plane: lane detection often uses "inverse perspective mapping" (IPM) to turn the front view into a bird's-eye view, which is a homography fixed by camera height, pitch and intrinsics; the horizon tells where the road ends. Attitude from vanishing points: in corridors and buildings (the Manhattan world) three orthogonal vanishing points give the camera rotation and even the focal length, useful for initializing uncalibrated cameras. Fiducial markers: between the corners of an AprilTag or ArUco marker and the marker plane there is a homography; OpenCV's findHomography, warpPerspective and solvePnP are all built on this homogeneous toolbox. SLAM front ends: the monocular initialization of ORB-SLAM-style systems estimates a homography and a fundamental matrix side by side and triangulates map points with a DLT, all of them this lesson's "null-space problems"; OpenCV offers convertPointsToHomogeneous / convertPointsFromHomogeneous, and Eigen has .homogeneous() / .hnormalized(). Graphics: the clip coordinates $(x,y,z,w)$ of a GPU and its "perspective divide" are the same thing (Lesson 12.3). tf / SE(3): the coordinate transforms in ROS's tf (often handled as 4×4 matrices) are the rigid transforms of Lesson 5.3, whose trailing 1 must not be rescaled; homogeneous coordinates here may be scaled freely. The forms look alike, the meanings differ.

7常见误区Common pitfalls

误区 1：随手除以 wPitfall 1: dividing by w without looking

$w=0$ 是合法的点（无穷远点），$w\approx0$ 的点除出来是一个巨大的数，误差被整体放大。尽量把运算留在齐次形式里，最后才除，并在除之前检查 $|w|$ 相对于 $\|\tilde{\mathbf x}\|$ 是否足够大（FIG 2 的判据就是这样写的）。更微妙的是：射影变换不保持"线段"。一条线段若跨过 $w=0$（比如 FIG 4 里把 $v_1$ 拉得很大），它的像不是一段，而是穿过无穷远的两段，画图时必须在 $w$ 变号处断开。

$w=0$ is a legitimate point (at infinity), and for $w\approx0$ the quotient is a huge number that amplifies every error. Keep computations homogeneous as long as possible, divide last, and before dividing check that $|w|$ is large relative to $\|\tilde{\mathbf x}\|$ (which is how FIG 2's test is written). More subtly, projective maps do not preserve "segments": a segment that crosses $w=0$ (pull $v_1$ far in FIG 4) has an image that is not one piece but two pieces joined through infinity, so plotting code must break the curve where $w$ changes sign.

误区 2：把"相等"当成"相等到比例"Pitfall 2: comparing homogeneous vectors for equality instead of equality up to scale

$(2,4,2)$ 与 $(1,2,1)$ 是同一个点，$(-1,-2,-1)$ 也是，所以不能直接逐元素比较，也不能用 allclose。正确的做法是各自归一化后比较（并处理符号），或检查 $\tilde{\mathbf a}\times\tilde{\mathbf b}\approx\mathbf 0$。但在相机里符号有物理意义：$P\tilde{\mathbf X}$ 的第三分量为负，说明点在相机后面，却会投影到画面里（针孔相机看不见，而射影几何并不知道）。需要区分前后时，要用"定向射影几何"的思路或单独检查深度（"手性检查"，11.1、11.4 节）。

$(2,4,2)$ and $(1,2,1)$ are the same point, and so is $(-1,-2,-1)$; comparing element by element, or with allclose, is wrong. Normalize both (and fix the sign) before comparing, or check $\tilde{\mathbf a}\times\tilde{\mathbf b}\approx\mathbf 0$. In a camera, however, the sign has physical meaning: a negative third component of $P\tilde{\mathbf X}$ means the point is behind the camera, yet it still lands somewhere in the image (a real pinhole cannot see it; projective geometry does not know). To tell front from back, think "oriented projective geometry" or check the depth separately (the "cheirality check" of Lessons 11.1 and 11.4).

