Part I · 第 1 章线性代数 · 1.9Part I · Ch. 1 Linear Algebra · 1.9

投影与正交分解Projection & Orthogonal Decomposition

任何向量都能拆成两份：一份沿着某个方向，一份垂直于它。沿着的那份就是投影，它落在离原向量最近的那个点上。机器人用它回答"我偏离路径多远"、"这股力有多少压进表面、有多少在滑"，最小二乘也只是它换了件衣服。 Any vector splits into two parts: one along a direction and one perpendicular to it. The part along is the projection, and it lands on the closest point. Robots use it to answer "how far off the path am I?" and "how much of this force presses into the surface and how much slides?", and least squares is just projection in disguise.

约 30 分钟~30 min 前置：1.3 矩阵即变换 · 1.7 点积Prereqs: 1.3 Matrices as transformations · 1.7 Dot product 8 个交互8 interactives

1为什么机器人需要它Why robots need this

一台移动机器人沿着规划好的路径行驶。控制器每个周期都要回答一个问题：我偏离路径多远，偏向哪一边？这里的"多远"不是离起点多远，而是离路径上最近的那一点多远。找这个最近点，就是把机器人的位置投影到路径上。

同样的拆分在机器人里随处可见：

接触力。手指按在物体表面，总的力有多少"压进"表面（法向力），有多少沿表面"滑动"（切向力）？摩擦锥 (friction cone) 规定：切向力不能超过 μ 倍的法向力，否则就打滑。抓取规划每时每刻都在做这个分解。
速度指令。规划器让机器人径直冲向一堵墙，安全层要把指令里"撞墙"的那一份去掉，只保留沿墙的那一份。被去掉的，就是指令在墙的法向上的投影。
最小二乘（1.13 节）。方程无解时，我们改找离 $\mathbf b$ 最近的 $A\mathbf x$。这恰好是把 $\mathbf b$ 投影到 $A$ 的列空间。

1.7 节里，$\hat{\mathbf u}\cdot\mathbf b$ 告诉我们 $\mathbf b$ 在 $\hat{\mathbf u}$ 方向上有多长。这一节把这个长度变成一支真正的向量，再把剩下的部分也一起管起来。

A mobile robot drives along a planned path. Every control cycle it must answer one question: how far off the path am I, and on which side? "How far" does not mean distance from the start; it means distance to the closest point on the path. Finding that closest point is projecting the robot's position onto the path.

The same split shows up all over robotics:

Contact forces. A finger presses on a surface. How much of the force presses into the surface (normal force) and how much slides along it (tangential force)? The friction cone says the tangential part may not exceed μ times the normal part, or the contact slips. Grasp planners do this split constantly.
Velocity commands. A planner sends the robot straight at a wall. A safety layer removes the "into the wall" part of the command and keeps the part along the wall. What it removes is the command's projection onto the wall's normal.
Least squares (Lesson 1.13). When a system has no solution we look for the $A\mathbf x$ closest to $\mathbf b$. That is exactly the projection of $\mathbf b$ onto the column space of $A$.

In Lesson 1.7, $\hat{\mathbf u}\cdot\mathbf b$ told us how long $\mathbf b$ is in the direction $\hat{\mathbf u}$. This lesson turns that length into a real vector, and takes care of the leftover part too.

2直觉：影子，和最近的那一点Intuition: a shadow, and the closest point

想象地上有一条直线，空中悬着一个点 $\mathbf b$。用手电筒从 $\mathbf b$ 的上方照下来，$\mathbf b$ 会在直线上留下一个影子。光从不同方向照，影子落在不同的地方。

问题是：哪一个影子离 $\mathbf b$ 最近？下面这张图里，拖动滑块改变光线的方向，看影子到 $\mathbf b$ 的距离怎么变。也可以拖动 $\mathbf b$。

Picture a straight line on the ground and a point $\mathbf b$ hanging in the air. Shine a flashlight down onto it and $\mathbf b$ casts a shadow on the line. Light from different directions puts the shadow in different places.

Which shadow is closest to $\mathbf b$? In the figure below, drag the slider to change the direction of the light and watch how the distance from the shadow to $\mathbf b$ changes. You can also drag $\mathbf b$.

FIG 1哪个影子最近？Which shadow is closest?

黄点是影子，青色空心圆是"垂线的脚"。光线斜着照时，影子到 $\mathbf b$ 的路径是一条斜边，一定比垂线更长。只有光线垂直于直线时，影子才和垂线的脚重合，这时出现直角标记。 The amber dot is the shadow and the hollow cyan circle is the "foot of the perpendicular". With slanted light the path from the shadow to $\mathbf b$ is a hypotenuse, so it is always longer than the perpendicular. Only when the light is perpendicular to the line does the shadow land on the foot, and the right-angle mark appears.

把光线调到垂直于直线（滑块读数 0°）。这时影子的距离有什么特别？Set the light perpendicular to the line (slider reads 0°). What is special about the shadow's distance then?

斜边永远比直角边长，所以垂直光线下的影子，就是直线上离 $\mathbf b$ 最近的点。这个点叫 $\mathbf b$ 在直线上的投影，记作 $\mathbf p$。从 $\mathbf p$ 指向 $\mathbf b$ 的那段叫残差 (residual) $\mathbf e = \mathbf b - \mathbf p$，它和直线成直角。于是 $\mathbf b$ 被拆成了两份：

A hypotenuse is always longer than a leg, so the shadow under perpendicular light is the point of the line closest to $\mathbf b$. That point is the projection of $\mathbf b$ onto the line, written $\mathbf p$. The segment from $\mathbf p$ to $\mathbf b$ is the residual $\mathbf e = \mathbf b - \mathbf p$, and it meets the line at a right angle. So $\mathbf b$ has been split into two parts:

$$ \mathbf b \;=\; \underbrace{\cv{\mathbf p}}_{\text{沿直线}} \;+\; \underbrace{\cw{\mathbf e}}_{\text{垂直于直线}} $$

$$ \mathbf b \;=\; \underbrace{\cv{\mathbf p}}_{\text{along the line}} \;+\; \underbrace{\cw{\mathbf e}}_{\text{perpendicular to it}} $$

核心洞见Key insight

投影 = 最近点 = 残差垂直。这三句话说的是同一件事。"最近"是一个要比较距离的优化问题，"垂直"是一个能用点积写出来的方程。把"最近"换成"垂直"，我们就能直接算出答案，不用去试。这个想法会一直用到最小二乘。

Projection = closest point = perpendicular residual. These three phrases say the same thing. "Closest" is an optimization problem that compares distances; "perpendicular" is an equation we can write with a dot product. Trading "closest" for "perpendicular" lets us compute the answer directly instead of searching. The same idea carries all the way to least squares.

3动手玩：投影实验室Play: the projection lab

拖动青色箭头 u 的尖端来转动那条过原点的直线，拖动 b 的尖端移动目标点。直线上那个空心的候选点 c 也能拖动，但它只能在直线上滑动。右下方的小图画出 $|\mathbf b-\mathbf c|$ 随 c 的位置的变化。

Drag the tip of the cyan arrow u to turn the line through the origin, and the tip of b to move the target. The hollow candidate point c on the line is draggable too, but it can only slide along the line. The small plot at the lower right shows $|\mathbf b-\mathbf c|$ against c's position.

