東京大学 新領域創成科学研究科 メディカル情報生命専攻 2014年8月実施 問題8

Author

zephyr

Description

Let be the set of non-zero complex two-dimensional vectors. Let be a 2 by 2 real symmetric matrix, and be the unit matrix.

(1) Find all the eigenvalues of .

(2) Under the assumption of , answer i) and ii).

• i) Let be the matrix whose first and second columns consist of the eigenvectors and for the eigenvalues and , respectively. Show that is invertible and satisfies .
• ii) Prove that the set and are equal.

(3) For each of the statements A), B), and C), answer the conditions on matrix elements for the statement to hold.

• A) Every can be expressed as with some .
• B) No can be expressed as with some .
• C) At least one can be expressed as with some .

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设 为非零复二维向量的集合。设 为一个 2×2 的实对称矩阵， 为单位矩阵。

(1) 找出 的所有特征值 。

(2) 在假设 的条件下，回答 i) 和 ii)。

• i) 设 为一个矩阵，其第一列和第二列分别由特征值 和 的特征向量 和 组成。证明 是可逆的，并且满足 。
• ii) 证明集合 和 是相等的。

(3) 对于每个陈述 A), B), 和 C)，回答矩阵元素 的条件使该陈述成立。

• A) 每个 都可以表示为 ，其中 。
• B) 没有 可以表示为 ，其中 。
• C) 至少有一个 可以表示为 ，其中 。

Kai

(1)

To find the eigenvalues of the matrix , we solve the characteristic equation:

The characteristic polynomial of is:

This simplifies to:

The eigenvalues and are the roots of this quadratic equation:

(2)

i) Showing is Invertible

Let and be the eigenvectors ==corresponding== to and , respectively. Define the matrix . Since , the eigenvectors and are ==linearly independent==, and thus is invertible.

To show that , consider the action of on the eigenvectors:

Therefore,

Thus, we have:

ii) Proving Set Equality

To prove that the set and are equal, consider any . Then if and only if . Since is invertible and consists of all non-zero complex vectors, applying to any vector in yields another non-zero complex vector, ensuring the sets are equal.

(3)

A) For every to be expressible as for some , must be invertible. This requires and , ensuring , , and .

B) No can be expressed as for some if is singular and its image does not cover . This happens when has a zero eigenvalue, i.e., and one of the eigenvalues is zero.

C) At least one can be expressed as for some if and do not both hold true. This requires to be invertible or have a non-trivial image, which is true if is not an eigenvalue of , ensuring .

Knowledge

特征值和特征向量 矩阵分解

解题技巧和信息

1. 特征值问题中，特征多项式是重要的工具，通过求解特征多项式可以得到特征值。
2. 当矩阵的特征值不同时，其特征向量是线性无关的，这使得特征向量矩阵是可逆的。
3. 在处理复杂矩阵时，注意到特征向量的规范性及其在不同基底下的表示。

重点词汇

eigenvalue 特征值

eigenvector 特征向量

invertible 可逆的

characteristic polynomial 特征多项式

quadratic equation 二次方程

参考资料

1. 《线性代数及其应用》 第 5 章 特征值和特征向量