误区 3：用 H 去变换线（或法向量）Pitfall 3: transforming lines (or normals) with H

线、平面、法向量要用 $H^{-\top}$。旋转矩阵碰巧满足 $R^{-\top}=R$，所以只做旋转、或者只变换法向量时这个错误不会暴露（带平移的齐次平面照样需要 $T^{-\top}$）；一旦 $H$ 含有缩放、剪切或透视（相机内参 $K$！），错误的线就不再过变换后的点。FIG 4 的"平行性"一行正是用 $H^{-\top}$ 变换左右两条边再求交。验证方法永远是：变换后检查 $\mathbf l'^\top\tilde{\mathbf x}'\approx0$。

Lines, planes and normals need $H^{-\top}$. A rotation happens to satisfy $R^{-\top}=R$, so the mistake stays hidden for pure rotations and for normal vectors (a homogeneous plane under a rigid motion with translation still needs $T^{-\top}$); once $H$ contains scale, shear or perspective (the intrinsics $K$!), a wrongly transformed line no longer passes through the transformed points. The "parallel" row of FIG 4 transforms the left and right sides by $H^{-\top}$ before intersecting them. The check is always: after transforming, verify $\mathbf l'^\top\tilde{\mathbf x}'\approx0$.

误区 4：条件数与几乎平行的线Pitfall 4: conditioning, and nearly parallel lines

像素坐标是 $10^3$ 量级而 $w=1$，叠成 $A\mathbf h=\mathbf 0$ 后各列数量级相差 $10^6$，SVD 解出的 $H$ 精度很差。DLT 之前要做 Hartley 归一化：把点平移到质心、缩放到平均距离 $\sqrt2$（10.5 节），做完再变换回来。另一个陷阱是两条几乎平行的线：它们的叉积很小，交点在很远处，位置对线的微小误差极度敏感（夹角为 $\epsilon$ 时误差放大 $1/\epsilon$ 倍）。消失点估计因此要用很多条线做最小二乘，而不是两条线求交。

Pixel coordinates are of order $10^3$ while $w=1$, so the columns of $A\mathbf h=\mathbf 0$ differ by about $10^6$ in magnitude and the SVD's $H$ is poor. Before a DLT, apply Hartley normalization: translate the points to their centroid and scale to mean distance $\sqrt2$ (Lesson 10.5), then undo it afterwards. Another trap is two nearly parallel lines: their cross product is small, the intersection is far away, and its position is extremely sensitive to tiny errors in the lines (an angle $\epsilon$ amplifies errors by $1/\epsilon$). Vanishing points are therefore estimated by least squares over many lines, not by intersecting two.

误区 5：把两种"齐次"混为一谈Pitfall 5: confusing the two kinds of "homogeneous"

5.3 节里 $T\in\mathrm{SE}(3)$ 的 4×4 矩阵，最后一个分量必须保持 1：把 $T$ 乘以 2，就不再是刚体变换。本节的 $H$ 乘以任何非零常数都是同一个变换。反过来，用 SE(3) 的求逆公式 $T^{-1}=[R^\top,-R^\top\mathbf t]$ 去求逆一个射影矩阵是错的，射影矩阵要用通用的 3×3 或 4×4 求逆。相机矩阵 $P$ 的左 3×3 块也不是旋转（它是 $KR$），用 RQ 分解（如 OpenCV 的 decomposeProjectionMatrix）来拆分，而不是把它当成位姿。

The 4×4 matrix of $T\in\mathrm{SE}(3)$ in Lesson 5.3 must keep its last entry equal to 1: doubling $T$ no longer gives a rigid transform. The $H$ of this lesson multiplied by any nonzero constant is the same transformation. Conversely, using the SE(3) inverse formula $T^{-1}=[R^\top,-R^\top\mathbf t]$ on a projective matrix is wrong; invert it with a general 3×3 or 4×4 inverse. And the left 3×3 block of the camera matrix $P$ is not a rotation (it is $KR$): split it with an RQ decomposition (e.g. OpenCV's decomposeProjectionMatrix), not by treating it as a pose.