FIG 2投影实验室Projection lab

|b − c| 随 c 的位置|b − c| vs. position of c

预设Presets

琥珀色 p 是 b 在直线上的投影，紫色虚线 e 是残差，小方块表示直角。把 c 拖到 p 上：右下小图里，距离曲线的最低点正好在那里，而且它的高度就是 $|\mathbf e|$。曲线是一条双曲线，$|\mathbf b-\mathbf c|^2 = |\mathbf e|^2 + (\text{c 到 p 的距离})^2$。 The amber p is the projection of b on the line, the dashed purple e is the residual, and the small square marks the right angle. Drag c onto p: in the small plot the lowest point of the distance curve sits exactly there, and its height is $|\mathbf e|$. The curve is a hyperbola, $|\mathbf b-\mathbf c|^2 = |\mathbf e|^2 + (\text{distance from c to p})^2$.

试一试Try this

把候选点 c 放到直线上离 b 最近的位置。小图里有什么标记告诉你对了？Put the candidate c at the spot on the line closest to b. What in the small plot confirms you got it?
在 $\mathbf b\ne\mathbf 0$ 的前提下让 p = 0。u 和 b 此时是什么关系？Make p = 0 while $\mathbf b\ne\mathbf 0$. How are u and b related then?
让 p 指向 u 的反方向。提示：看 $\mathbf u\cdot\mathbf b$ 的符号。Make p point opposite to u. Hint: look at the sign of $\mathbf u\cdot\mathbf b$.
让残差的长度正好是 b 的一半：$|\mathbf e| = |\mathbf b|/2$。这时 u 与 b 的夹角是多少度？Make the residual exactly half as long as b: $|\mathbf e| = |\mathbf b|/2$. What angle between u and b does that take?

注意 $\mathbf u$ 只是用来指方向的：把 $\mathbf u$ 拉长或缩短，直线、投影 $\mathbf p$、残差 $\mathbf e$ 都不变。公式里 $\mathbf u$ 出现了两次（分子一次、分母一次），长度正好抵消。

Note that $\mathbf u$ only supplies a direction: stretch or shrink it and the line, the projection $\mathbf p$ and the residual $\mathbf e$ do not change. In the formula $\mathbf u$ appears twice (once on top, once below), so its length cancels.

4第二种视角：投影是一个矩阵A second view: projection is a matrix

投影对 $\mathbf b$ 是线性的：把 $\mathbf b$ 放大 2 倍，投影也放大 2 倍；两个向量之和的投影，等于两个投影之和。根据 1.3 节，每个线性变换都对应一个矩阵，这个矩阵叫投影矩阵 $P$。它对整个平面做同一件事：把每一个点推到直线上最近的位置。

下面用 1.3 节的方式看它。拖动 $\mathbf u$ 的尖端，点"作用一次"，整张网格会被压扁到那条直线上。接着再点一次"再作用一次"，什么都没有发生。

Projection is linear in $\mathbf b$: double $\mathbf b$ and the projection doubles; the projection of a sum is the sum of the projections. By Lesson 1.3 every linear map is a matrix, and this one is called the projection matrix $P$. It does the same thing to the whole plane: push every point to the closest spot on the line.

Let us watch it the way we did in Lesson 1.3. Drag the tip of $\mathbf u$, press "apply once", and the whole grid gets squashed onto the line. Then press "apply again" and nothing happens.

FIG 3把整个平面压到直线上Squashing the plane onto the line

淡色网格是原来的格点，亮色是当前的像。琥珀色点是 b 的像，不带颜色的普通圆点是 b 的原位置（可拖动）。切到 $I-P$，平面被压到垂直于 u 的那条直线上。两种情况下，作用第二次都不会再移动任何点，这就是 $P^2=P$。 The faint grid is the original lattice and the bright one is the current image. The amber dot is the image of b; the plain, uncoloured dot is b's original position (draggable). Switch to $I-P$ and the plane is squashed onto the line perpendicular to u. In both cases a second application moves nothing further: that is $P^2=P$.

先"作用一次"，再点"再作用一次"。数一数有几个点动了。Press "apply once", then "apply again". Count how many points moved.
转动 u 让 $P = \tfrac12\begin{bmatrix}1&1\\1&1\end{bmatrix}$（直线 y = x）。u 的方向是什么？Turn u so that $P = \tfrac12\begin{bmatrix}1&1\\1&1\end{bmatrix}$ (the line y = x). What direction is u?

	$P = \dfrac{\mathbf u\mathbf u^\top}{\mathbf u^\top\mathbf u}$	$I - P$
把平面压到Squashes the plane onto	u 所在的直线the line along u	垂直于 u 的直线the line perpendicular to u
被送到 0 的向量Vectors sent to 0	垂直于 u 的所有向量everything perpendicular to u	u 方向的所有向量everything along u
作用在 b 上得到Applied to b gives	$\cv{\mathbf p}$	$\cw{\mathbf e}$
$P^2$	$P$	$I-P$
$\det$, 秩rank	0, 1	0, 1

把这些性质连在一起想：$P$ 把 $\mathbf b$ 压到直线上之后，信息已经丢掉了垂直方向的那一份，再压一次当然没有新东西可丢。所以 $P^2=P$（幂等，idempotent）。$P$ 还是对称的（$P^\top=P$），这一点会在下一步"垂直"里起作用。最后，平面被压成一条线，面积变成 0，所以 $\det P = 0$，$P$ 的秩是 1，不可逆（1.5 节、1.6 节）：知道了影子，你无法还原出原来的点。

Put these facts together. After $P$ squashes $\mathbf b$ onto the line, the perpendicular part has already been thrown away, so a second squash has nothing new to discard. That is $P^2=P$ (idempotent). $P$ is also symmetric ($P^\top=P$), which matters for the "perpendicular" step in the next sections. Finally, a plane squashed to a line has zero area, so $\det P = 0$, the rank of $P$ is 1, and $P$ is not invertible (Lessons 1.5 and 1.6): given the shadow, you cannot recover the original point.

