東京大学 情報理工学研究科 2023年8月実施 数学 第1問

Author

zephyr

Description

Let be the set of the three-dimensional real column vectors and be the set of the three-by-three real matrices. Let , , and be linearly independent unit-length vectors and be a unit-length vector not parallel to , , or . Let and be square matrices defined as

Here, and denote the transpose of a matrix and a vector , respectively. Answer the following questions.

(1) Find the condition for such that the rank of is three.

(2) In the three-dimensional Euclidean space , consider four planes is a real number, and that satisfy the following three conditions: (i) the rank of is three, (ii) is not the empty set, and (iii) there exists a sphere to which are tangent. The position vector of the center of is represented by using a vector . Express using .

(3) Show that is a positive definite symmetric matrix.

(4) Consider the point from which the sum of squared distances to four planes is a real number, and is minimized. The position vector of is represented by using a vector . Express using and .

(5) Let be a straight line through a point , the position vector of which is , parallel to in . Let be the orthogonal projection of an arbitrary point , the position vector of which is , onto . The position vector of is represented by using a matrix . The identity matrix is denoted by .

• (a) Express using and .
• (b) Show that .
• () Consider a plane is a non-zero vector, and is a real number). Let be the point from which the sum of squared distances to , , and is minimized. When , , and are orthogonal to each other, the position vector of is represented by . using a vector which is independent of and . Express using and .

Kai

(1)

Given the matrix :

we need to determine the conditions on that ensure has a rank of three.

Let's assume that the third row of matrix can be written as a linear combination of the first two rows. Thus, we assume:

where and are some scalars. Substituting as a linear combination of , , and , we have:

Expanding and rearranging the equation, we get:

Grouping like terms:

For this equation to hold for arbitrary vectors , , and , the coefficients of each vector must match:

1. For :

2. For :

3. For :

Thus, we have the following system of equations:

Substituting into the last equation:

So, the conditions under which the third row of can be written as a linear combination of the first two rows (i.e., the matrix would not have full rank) are:

Conclusion for Full Rank

For the matrix to have full rank (rank 3), must be such that the above condition does not hold. Therefore, the condition for the rank of to be three is:

(2)

Problem Setup

We are given four planes in :

where , and are unit vectors, and and are real numbers. The center of a sphere tangent to all four planes is represented as , where is a 3x3 matrix, and is what we need to find.

Conditions

For the sphere to be tangent to each plane, the distance from the center of the sphere to each plane must satisfy:

We subtract the equations pairwise to eliminate , yielding:

These can be rewritten as:

Matrix Representation

The matrix is defined by the differences between the normals:

Thus, we can write the system of equations in matrix form as:

Simplifying, we find:

(3)

The matrix is defined as:

First, we show that is symmetric. Since each term is symmetric (as the outer product of a vector with itself is symmetric), their sum is also symmetric.

Next, to prove that is positive definite, we need to show that for any non-zero vector , the quadratic form .

Since are unit vectors and linearly independent, the sum is strictly positive for any non-zero , proving that is positive definite.

(4)

The sum of squared distances from a point to the four planes is minimized when is the point of orthogonal projection of the origin onto these planes. The squared distance from a point with position vector to the plane is given by:

Since are unit vectors (), this simplifies to:

The sum of squared distances to all four planes is:

To minimize , we take the gradient with respect to and set it equal to zero:

This equation can be rearranged into the form:

The matrix is defined as:

which is a matrix. Therefore, the position vector that minimizes the sum of squared distances can be expressed as:

Given that is the point minimizing the sum of squared distances, its position vector is , where is defined by:

(5)

(a)

Orthogonal Projection onto a Line

Given a vector , the orthogonal projection of onto a line in the direction of a unit vector is calculated as follows:

1. Dot Product Calculation:

   • The scalar projection of onto is given by the dot product .

2. Vector Projection:

   • The vector projection of onto the line parallel to is then:

3. Matrix Representation:
   • This vector projection can be represented in matrix form. Specifically, if we want to express this operation as a matrix multiplication where gives the projection, then the matrix must satisfy:

• To express this operation using a matrix, note that:

• This is because when acts on a vector , it first computes the dot product (which is a scalar), and then multiplies this scalar by to produce the vector projection.

(b)

We calculate:

Since , we have:

()

We need to find the point that minimizes the sum of squared distances to three orthogonal lines.

The projection of a point onto the line parallel to is given by:

where is the orthogonal projection matrix onto .

Sum of Squared Distances

The distance from to the line is:

The sum of squared distances from to the three lines is:

Objective Function for Minimization

The objective is to minimize the sum:

Since (as derived in part (b)), this simplifies to:

Setting Up the Minimization Problem

Expanding the quadratic form:

To minimize , we take the derivative with respect to and set it to zero:

Simplifying:

Solving for

Since , , and are orthogonal, their corresponding matrices satisfy:

Thus:

This minimizes the sum of squared distances to the three lines, and we denote this as :

Knowledge

矩阵秩 正定矩阵 最小二乘法 正交投影

难点思路

题目较难的部分是处理涉及到多平面的几何关系和正定矩阵的性质证明。特别是第 4 问中的最小二乘问题，需要对平面到点的距离公式有深刻理解。

解题技巧和信息

在解答此类问题时，明确矩阵的几何意义和代数性质非常关键。利用向量投影和最小二乘法的基本原理，可以有效地处理平面、直线和点之间的距离问题。

重点词汇

• Rank of a matrix: 矩阵的秩
• Positive definite matrix: 正定矩阵
• Orthogonal projection: 正交投影
• Least squares: 最小二乘法

参考资料

1. Gilbert Strang, Linear Algebra and Its Applications, 4th Edition, Section 6.5.
2. David C. Lay, Linear Algebra and Its Applications, 5th Edition, Chapter 7.