東京大学 情報理工学系研究科 創造情報学専攻 2012年8月実施 筆記試験 第3問

标签：

• Tokyo-University
• Control-Theory
• Inner-Product-and-Orthogonality
• Projection-Operator
• Rotation-Matrix-and-Axis-Angle

Author

itsuitsuki

Description

Official examination, archived Japanese PDF. In kinematic calculations and visual computations in robotics, vector operations are expressed with matrices sometimes. Answer the following questions on inner-product, outer-product, projection and rotation of three dimensional vectors. $\boldsymbol{I}$ is the $3\times3$ identity matrix. The three-dimensional vectors $\boldsymbol{x}$, $\boldsymbol{a}$, $\boldsymbol{b}$ and $\boldsymbol{n}$ are $3\times1$ column vectors: $\boldsymbol{x}=\begin{bmatrix}x_x\\x_y\\x_z\end{bmatrix},\quad\boldsymbol{a}=\begin{bmatrix}a_x\\a_y\\a_z\end{bmatrix},\quad\boldsymbol{b}=\begin{bmatrix}b_x\\b_y\\b_z\end{bmatrix},\quad\boldsymbol{n}=\begin{bmatrix}n_x\\n_y\\n_z\end{bmatrix}.$ $\boldsymbol{x}^T$, which is a $1\times3$ row vector, shows the transpose of $\boldsymbol{x}$.

(1) On the inner product $(\boldsymbol{a},\boldsymbol{b})$ between vectors $\boldsymbol{a}$ and $\boldsymbol{b}$, describe i) the value of $(\boldsymbol{a},\boldsymbol{b})$, ii) a $3\times3$ matrix $\boldsymbol{A}$ which satisfies $(\boldsymbol{a},\boldsymbol{b})\boldsymbol{a}=\boldsymbol{A}\boldsymbol{b}$, and iii) $\boldsymbol{A}$ with vector $\boldsymbol{a}$ and its transpose $\boldsymbol{a}^T$.

(2) On the outer product $\boldsymbol{a}\times\boldsymbol{b}$ from $\boldsymbol{a}$ to $\boldsymbol{b}$, describe i) $3\times1$ expression of $\boldsymbol{a}\times\boldsymbol{b}$, ii) $3\times3$ expression of the matrix $\boldsymbol{A}$ which satisfies $\boldsymbol{a}\times\boldsymbol{b}=\boldsymbol{A}\boldsymbol{b}$, iii) the $3\times3$ matrix $\boldsymbol{Q}$ which satisfies $\boldsymbol{A}=\boldsymbol{a}\times\boldsymbol{Q}$ where $\boldsymbol{x}\times\boldsymbol{Q}$ between a vector $\boldsymbol{x}$ and a $3\times3$ matrix $\boldsymbol{Q}$ means a $3\times3$ matrix whose column vectors are three outer products from the vector $\boldsymbol{x}$ to each column vector in the matrix $\boldsymbol{Q}$ respectively.

(3) As Figure 1 shows, a vector $\boldsymbol{x}$ is vertically projected to a vector $\boldsymbol{y}$ on a plane whose normal vector is a unit vector $\boldsymbol{n}$. If the vector $\boldsymbol{y}$ is described as $\boldsymbol{y}=\boldsymbol{P}\boldsymbol{x}$, show that the matrix $\boldsymbol{P}$ becomes $\boldsymbol{P}=\boldsymbol{I}-\boldsymbol{n}\boldsymbol{n}^T$.

(4) Three rotational matrices $\boldsymbol{R}_x(\theta_x)$, $\boldsymbol{R}_y(\theta_y)$ and $\boldsymbol{R}_z(\theta_z)$ are rotational matrices which rotate a vector $\boldsymbol{x}$ around the X-axis, Y-axis and Z-axis with $\theta_x$, $\theta_y$ and $\theta_z$ respectively, where the direction of the rotation for plus is clock-wise around the axis from the origin to infinity.

i) Describe $3\times3$ expression of the matrix $\boldsymbol{R}_x(\theta_x)$, ii) As Figure 2 shows, the $3\times3$ matrix $\boldsymbol{R}_n(\theta_n)$ is defined as the rotation matrix around a unit orientation vector $\boldsymbol{n}$ with $\theta_n$. $\boldsymbol{R}_n(\theta_n)$ is described as $\boldsymbol{R}_n(\theta_n)=\boldsymbol{R}_x(-\alpha)\boldsymbol{R}_y(\beta)\boldsymbol{R}_z(\theta_n)\boldsymbol{R}_y(-\beta)\boldsymbol{R}_x(\alpha).$ Explain what the variables $\alpha$ and $\beta$ become and explain why the expression is satisfied.