5第三种视角：投影到平面，以及为什么要正交基A third view: projecting onto a plane, and why orthogonal bases help

在三维里，可以把 $\mathbf b$ 投影到一张过原点的平面上。想法完全一样：平面上离 $\mathbf b$ 最近的点是 $\mathbf p$，残差 $\mathbf e=\mathbf b-\mathbf p$ 垂直于整张平面，也就是和平面的法向量 $\mathbf n$ 平行。设 $\mathbf n$ 是单位向量，残差就是 $\mathbf b$ 在法向上的那一份 $(\mathbf n\cdot\mathbf b)\,\mathbf n$，所以

In 3D we can project $\mathbf b$ onto a plane through the origin. The idea is identical: the closest point of the plane to $\mathbf b$ is $\mathbf p$, and the residual $\mathbf e=\mathbf b-\mathbf p$ is perpendicular to the whole plane, so it is parallel to the plane's normal $\mathbf n$. If $\mathbf n$ is a unit vector, the residual is the part of $\mathbf b$ along the normal, $(\mathbf n\cdot\mathbf b)\,\mathbf n$, so

$$ \cv{\mathbf p} = \mathbf b - (\mathbf n\cdot\mathbf b)\,\mathbf n, \qquad \cw{\mathbf e} = (\mathbf n\cdot\mathbf b)\,\mathbf n . $$

另一种写法是用平面里的两个互相垂直的单位向量 $\mathbf q_1,\mathbf q_2$（正交基）。此时投影就是把两条"沿线投影"加起来：$\mathbf p=(\mathbf q_1\cdot\mathbf b)\mathbf q_1+(\mathbf q_2\cdot\mathbf b)\mathbf q_2$。用滑块倾斜平面、移动 $\mathbf b$，切换两种写法，看它们给出同一个 $\mathbf p$。拖动画面可以转动视角。

The other way to write it uses two perpendicular unit vectors $\mathbf q_1,\mathbf q_2$ in the plane (an orthonormal basis). Then the projection is just two line-projections added together: $\mathbf p=(\mathbf q_1\cdot\mathbf b)\mathbf q_1+(\mathbf q_2\cdot\mathbf b)\mathbf q_2$. Tilt the plane and move $\mathbf b$ with the sliders, switch between the two formulas, and see that they give the same $\mathbf p$. Drag the view to orbit.

FIG 4投影到一张平面Projecting onto a plane

半透明青色方块是平面，紫色箭头 $\mathbf n$ 是它的单位法向量。"法向量"模式画出残差 $\cw{\mathbf e}$ 沿 $\mathbf n$ 的方向；"正交基"模式画出平面内的 $\mathbf q_1,\mathbf q_2$ 以及两段虚线 $(\mathbf q_i\cdot\mathbf b)\mathbf q_i$，它们首尾相接，正好是 $\cv{\mathbf p}$。右侧的勾股数 $|\mathbf b|^2=|\mathbf p|^2+|\mathbf e|^2$ 在任何姿态下都成立。 The translucent cyan square is the plane and the purple arrow $\mathbf n$ is its unit normal. "Normal" mode draws the residual $\cw{\mathbf e}$ along $\mathbf n$; "Orthonormal basis" mode draws $\mathbf q_1,\mathbf q_2$ in the plane and the two dashed legs $(\mathbf q_i\cdot\mathbf b)\mathbf q_i$, which chain head to tail into $\cv{\mathbf p}$. The Pythagoras numbers $|\mathbf b|^2=|\mathbf p|^2+|\mathbf e|^2$ on the right hold for every pose.

让 b 正好躺在平面里（$\mathbf e=\mathbf 0$）。这时 $\mathbf n\cdot\mathbf b$ 是多少？Make b lie exactly in the plane ($\mathbf e=\mathbf 0$). What is $\mathbf n\cdot\mathbf b$ then?
让投影 p = 0（b 不为零）。提示：最容易的情形是水平的平面。Make the projection p = 0 (with b nonzero). Hint: the easiest case is a horizontal plane.

核心洞见Key insight

基向量互相垂直，投影就能一个方向一个方向地分开算。每个 $\mathbf q_i$ 只管自己那条线，互不干扰，把各个投影加起来就行。如果基向量不互相垂直，一个方向上的分量会"漏"到另一个方向里，这个简单的加法就不对了（第 8 节的误区 3 会让你亲眼看到）。所以我们想要正交基。手里只有一组斜着的基怎么办？用 Gram–Schmidt 把它们掰正。

When the basis vectors are perpendicular, projection can be computed one direction at a time. Each $\mathbf q_i$ minds its own line and does not disturb the others, so you just add the projections. If the basis vectors are not perpendicular, the component along one direction "leaks" into the other and this simple sum is wrong (Pitfall 3 below lets you see it happen). That is why we want an orthonormal basis. What if all we have is a slanted pair? Gram–Schmidt straightens it.

Gram–Schmidt：把斜基掰成正交基Gram–Schmidt: straightening a slanted basis

给你两个不垂直的向量 $\mathbf v_1,\mathbf v_2$。做法只有三步，而且每一步都用到本节的投影：

保留 $\mathbf v_1$（方向不变）。
从 $\mathbf v_2$ 里减去它在 $\mathbf v_1$ 上的投影。剩下的残差 $\mathbf w$ 垂直于 $\mathbf v_1$。
把两个向量单位化：$\mathbf q_1=\mathbf v_1/|\mathbf v_1|$，$\mathbf q_2=\mathbf w/|\mathbf w|$。

用步骤按钮一步步走一遍，拖动 $\mathbf v_1,\mathbf v_2$ 的尖端换一对向量。

Given two non-perpendicular vectors $\mathbf v_1,\mathbf v_2$, there are only three steps, and each uses this lesson's projection:

Keep $\mathbf v_1$ (same direction).
Subtract from $\mathbf v_2$ its projection onto $\mathbf v_1$. The residual $\mathbf w$ that is left is perpendicular to $\mathbf v_1$.
Normalise both: $\mathbf q_1=\mathbf v_1/|\mathbf v_1|$, $\mathbf q_2=\mathbf w/|\mathbf w|$.

Step through it with the buttons, and drag the tips of $\mathbf v_1,\mathbf v_2$ to try another pair.

FIG 5Gram–Schmidt 步进器Gram–Schmidt stepper

青色是 $\mathbf v_1$，琥珀色是 $\mathbf v_2$。第 ③ 步里，琥珀色虚线是 $\mathbf v_2$ 在 $\mathbf v_1$ 上的投影，紫色 $\mathbf w$ 是减掉它之后的残差。第 ④ 步里两个单位向量 $\mathbf q_1,\mathbf q_2$ 站在单位圆上，成直角。 Cyan is $\mathbf v_1$ and amber is $\mathbf v_2$. In step ③ the dashed amber segment is the projection of $\mathbf v_2$ onto $\mathbf v_1$, and the purple $\mathbf w$ is what is left after subtracting it. In step ④ the two unit vectors $\mathbf q_1,\mathbf q_2$ stand on the unit circle at a right angle.

走完全部 4 步，得到一对正交单位向量 $\mathbf q_1,\mathbf q_2$。Walk through all 4 steps to get an orthonormal pair $\mathbf q_1,\mathbf q_2$.
让 Gram–Schmidt 失败：把 $\mathbf v_2$ 拖到和 $\mathbf v_1$ 共线，使 $\mathbf w\to\mathbf 0$。为什么这时没法单位化？Make Gram–Schmidt fail: drag $\mathbf v_2$ onto the line of $\mathbf v_1$ so that $\mathbf w\to\mathbf 0$. Why can't it be normalised then?

6再看数学The math

提示：把鼠标悬停在公式中带底色的符号上，FIG 2、FIG 3、FIG 4、FIG 5 中对应的元素会被加粗。Tip: hover the highlighted symbols in the formulas; the matching element in FIG 2, FIG 3, FIG 4 or FIG 5 is thickened.