8小练习Check yourself

1. 点 $(1,2)$ 与 $(3,1)$ 的连线（齐次形式，差一个比例）是？1. What is the line (homogeneous, up to scale) through the points $(1,2)$ and $(3,1)$?

$(1,2,1)\times(3,1,1)=(2\cdot1-1\cdot1,\ 1\cdot3-1\cdot1,\ 1\cdot1-2\cdot3)=(1,2,-5)$，即 $x+2y-5=0$。验证：$1+4-5=0$，$3+2-5=0$。选项 A 是把前两个分量写反了：$(2,1,-5)$ 过 $(1,3)$，不过 $(1,2)$。$(1,2,1)\times(3,1,1)=(2\cdot1-1\cdot1,\ 1\cdot3-1\cdot1,\ 1\cdot1-2\cdot3)=(1,2,-5)$, i.e. $x+2y-5=0$. Check: $1+4-5=0$ and $3+2-5=0$. Option A swaps the first two components: $(2,1,-5)$ passes through $(1,3)$, not $(1,2)$.

2. 直线 $x+2y-5=0$ 与 $x+2y+1=0$ 的交点（齐次坐标，差一个比例）是？2. What is the intersection (homogeneous, up to scale) of the lines $x+2y-5=0$ and $x+2y+1=0$?

$(1,2,-5)\times(1,2,1)=(2\cdot1-(-5)\cdot2,\ (-5)\cdot1-1\cdot1,\ 1\cdot2-2\cdot1)=(12,-6,0)\sim(2,-1,0)$：无穷远点，方向 $(2,-1)$ 正是两条线的公共方向（沿线移动时 $x+2y$ 不变）。选项 B 给出的是法向 $(1,2)$ 而不是方向；选项 D 偷偷把 $w$ 改成了 1。$(1,2,-5)\times(1,2,1)=(2\cdot1-(-5)\cdot2,\ (-5)\cdot1-1\cdot1,\ 1\cdot2-2\cdot1)=(12,-6,0)\sim(2,-1,0)$: a point at infinity whose direction $(2,-1)$ is the common direction of both lines (moving along them keeps $x+2y$ constant). Option B gives the normal $(1,2)$, not the direction; option D quietly changed $w$ to 1.

3. 变换 $H=\operatorname{diag}(2,1,1)$（x 方向放大 2 倍）作用于直线 $x+y-1=0$，即 $\mathbf l=(1,1,-1)$。像是？3. The transformation $H=\operatorname{diag}(2,1,1)$ (stretch $x$ by 2) acts on the line $x+y-1=0$, i.e. $\mathbf l=(1,1,-1)$. What is its image?

$\mathbf l'=H^{-\top}\mathbf l=\operatorname{diag}(\tfrac12,1,1)(1,1,-1)=(\tfrac12,1,-1)$，即 $x/2+y-1=0$。验证：点 $(1,0)\to(2,0)$ 与 $(0,1)\to(0,1)$ 都满足。选项 A 是 $H\mathbf l$，那是把线当成点来变换的典型错误：它不过 $(2,0)$。$\mathbf l'=H^{-\top}\mathbf l=\operatorname{diag}(\tfrac12,1,1)(1,1,-1)=(\tfrac12,1,-1)$, i.e. $x/2+y-1=0$. Check: the points $(1,0)\to(2,0)$ and $(0,1)\to(0,1)$ both satisfy it. Option A is $H\mathbf l$, the classic mistake of transforming a line like a point: it misses $(2,0)$.

4. 相机 $f=500$ px，主点 $(320,240)$。相机坐标系中方向为 $(1,0,2)$ 的一族平行线，消失点的像素坐标是？4. A camera has $f=500$ px and principal point $(320,240)$. A family of parallel lines has direction $(1,0,2)$ in the camera frame. What are the vanishing point's pixel coordinates?