题目描述

机器人运动学和视觉计算常用矩阵表示向量运算。回答三维向量的内积、外积、投影与旋转问题。令 $\boldsymbol I$ 为 $3\times3$ 单位矩阵，并令

$$\boldsymbol{x}=\begin{bmatrix}x_x\\x_y\\x_z\end{bmatrix},\quad \boldsymbol{a}=\begin{bmatrix}a_x\\a_y\\a_z\end{bmatrix},\quad \boldsymbol{b}=\begin{bmatrix}b_x\\b_y\\b_z\end{bmatrix},\quad \boldsymbol{n}=\begin{bmatrix}n_x\\n_y\\n_z\end{bmatrix}$$

均为 $3\times1$ 列向量；$\boldsymbol x^T$ 是其 $1\times3$ 转置行向量。

1. 对 $\boldsymbol a,\boldsymbol b$ 的内积 $(\boldsymbol a,\boldsymbol b)$，分别写出：a. 内积数值；b. 满足

   $$(\boldsymbol a,\boldsymbol b)\boldsymbol a=\boldsymbol A\boldsymbol b$$

   的 $3\times3$ 矩阵 $\boldsymbol A$；c. 用 $\boldsymbol a,\boldsymbol a^T$ 表示该 $\boldsymbol A$。

2. 对从 $\boldsymbol a$ 到 $\boldsymbol b$ 的外积 $\boldsymbol a\times\boldsymbol b$，分别写出：a. 其 $3\times1$ 分量表达式；b. 满足 $\boldsymbol a\times\boldsymbol b=\boldsymbol A\boldsymbol b$ 的 $3\times3$ 矩阵 $\boldsymbol A$；c. 满足 $\boldsymbol A=\boldsymbol a\times\boldsymbol Q$ 的 $3\times3$ 矩阵 $\boldsymbol Q$。这里 $\boldsymbol x\times\boldsymbol Q$ 定义为 $3\times3$ 矩阵，其每一列分别是 $\boldsymbol x$ 与 $\boldsymbol Q$ 对应列向量的外积。

3. 如图 1，把 $\boldsymbol x$ 正交投影到法向量为单位向量 $\boldsymbol n$ 的平面，得到 $\boldsymbol y$。若 $\boldsymbol y=\boldsymbol P\boldsymbol x$，证明

   $$\boldsymbol P=\boldsymbol I-\boldsymbol n\boldsymbol n^T.$$

4. $\boldsymbol R_x(\theta_x),\boldsymbol R_y(\theta_y),\boldsymbol R_z(\theta_z)$ 分别把向量绕 $X,Y,Z$ 轴旋转 $\theta_x,\theta_y,\theta_z$；从原点沿轴正方向看去，顺时针规定为正。

   1. 写出 $\boldsymbol R_x(\theta_x)$ 的 $3\times3$ 形式。

   2. 如图 2，$\boldsymbol R_n(\theta_n)$ 表示绕单位方向向量 $\boldsymbol n$ 旋转 $\theta_n$，且

      $$\boldsymbol R_n(\theta_n)= \boldsymbol R_x(-\alpha)\boldsymbol R_y(\beta)\boldsymbol R_z(\theta_n) \boldsymbol R_y(-\beta)\boldsymbol R_x(\alpha).$$

      求 $\alpha,\beta$，并解释该矩阵分解成立的原因。

Kai

Here “outer product” in part (2) means the three-dimensional cross product $\boldsymbol a\times\boldsymbol b$. It should be distinguished from the matrix outer product $\boldsymbol a\boldsymbol b^T$.

(1) Inner product

i) The scalar inner product is

$$(\boldsymbol a,\boldsymbol b)=a_xb_x+a_yb_y+a_zb_z=\boldsymbol a^T\boldsymbol b.$$

ii) A matrix that gives $(\boldsymbol a,\boldsymbol b)\boldsymbol a$ for every $\boldsymbol b$ is

$$\boxed{\boldsymbol A=\begin{pmatrix} a_x^2&a_xa_y&a_xa_z\\ a_ya_x&a_y^2&a_ya_z\\ a_za_x&a_za_y&a_z^2 \end{pmatrix}.}$$

iii) In vector notation, $\boxed{\boldsymbol A=\boldsymbol a\boldsymbol a^T}$, because $\boldsymbol A\boldsymbol b=\boldsymbol a(\boldsymbol a^T\boldsymbol b)$.