投影到一条直线：从"残差垂直"推出公式Projection onto a line: from "the residual is perpendicular"

直线上的点都写成 $t\,\mathbf u$。投影 $\mathbf p=t\,\mathbf u$ 的残差 $\mathbf e=\mathbf b-t\,\mathbf u$ 要垂直于 $\mathbf u$，也就是点积为零：

Every point of the line is $t\,\mathbf u$. The residual $\mathbf e=\mathbf b-t\,\mathbf u$ of the projection $\mathbf p=t\,\mathbf u$ must be perpendicular to $\mathbf u$, i.e. have zero dot product:

$$ \hl{u}{\cx{\mathbf u}}^\top\big(\hl{b}{\mathbf b} - t\,\hl{u}{\cx{\mathbf u}}\big) = 0 \quad\Longrightarrow\quad t = \frac{\cx{\mathbf u}^\top\mathbf b}{\cx{\mathbf u}^\top\cx{\mathbf u}} \quad\Longrightarrow\quad \hl{p}{\cv{\mathbf p}} = \frac{\cx{\mathbf u}\cdot\mathbf b}{\cx{\mathbf u}\cdot\cx{\mathbf u}}\,\cx{\mathbf u}, \qquad \hl{e}{\cw{\mathbf e}} = \mathbf b - \cv{\mathbf p}. $$

分母 $\mathbf u\cdot\mathbf u=|\mathbf u|^2$ 保证了 $\mathbf u$ 的长度不影响结果。当 $\mathbf u=\hat{\mathbf u}$ 是单位向量时，分母为 1，公式简化成 1.7 节熟悉的样子：$\mathbf p=(\hat{\mathbf u}\cdot\mathbf b)\,\hat{\mathbf u}$，其中 $\hat{\mathbf u}\cdot\mathbf b$ 是投影的（带符号的）长度。

The denominator $\mathbf u\cdot\mathbf u=|\mathbf u|^2$ makes the length of $\mathbf u$ irrelevant. When $\mathbf u=\hat{\mathbf u}$ is a unit vector the denominator is 1 and we get the familiar form from Lesson 1.7: $\mathbf p=(\hat{\mathbf u}\cdot\mathbf b)\,\hat{\mathbf u}$, where $\hat{\mathbf u}\cdot\mathbf b$ is the signed length of the projection.

投影矩阵The projection matrix

把标量 $\mathbf u^\top\mathbf b$ 挪到 $\mathbf u$ 的右边，立刻看出一个作用在 $\mathbf b$ 上的矩阵：

Move the scalar $\mathbf u^\top\mathbf b$ to the other side of $\mathbf u$ and a matrix acting on $\mathbf b$ appears at once:

$$ \hl{p}{\cv{\mathbf p}} = \frac{\cx{\mathbf u}\,(\cx{\mathbf u}^\top\mathbf b)}{\cx{\mathbf u}^\top\cx{\mathbf u}} = \underbrace{\frac{\cx{\mathbf u}\,\cx{\mathbf u}^\top}{\cx{\mathbf u}^\top\cx{\mathbf u}}}_{\hl{P}{P}}\;\mathbf b, \qquad P^2 = P,\quad P^\top = P,\quad \operatorname{rank}P=1 . $$

$\mathbf u\mathbf u^\top$ 是"列乘行"得到的 $n\times n$ 矩阵（外积）。$P^2=P$ 的证明只有一行：$(\mathbf u\mathbf u^\top)(\mathbf u\mathbf u^\top)=\mathbf u(\mathbf u^\top\mathbf u)\mathbf u^\top=(\mathbf u^\top\mathbf u)\,\mathbf u\mathbf u^\top$，除以 $(\mathbf u^\top\mathbf u)^2$ 就得到 $P$。一个具体的例子：$\mathbf u=(2,1)$ 时

$\mathbf u\mathbf u^\top$ is the $n\times n$ matrix of "a column times a row" (an outer product). The proof of $P^2=P$ is one line: $(\mathbf u\mathbf u^\top)(\mathbf u\mathbf u^\top)=\mathbf u(\mathbf u^\top\mathbf u)\mathbf u^\top=(\mathbf u^\top\mathbf u)\,\mathbf u\mathbf u^\top$, and dividing by $(\mathbf u^\top\mathbf u)^2$ gives $P$. A concrete case: for $\mathbf u=(2,1)$,

$$ P = \frac{1}{5}\begin{bmatrix}4&2\\2&1\end{bmatrix}, \qquad P\begin{bmatrix}3\\4\end{bmatrix} = \frac15\begin{bmatrix}20\\10\end{bmatrix} = \begin{bmatrix}4\\2\end{bmatrix},\qquad (I-P)\begin{bmatrix}3\\4\end{bmatrix} = \begin{bmatrix}-1\\2\end{bmatrix}. $$

检查：$\mathbf p+\mathbf e=(4,2)+(-1,2)=(3,4)=\mathbf b$，而且 $\mathbf e\cdot\mathbf u=-2+2=0$。$P$ 的特征值是 1（沿 $\mathbf u$）和 0（垂直于 $\mathbf u$），1.11 节会回到这一点。

Check: $\mathbf p+\mathbf e=(4,2)+(-1,2)=(3,4)=\mathbf b$, and $\mathbf e\cdot\mathbf u=-2+2=0$. The eigenvalues of $P$ are 1 (along $\mathbf u$) and 0 (perpendicular to $\mathbf u$); Lesson 1.11 returns to this.

正交分解与勾股定理Orthogonal decomposition and Pythagoras

每个 $\mathbf b$ 都能唯一地写成"直线内的一份 + 垂直于直线的一份"。因为 $\mathbf p\cdot\mathbf e=0$，两份的平方长度直接相加：

Every $\mathbf b$ can be written uniquely as "a part in the line plus a part perpendicular to it". Because $\mathbf p\cdot\mathbf e=0$, the squared lengths just add:

$$ \mathbf b = \cv{\mathbf p} + \cw{\mathbf e}, \qquad |\mathbf b|^2 = |\cv{\mathbf p}|^2 + |\cw{\mathbf e}|^2, \qquad |\cw{\mathbf e}|^2 = |\mathbf b|^2 - (\hat{\mathbf u}\cdot\mathbf b)^2. $$

$|\mathbf e|$ 就是 $\mathbf b$ 到直线的距离。在二维里还有一种算法：$|\mathbf e| = |\hat{\mathbf u}\times\mathbf b|$，即 1.8 节里的二维叉积的绝对值（平行四边形面积除以底）。叉积的符号给出 $\mathbf b$ 在直线左边还是右边，这正是 7 节里"有符号的横向误差"。

$|\mathbf e|$ is the distance from $\mathbf b$ to the line. In 2D there is another way to get it: $|\mathbf e| = |\hat{\mathbf u}\times\mathbf b|$, the absolute value of the 2D cross product of Lesson 1.8 (parallelogram area divided by the base). The sign of the cross product tells whether $\mathbf b$ is on the left or the right of the line, which is the "signed cross-track error" of Section 7.