$\tilde{\mathbf v}=K\mathbf d=(500\cdot1+320\cdot2,\ 240\cdot2,\ 2)=(1140,480,2)$，除以 $w=2$：$(570,240)$。选项 C 忘了除以 $w$；选项 B 只除了一半：$500\cdot1+320=820$ 没有除以深度 2；选项 D 只在 $d_z=0$（方向平行于像平面）时才成立，此时 $w=0$。$\tilde{\mathbf v}=K\mathbf d=(500\cdot1+320\cdot2,\ 240\cdot2,\ 2)=(1140,480,2)$, divided by $w=2$: $(570,240)$. Option C forgot to divide by $w$; option B forgot to divide by the depth: $500\cdot1+320=820$; option D holds only when $d_z=0$ (direction parallel to the image plane), where $w=0$.

5. 用 DLT 求平面单应 $H$（8 个自由度），至少需要几对一般位置的对应点？为什么？5. How many point correspondences in general position does the DLT need to find a planar homography $H$ (8 degrees of freedom), and why?

叉积的三个分量只有两个独立，每对点 2 个方程；8 个自由度需要 8 个方程，所以 4 对点（任意三点不共线）。3 对点只有 6 个方程，$A$ 的零空间是 3 维的，$H$ 不唯一；更多的点用于最小二乘抗噪（SVD 的最小奇异值向量）。The three components of the cross product contain only two independent equations, so each pair contributes 2; 8 degrees of freedom need 8 equations, hence 4 pairs (no three collinear). Three pairs give 6 equations, the null space of $A$ is 3-dimensional, and $H$ is not unique; extra points are used by least squares for noise robustness (the smallest-singular-value vector of the SVD).

9延伸阅读Further reading

R. Hartley & A. Zisserman, Multiple View Geometry in Computer Vision, 2nd ed., Cambridge University Press, 2004, Ch. 2 and Ch. 3 — 本课内容的标准出处：射影平面、对偶、变换层级、交比、三维射影几何；第 4 章的 DLT 与归一化the standard source for this lesson: the projective plane, duality, the hierarchy of transformations, the cross-ratio, 3D projective geometry; Ch. 4 for the DLT and normalization
Y. Ma, S. Soatto, J. Košecká & S. S. Sastry, An Invitation to 3-D Vision, Springer, 2004, Ch. 3 — 图像形成与透视投影的齐次坐标表述image formation and the homogeneous-coordinate form of perspective projection
R. Szeliski, Computer Vision: Algorithms and Applications, 2nd ed., Springer, 2022, §2.1 — 几何基元与变换的层级，图文并茂的简明版本，可免费在线阅读geometric primitives and the hierarchy of 2D transformations, a concise illustrated version that is free online
J. Richter-Gebert, Perspectives on Projective Geometry, Springer, 2011 — 纯几何视角的射影几何，适合想理解"为什么对偶、交比成立"的读者projective geometry from the pure-geometry side, for readers who want to see why duality and the cross-ratio hold
J. Stolfi, Oriented Projective Geometry: A Framework for Geometric Computations, Academic Press, 1991 — 区分 $\tilde{\mathbf x}$ 与 $-\tilde{\mathbf x}$ 的"定向"射影几何，对应上面的误区 2the "oriented" version that distinguishes $\tilde{\mathbf x}$ from $-\tilde{\mathbf x}$, related to Pitfall 2 above

类别class	H	自由度DoF	保持不变的量invariants
欧氏Euclidean	\(\begin{bmatrix}R&\mathbf t\\ \mathbf 0^\top&1\end{bmatrix}\)	3	长度、角度、面积length, angle, area
相似Similarity	\(\begin{bmatrix}sR&\mathbf t\\ \mathbf 0^\top&1\end{bmatrix}\)	4	角度、长度之比angle, ratio of lengths
仿射Affine	\(\begin{bmatrix}A&\mathbf t\\ \mathbf 0^\top&1\end{bmatrix}\)	6	平行性、平行线段的长度比、面积比；\(\mathbf l_\infty\) 映到自身parallelism, ratio of lengths along parallel lines, area ratio; \(\mathbf l_\infty\) maps to itself
射影Projective	\(\begin{bmatrix}A&\mathbf t\\ \mathbf v^\top&1\end{bmatrix}\)	8	关联性（共线、共点）、交比incidence (collinearity, concurrency), cross-ratio