(2) Cross product

i)

$$\boxed{\boldsymbol a\times\boldsymbol b= \begin{pmatrix}a_yb_z-a_zb_y\\a_zb_x-a_xb_z\\a_xb_y-a_yb_x\end{pmatrix}.}$$

ii) The corresponding skew-symmetric matrix is

$$\boxed{\boldsymbol A=[\boldsymbol a]_\times= \begin{pmatrix}0&-a_z&a_y\\a_z&0&-a_x\\-a_y&a_x&0\end{pmatrix}.}$$

iii) Taking $\boxed{\boldsymbol Q=\boldsymbol I}$ works: its columns are the coordinate basis vectors, and the columns $\boldsymbol a\times\boldsymbol e_i$ are exactly those of $[\boldsymbol a]_\times$. This choice is not unique for a fixed nonzero $\boldsymbol a$: the general solution is $\boldsymbol Q=\boldsymbol I+\boldsymbol a\boldsymbol v^T$ for any vector $\boldsymbol v$, because each column may be changed by a multiple of $\boldsymbol a$. If $\boldsymbol a=0$, every $\boldsymbol Q$ works.

(3) Orthogonal projection

The plane of vectors is $\boldsymbol n^T\boldsymbol y=0$, through the origin. Decompose $\boldsymbol x$ into its normal component $(\boldsymbol n^T\boldsymbol x)\boldsymbol n$ and its component in the plane. Because $\boldsymbol n^T\boldsymbol n=1$,

$$\boldsymbol y=\boldsymbol x-(\boldsymbol n^T\boldsymbol x)\boldsymbol n =(\boldsymbol I-\boldsymbol n\boldsymbol n^T)\boldsymbol x.$$

Thus $\boxed{\boldsymbol P=\boldsymbol I-\boldsymbol n\boldsymbol n^T}$. Indeed $\boldsymbol n^T\boldsymbol P=0$, so the result lies in the plane, and $\boldsymbol x-\boldsymbol P\boldsymbol x$ is parallel to its normal. Also $\boldsymbol P^T=\boldsymbol P$ and $\boldsymbol P^2=\boldsymbol P$. For an affine plane not through the origin, an additional translation term would be necessary.

(4) Rotation about an arbitrary axis

i) “Clockwise looking from the origin toward the positive axis” is the right-hand rotation convention; looking back toward the origin would reverse the apparent direction. In particular a positive rotation about $X$ sends $\boldsymbol e_y$ toward $\boldsymbol e_z$. Therefore

$$\boxed{\boldsymbol R_x(\theta)= \begin{pmatrix}1&0&0\\0&\cos\theta&-\sin\theta\\0&\sin\theta&\cos\theta\end{pmatrix}.}$$

For clarity the other matrices in the factorization use the same active-rotation convention:

$$\boldsymbol R_y(\theta)=\begin{pmatrix}\cos\theta&0&\sin\theta\\0&1&0\\-\sin\theta&0&\cos\theta\end{pmatrix},\qquad \boldsymbol R_z(\theta)=\begin{pmatrix}\cos\theta&-\sin\theta&0\\\sin\theta&\cos\theta&0\\0&0&1\end{pmatrix}.$$

ii) Let $r=\sqrt{n_y^2+n_z^2}$. If $r>0$, choose

$$\boxed{\alpha=\operatorname{atan2}(n_y,n_z),\qquad \beta=\operatorname{atan2}(n_x,r).}$$

Thus $\cos\alpha=n_z/r$, $\sin\alpha=n_y/r$, $\cos\beta=r$ and $\sin\beta=n_x$, using the unit length of $\boldsymbol n$. These choices successively give

$$\boldsymbol R_x(\alpha)\boldsymbol n=(n_x,0,r)^T,\qquad \boldsymbol R_y(-\beta)(n_x,0,r)^T=(0,0,1)^T.$$

Define $\boldsymbol C=\boldsymbol R_y(-\beta)\boldsymbol R_x(\alpha)$. First $\boldsymbol C$ aligns the rotation axis with $Z$, then $\boldsymbol R_z(\theta_n)$ rotates around that aligned axis, and finally $\boldsymbol C^{-1}$ restores the original coordinates. Reading matrix actions from right to left,

$$\boxed{\boldsymbol R_n(\theta_n)=\boldsymbol C^{-1}\boldsymbol R_z(\theta_n)\boldsymbol C =\boldsymbol R_x(-\alpha)\boldsymbol R_y(\beta)\boldsymbol R_z(\theta_n) \boldsymbol R_y(-\beta)\boldsymbol R_x(\alpha).}$$

If $r=0$, then $\boldsymbol n=(\pm1,0,0)^T$; choose $\alpha=0$ and $\beta=\pm\pi/2$. This avoids an undefined $\operatorname{atan2}(0,0)$ and yields the same factorization. An independent equivalent formula is Rodrigues' expression

$$\boldsymbol R_n(\theta)=\cos\theta\,\boldsymbol I+(1-\cos\theta)\boldsymbol n\boldsymbol n^T +\sin\theta\,[\boldsymbol n]_\times,$$

which also makes the fixed axis and rotation sign explicit.