投影到子空间Projection onto a subspace

设 $Q=[\,\mathbf q_1\;\cdots\;\mathbf q_k\,]$ 的列是子空间的一组标准正交基（$Q^\top Q=I$，1.10 节）。把每条线的投影加起来：

Let the columns of $Q=[\,\mathbf q_1\;\cdots\;\mathbf q_k\,]$ be an orthonormal basis of the subspace ($Q^\top Q=I$, Lesson 1.10). Adding up the line projections gives

$$ \hl{p}{\cv{\mathbf p}} = \sum_{i=1}^{k}(\hl{q}{\cx{\mathbf q_i}}\cdot\mathbf b)\,\cx{\mathbf q_i} = QQ^\top\mathbf b, \qquad \hl{P}{P} = QQ^\top, \qquad P^2 = Q(Q^\top Q)Q^\top = P . $$

平面还有一种写法：设 $\hl{n}{\mathbf n}$ 是平面的单位法向量，那么 $P = I-\mathbf n\mathbf n^\top$，即 $\mathbf p=\mathbf b-(\mathbf n\cdot\mathbf b)\,\mathbf n$（FIG 4）。这在机器人里最常用：给一个接触面的法向量，去掉速度或力里"进出表面"的那份。

预告：一般的基。如果子空间的基是矩阵 $A$ 的列，但它们不垂直，投影矩阵变成

For a plane there is also this form: if $\hl{n}{\mathbf n}$ is the plane's unit normal, then $P = I-\mathbf n\mathbf n^\top$, i.e. $\mathbf p=\mathbf b-(\mathbf n\cdot\mathbf b)\,\mathbf n$ (FIG 4). This is the one robots use most: given a contact surface's normal, remove the "into or out of the surface" part of a velocity or force.

Preview: a general basis. If the subspace is spanned by the columns of a matrix $A$ that are not perpendicular, the projection matrix becomes

$$ P = A\,(A^\top A)^{-1}A^\top . $$

当 $A$ 只有一列 $\mathbf u$ 时，$A^\top A=\mathbf u^\top\mathbf u$，回到 $\mathbf u\mathbf u^\top/\mathbf u^\top\mathbf u$；当 $A=Q$ 有标准正交列时，$A^\top A=I$，回到 $QQ^\top$。中间那个 $(A^\top A)^{-1}$ 就是"基不垂直"要付的代价，它会在 1.13 节的最小二乘里成为主角。

With a single column $\mathbf u$, $A^\top A=\mathbf u^\top\mathbf u$ and we recover $\mathbf u\mathbf u^\top/\mathbf u^\top\mathbf u$; with orthonormal columns $A=Q$, $A^\top A=I$ and we recover $QQ^\top$. The middle factor $(A^\top A)^{-1}$ is the price of a basis that is not perpendicular, and it takes center stage in the least squares of Lesson 1.13.

Gram–Schmidt

对线性无关的 $\mathbf v_1,\dots,\mathbf v_k$，依次做：

For linearly independent $\mathbf v_1,\dots,\mathbf v_k$, proceed in order:

$$ \mathbf q_1 = \frac{\mathbf v_1}{|\mathbf v_1|}, \qquad \hl{w}{\cw{\mathbf w_k}} = \mathbf v_k - \sum_{j<k}(\mathbf q_j\cdot\mathbf v_k)\,\mathbf q_j, \qquad \mathbf q_k = \frac{\mathbf w_k}{|\mathbf w_k|}. $$

第 $k$ 步的意思是：先把 $\mathbf v_k$ 投影到已经做好的 $\mathbf q_1,\dots,\mathbf q_{k-1}$ 张成的子空间（正交基，所以逐条相加），再取残差。残差垂直于前面所有的 $\mathbf q_j$。如果 $\mathbf w_k=\mathbf 0$，说明 $\mathbf v_k$ 本来就在前面的张成空间里，向量线性相关（1.2 节），无法继续。把每一步的系数 $r_{jk}=\mathbf q_j\cdot\mathbf v_k$ 和 $r_{kk}=|\mathbf w_k|$ 记下来，就得到 $A=QR$，其中 $R$ 是上三角矩阵，这就是 QR 分解。

Step $k$ says: project $\mathbf v_k$ onto the subspace spanned by the $\mathbf q_1,\dots,\mathbf q_{k-1}$ already built (an orthonormal basis, so we add line by line), then keep the residual. The residual is perpendicular to every earlier $\mathbf q_j$. If $\mathbf w_k=\mathbf 0$, then $\mathbf v_k$ was already in the span of the earlier vectors, the vectors are linearly dependent (Lesson 1.2), and we cannot continue. Record the coefficients $r_{jk}=\mathbf q_j\cdot\mathbf v_k$ and $r_{kk}=|\mathbf w_k|$ and you have $A=QR$ with $R$ upper triangular: the QR decomposition.

7机器人中的它：循迹误差与摩擦锥In robotics: cross-track error and friction cones

路径跟踪：最近点、横向误差、沿路进度Path following: closest point, cross-track error, along-track progress

路径通常是一条由若干线段连成的折线。要知道机器人 $\mathbf r$ 在路径上"走到了哪"，对每一段 $\mathbf A\to\mathbf B$（方向 $\mathbf d=\mathbf B-\mathbf A$）做投影：

A path is usually a polyline made of segments. To find where the robot $\mathbf r$ is "up to" along it, project onto every segment $\mathbf A\to\mathbf B$ (direction $\mathbf d=\mathbf B-\mathbf A$):

$$ t = \operatorname{clamp}\!\Big(\frac{(\mathbf r-\mathbf A)\cdot\mathbf d}{\mathbf d\cdot\mathbf d},\,0,\,1\Big), \qquad \mathbf c = \mathbf A + t\,\mathbf d, \qquad e_{\text{ct}} = \hat{\mathbf d}\times(\mathbf r-\mathbf A) . $$

线段不是无限长的直线，所以要把 $t$ 夹到 $[0,1]$：投影落到线段外面时，最近点就是端点。取所有线段里离 $\mathbf r$ 最近的那个 $\mathbf c$，得到最近点。$e_{\text{ct}}$ 是有符号的横向误差（cross-track error）：用 1.8 节的二维叉积，机器人在路径前进方向的左边为正，右边为负。沿路进度（along-track）是前面各段的长度之和，加上 $t\,|\mathbf d|$。拖动机器人，看每一段的候选最近点，还有哪一段赢了。

A segment is not an infinite line, so $t$ must be clamped to $[0,1]$: when the projection falls outside the segment, the closest point is an endpoint. Take the $\mathbf c$ with the smallest distance over all segments and you have the closest point. $e_{\text{ct}}$ is the signed cross-track error, using the 2D cross product of Lesson 1.8: positive when the robot is on the left of the direction of travel, negative on the right. The along-track progress is the total length of the earlier segments plus $t\,|\mathbf d|$. Drag the robot and watch each segment's candidate closest point, and which segment wins.

FIG 6折线路径上的最近点Closest point on a polyline path

灰色空心圆是每一段上的候选最近点（虚线连到机器人），赢的那一段变成青色，它的最近点是琥珀色实心点。路径上从起点到 c 的那段淡琥珀色是沿路进度。点"出发"，机器人会按 Stanley 式的控制律自己开：转向 = 航向误差 − arctan(k · 横向误差 / 速度)。把增益 k 调到 0 看看，只对齐航向、不管横向误差，会怎样？ Hollow gray circles are each segment's candidate closest point (dashed lines to the robot). The winning segment turns cyan and its closest point is the solid amber dot. The pale amber band along the path from the start up to c is the along-track progress. Press "Drive" and the robot steers itself with a Stanley-style law: steering = heading error − arctan(k · cross-track error / speed). Set the gain k to 0 and see what happens when it aligns its heading but ignores the lateral error.

把机器人放在最近点从一段切换到下一段的位置：两段的距离几乎相等。这些位置连成什么形状？Place the robot where the closest point switches from one segment to the next: the two segments are almost equally near. What shape do these places form?
把机器人拖到离路径至少 0.8 的地方，点"出发"，看横向误差收敛到 0.1 以内。Drag the robot at least 0.8 away from the path, press "Drive", and watch the cross-track error shrink below 0.1.

真实系统里In real systems

路径跟踪控制器（纯追踪 pure pursuit、Stanley、MPC）的输入几乎都是这两个量：横向误差和航向误差，二者都由"最近点"得到。实现里要注意：最近点需要做"单调前进"（只在上次所在线段附近搜索），否则路径自交或往返时最近点会跳到另一段；在路径端点要夹 $t$；在拐角的外侧，最近点是顶点本身，此时横向误差通常改用相邻两段方向的平均来定义。Snider (2009) 对这些控制律做了系统比较。

Path-tracking controllers (pure pursuit, Stanley, MPC) take almost exactly these two numbers as input: cross-track error and heading error, both derived from the "closest point". In practice the closest-point search is made "monotone" (only look near the previous segment), otherwise a self-crossing or out-and-back path makes the closest point jump to another segment; $t$ is clamped at the path ends; and on the outside of a corner the closest point is the vertex itself, where the cross-track error is usually defined with the average of the two neighbouring directions. Snider (2009) compares these control laws systematically.

接触力：法向力、切向力与摩擦锥Contact force: normal, tangential, and the friction cone

手指（或一个箱子）压在一个表面上，表面的单位法向量是 $\mathbf n$（指向表面外），沿表面的单位方向是 $\hat{\mathbf t}$。受力 $\mathbf F$ 是一个向量，它在两个方向上的投影就是

A finger (or a box) presses on a surface with outward unit normal $\mathbf n$ and unit tangent $\hat{\mathbf t}$. The contact force $\mathbf F$ is a vector, and its projections on the two directions are

$$ f_n = -\,\mathbf F\cdot\mathbf n, \qquad f_t = \mathbf F\cdot\hat{\mathbf t}, \qquad \mathbf F = f_n(-\mathbf n) + f_t\,\hat{\mathbf t}. $$

$f_n>0$ 表示在压向表面，$f_t$ 的正负表示沿表面的哪个方向。

$f_n>0$ means pressing into the surface; the sign of $f_t$ says which way along the surface.

库仑摩擦说：只要 $f_n>0$（确实在压着）且 $|f_t|\le\mu f_n$，接触就粘住，否则打滑。把条件翻译成角度：$\mathbf F$ 和"压进表面"方向 $-\mathbf n$ 的夹角不超过 $\arctan\mu$。这些方向组成一个楔形，也就是二维里的摩擦锥。下面的图让你拖动力的箭头尖端，也可以点"箱子"预设，把力换成重力，用滑块倾斜斜面。

Coulomb friction says: as long as $f_n>0$ (it really is pressing) and $|f_t|\le\mu f_n$, the contact sticks; otherwise it slips. In terms of angles: $\mathbf F$ makes an angle of at most $\arctan\mu$ with the "into the surface" direction $-\mathbf n$. These directions form a wedge, the friction cone (a wedge in 2D). In the figure, drag the tip of the force arrow, or press the "box" preset to swap the force for gravity and tilt the slope with the slider.

FIG 7摩擦锥The friction cone

箭头是总力 $\mathbf F$（作用在表面上，从接触点指出；绿色表示粘住，红色表示打滑）。它在表面上的投影 $f_t$（琥珀色）和垂直于表面的残差 $f_n$（紫色虚线）正好是本节的投影与残差，表面这条线就是"那条直线"。青色楔形是摩擦锥：箭头尖落在锥内，接触粘住；落在锥外，打滑。 The arrow is the total force $\mathbf F$ (acting on the surface, drawn from the contact point; green means it sticks, red that it slips). Its projection onto the surface, $f_t$ (amber), and the residual perpendicular to the surface, $f_n$ (dashed purple), are exactly this lesson's projection and residual, with the surface as "the line". The cyan wedge is the friction cone: with the arrow tip inside the cone the contact sticks; outside, it slips.

找到最陡但不打滑的推力方向：箭头方向正好贴着锥的边缘（偏差不超过 3°）。Find the steepest push that does not slip: the arrow lies right at the edge of the cone (within 3°).
点"箱子"预设，倾斜斜面，找到箱子刚好不下滑的最大角度（$\tan\alpha$ 在 $0.9\mu$ 与 $\mu$ 之间）。Press the "box" preset and tilt the slope to find the largest angle at which the box just does not slide ($\tan\alpha$ between $0.9\mu$ and $\mu$).

箱子在斜面上的结论很漂亮：重力 $mg$ 的法向分量是 $mg\cos\alpha$，切向分量是 $mg\sin\alpha$，粘住的条件 $mg\sin\alpha\le\mu\,mg\cos\alpha$ 化简成 $\tan\alpha\le\mu$，质量约掉了。

The box on a slope gives a neat result: gravity $mg$ has normal part $mg\cos\alpha$ and tangential part $mg\sin\alpha$, so the sticking condition $mg\sin\alpha\le\mu\,mg\cos\alpha$ reduces to $\tan\alpha\le\mu$, and the mass cancels.

真实系统里In real systems

抓取规划把每个接触点的摩擦锥当作约束：手指施加的力必须落在锥里，抓取才稳。三维里锥是圆锥，常被近似成四棱锥，好让条件变成线性不等式（本节的法向分量加上几个线性约束），交给二次规划去解（6.8 节）。Lynch & Park 第 12 章对此有完整讨论。

Grasp planners treat each contact's friction cone as a constraint: the force a finger applies must lie inside the cone for the grasp to hold. In 3D the cone is round and is often approximated by a pyramid so the condition becomes linear inequalities (this lesson's normal component plus a few linear constraints) that a quadratic program can handle (Lesson 6.8). Chapter 12 of Lynch & Park discusses this fully.

8常见误区Common pitfalls

误区 1：$\mathbf u$ 不是单位向量，却忘了除以 $|\mathbf u|^2$Pitfall 1: $\mathbf u$ is not a unit vector and you forgot to divide by $|\mathbf u|^2$

公式 $\mathbf p=(\mathbf u\cdot\mathbf b)\,\mathbf u$ 只在 $|\mathbf u|=1$ 时成立。对 $\mathbf u=(2,1)$、$\mathbf b=(3,4)$，漏掉分母会得到 $(20,10)$，比 $\mathbf b$ 本身还长，显然不是影子；正确答案是 $(4,2)$。代码里最稳的做法：要么先 u = u / norm(u)，要么直接写 u * dot(u,b) / dot(u,u)，并在 $\mathbf u\approx\mathbf 0$ 时提前返回。

The formula $\mathbf p=(\mathbf u\cdot\mathbf b)\,\mathbf u$ holds only when $|\mathbf u|=1$. For $\mathbf u=(2,1)$ and $\mathbf b=(3,4)$, dropping the denominator gives $(20,10)$, longer than $\mathbf b$ itself and obviously not a shadow; the right answer is $(4,2)$. In code, either normalise first (u = u / norm(u)) or write u * dot(u,b) / dot(u,u), and return early when $\mathbf u\approx\mathbf 0$.

误区 2：把"到线段的距离"当成"到直线的距离"Pitfall 2: treating distance to a segment as distance to a line

直线无限长，线段有两个端点。点投影到线段所在直线上的位置如果落在线段外，最近点就是端点，距离会比到直线的距离更大。漏掉 clip(t, 0, 1) 是最经典的"点到线段距离"bug，症状是机器人在路径末端之外也"贴着路径"，或者折线上两段的最近点判断错误。FIG 6 里把机器人拖到线段两端的外侧就能看到。

A line is infinite; a segment has two ends. If the projection falls outside the segment, the closest point is an endpoint and the distance is larger than the distance to the line. Forgetting clip(t, 0, 1) is the classic "distance to segment" bug; the symptom is a robot that still looks "on the path" beyond its end, or a polyline whose segments pick the wrong closest point. Drag the robot beyond either end of a segment in FIG 6 to see it.

误区 3：把投影到几个非正交向量上的结果加起来Pitfall 3: adding projections onto several non-orthogonal vectors

在平面里，任意两个不共线的向量 $\mathbf u_1,\mathbf u_2$ 都张成整个平面，所以 $\mathbf b$ 在它们张成的空间上的投影就是 $\mathbf b$ 自己。然而把 $\mathbf b$ 分别投影到 $\mathbf u_1$ 和 $\mathbf u_2$ 再相加，一般得不到 $\mathbf b$。只有 $\mathbf u_1\perp\mathbf u_2$ 时才对。拖动两个基向量试试。

In the plane, any two non-collinear vectors $\mathbf u_1,\mathbf u_2$ span the whole plane, so the projection of $\mathbf b$ onto their span is $\mathbf b$ itself. Yet projecting $\mathbf b$ onto $\mathbf u_1$ and onto $\mathbf u_2$ separately and adding the results does not give back $\mathbf b$ in general. It does only when $\mathbf u_1\perp\mathbf u_2$. Drag the two basis vectors and see.

FIG 8投影之和 ≠ 投影Sum of projections ≠ projection

青色是 $\mathbf u_1,\mathbf u_2$，各自的投影 $\mathbf p_1,\mathbf p_2$ 是琥珀色虚线。红色箭头是它们的和 $\mathbf p_1+\mathbf p_2$，不带颜色的普通箭头是真正该得到的 $\mathbf b$。两个基向量成直角时，红箭头才会盖住 $\mathbf b$。夹角离 90° 越远，和通常偏离 $\mathbf b$ 越多，可能太长，也可能太短。 Cyan are $\mathbf u_1,\mathbf u_2$, and their projections $\mathbf p_1,\mathbf p_2$ are the dashed amber segments. The red arrow is their sum $\mathbf p_1+\mathbf p_2$ and the plain, uncoloured arrow is the $\mathbf b$ we should get. The red arrow covers $\mathbf b$ only when the basis vectors meet at a right angle. The further the angle is from 90°, the further the sum usually lands from $\mathbf b$, sometimes too long, sometimes too short.

让两个基向量垂直，使"和"与 $\mathbf b$ 重合（误差小于 0.1）。Make the two basis vectors perpendicular so the "sum" coincides with $\mathbf b$ (error below 0.1).

误区 4：经典 Gram–Schmidt 在数值上会丢失正交性Pitfall 4: classical Gram–Schmidt loses orthogonality numerically

上面的公式对精确算术是对的，但在浮点数里，当输入向量几乎共线时，相减会抵消掉几乎所有有效数字，剩下的是舍入误差，于是 $\mathbf q_k$ 不再垂直于前面的向量。取四行三列的矩阵，第一行全是 1，下面是对角线上为 $\varepsilon=10^{-8}$ 的 $3\times3$ 矩阵（Läuchli 矩阵）：经典 Gram–Schmidt 算出 $\mathbf q_2\cdot\mathbf q_3=0.5$，完全不垂直；改良版 (modified) Gram–Schmidt 把每次减法立刻用在更新后的向量上，最差的 $|\mathbf q_i\cdot\mathbf q_j|$（$i\ne j$）约为 $10^{-8}$；Householder QR（numpy.linalg.qr 用的方法）约 $10^{-16}$。实际代码请用 QR 库函数，不要手写经典 Gram–Schmidt。

The formulas are right in exact arithmetic, but in floating point, when the input vectors are nearly collinear, the subtraction cancels almost all significant digits and what is left is rounding error, so $\mathbf q_k$ is no longer perpendicular to the earlier vectors. Take the 4-by-3 Läuchli matrix: a first row of ones above a $3\times3$ diagonal with $\varepsilon=10^{-8}$. Classical Gram–Schmidt gives $\mathbf q_2\cdot\mathbf q_3=0.5$, nowhere near perpendicular; modified Gram–Schmidt, which applies each subtraction to the already-updated vector, keeps the worst $|\mathbf q_i\cdot\mathbf q_j|$ ($i\ne j$) at about $10^{-8}$; Householder QR (what numpy.linalg.qr uses) about $10^{-16}$. In real code call a QR routine rather than hand-writing classical Gram–Schmidt.

误区 5：以为投影可以"撤销"Pitfall 5: thinking a projection can be undone

$P$ 的秩比维数低，$\det P=0$，所以 $P^{-1}$ 不存在：直线上每一个点，都是垂直于直线的整条线上所有点的影子，只知道影子，没法判断原来的点在哪。需要"还原"时，要补上被丢掉的那一份 $\mathbf e$，或者像 1.13 节那样退而求其次，只求"最好的"答案。

$P$ has rank below the dimension and $\det P=0$, so $P^{-1}$ does not exist: every point of the line is the shadow of a whole perpendicular line of points, and from the shadow alone you cannot tell where the original was. To "undo" it you must bring back the discarded part $\mathbf e$, or, as in Lesson 1.13, settle for the "best" answer.

9小练习Check yourself

1. 把 $\mathbf b=(3,4)$ 投影到过原点、方向为 $\mathbf u=(2,1)$ 的直线上，投影 $\mathbf p$ 是？1. Project $\mathbf b=(3,4)$ onto the line through the origin with direction $\mathbf u=(2,1)$. What is $\mathbf p$?

$\mathbf u\cdot\mathbf b=6+4=10$，$\mathbf u\cdot\mathbf u=5$，所以 $\mathbf p=\tfrac{10}{5}\mathbf u=(4,2)$。(20, 10) 是忘了除以 $\mathbf u\cdot\mathbf u$ 的结果。检查：$\mathbf e=\mathbf b-\mathbf p=(-1,2)$，$\mathbf e\cdot\mathbf u=-2+2=0$，垂直。$\mathbf u\cdot\mathbf b=6+4=10$ and $\mathbf u\cdot\mathbf u=5$, so $\mathbf p=\tfrac{10}{5}\mathbf u=(4,2)$. (20, 10) is what you get by forgetting to divide by $\mathbf u\cdot\mathbf u$. Check: $\mathbf e=\mathbf b-\mathbf p=(-1,2)$ and $\mathbf e\cdot\mathbf u=-2+2=0$, so it is perpendicular.

2. 关于投影矩阵 $P=\mathbf u\mathbf u^\top/\mathbf u^\top\mathbf u$，下面哪一句是对的？2. Which statement about the projection matrix $P=\mathbf u\mathbf u^\top/\mathbf u^\top\mathbf u$ is true?

第二项。投影之后，垂直的那份已经丢掉，再投影一次没有东西可丢：$P^2=P$。$P$ 把平面压成一条线，$\det P=0$，不可逆（所以第一、三项都错）。第四项也不对：直线上的向量完全不变，长度不缩短。The second one. After projecting, the perpendicular part is already gone, so a second projection has nothing to remove: $P^2=P$. $P$ squashes the plane onto a line, so $\det P=0$ and it is not invertible (which kills the first and third options). The fourth is wrong too: vectors on the line are left completely unchanged.

3. 路径的一段是从 $\mathbf A=(0,0)$ 到 $\mathbf B=(2,0)$。机器人在 $\mathbf r=(4,3)$。它到这条线段的距离是？3. One path segment runs from $\mathbf A=(0,0)$ to $\mathbf B=(2,0)$. The robot is at $\mathbf r=(4,3)$. What is its distance to this segment?

先算 $t=\frac{(\mathbf r-\mathbf A)\cdot\mathbf d}{\mathbf d\cdot\mathbf d}=\frac{8}{4}=2$，超出 $[0,1]$，夹到 $t=1$，最近点是端点 $\mathbf B=(2,0)$。距离是 $|(4,3)-(2,0)|=\sqrt{4+9}=\sqrt{13}\approx3.61$。答案 3 是到整条直线（x 轴）的距离，正是误区 2。First $t=\frac{(\mathbf r-\mathbf A)\cdot\mathbf d}{\mathbf d\cdot\mathbf d}=\frac{8}{4}=2$, outside $[0,1]$, so clamp to $t=1$ and the closest point is the endpoint $\mathbf B=(2,0)$. The distance is $|(4,3)-(2,0)|=\sqrt{4+9}=\sqrt{13}\approx3.61$. The answer 3 is the distance to the whole line (the x-axis), which is Pitfall 2.

4. 手指压在表面上，法向力 $f_n=4$ N，切向力 $f_t=3$ N，摩擦系数 $\mu=0.5$。接触会怎样？4. A finger presses on a surface with normal force $f_n=4$ N and tangential force $f_t=3$ N, friction coefficient $\mu=0.5$. What happens at the contact?

粘住的条件是 $|f_t|\le\mu f_n$。这里 $\mu f_n=2$ N，而切向力有 3 N，超出了，所以打滑。等价地：力与法向的夹角是 $\arctan(3/4)\approx36.9^\circ$，大于摩擦锥半角 $\arctan0.5\approx26.6^\circ$。$f_n$ 和 $f_t$ 已经包含了方向信息，所以能判断。Sticking requires $|f_t|\le\mu f_n$. Here $\mu f_n=2$ N while the tangential force is 3 N, which exceeds it, so the contact slips. Equivalently, the force makes an angle $\arctan(3/4)\approx36.9^\circ$ with the normal, more than the cone half-angle $\arctan0.5\approx26.6^\circ$. $f_n$ and $f_t$ already carry the direction, so we can decide.

5. 对 $\mathbf v_1=(3,0)$、$\mathbf v_2=(2,2)$ 做 Gram–Schmidt，得到的 $\mathbf q_2$ 是？5. Apply Gram–Schmidt to $\mathbf v_1=(3,0)$ and $\mathbf v_2=(2,2)$. What is $\mathbf q_2$?

$\mathbf q_1=(1,0)$。$\mathbf v_2$ 在 $\mathbf q_1$ 上的投影是 $(\mathbf q_1\cdot\mathbf v_2)\mathbf q_1=(2,0)$，残差 $\mathbf w=(2,2)-(2,0)=(0,2)$，再单位化得 $\mathbf q_2=(0,1)$。选项 (0, 2) 忘了最后一步单位化；$(2,2)/\sqrt8$ 是只单位化、没有减去投影的结果。$\mathbf q_1=(1,0)$. The projection of $\mathbf v_2$ onto $\mathbf q_1$ is $(\mathbf q_1\cdot\mathbf v_2)\mathbf q_1=(2,0)$, so the residual is $\mathbf w=(2,2)-(2,0)=(0,2)$, and normalising gives $\mathbf q_2=(0,1)$. The option (0, 2) skips the final normalisation; $(2,2)/\sqrt8$ normalises without subtracting the projection.

10延伸阅读Further reading

G. Strang, Introduction to Linear Algebra, §4.2 and §4.4 — "投影"与"标准正交基和 Gram–Schmidt""Projections" and "Orthonormal bases and Gram–Schmidt"
G. Strang, MIT 18.06 Linear Algebra (OCW) — 第 15、16、17 讲：子空间上的投影、投影矩阵、正交矩阵与 Gram–SchmidtLectures 15, 16, 17: projections onto subspaces, projection matrices, orthogonal matrices and Gram–Schmidt
K. Lynch & F. Park, Modern Robotics, Ch. 12 — 抓取与操作：接触、摩擦锥与力封闭Grasping and manipulation: contacts, friction cones and force closure
J. M. Snider, “Automatic Steering Methods for Autonomous Automobile Path Tracking”, CMU-RI-TR-09-08, 2009 — 横向误差与纯追踪、Stanley 等路径跟踪控制律的比较Cross-track error and a comparison of pure pursuit, Stanley and other path-tracking laws
L. N. Trefethen & D. Bau, Numerical Linear Algebra, Lectures 6–10 — 投影矩阵、QR、经典与改良 Gram–Schmidt、HouseholderProjectors, QR, classical vs. modified Gram–Schmidt, Householder