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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 \(Z^* = \left\{ \begin{pmatrix} z_1 \\ z_2 \end{pmatrix} : \text{complex number}, |z_1|^2 + |z_2|^2 \neq 0 \right\}\) be the set of non-zero complex two-dimensional vectors. Let \(M = \begin{pmatrix} a & b \\ b & d \end{pmatrix}\) be a 2 by 2 real symmetric matrix, and \(I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) be the unit matrix.

(1) Find all the eigenvalues \(\lambda_1, \lambda_2\) of \(M\).

(2) Under the assumption of \(\lambda_1 \neq \lambda_2\), answer i) and ii).

  • i) Let \(U = (v_1, v_2)\) be the matrix whose first and second columns consist of the eigenvectors \(v_1\) and \(v_2\) for the eigenvalues \(\lambda_1\) and \(\lambda_2\), respectively. Show that \(U\) is invertible and satisfies \(M = U \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} U^{-1}\).
  • ii) Prove that the set \(\{ U^{-1} x | x \in Z^* \}\) and \(Z^*\) are equal.

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

  • A) Every \(y \in Z^*\) can be expressed as \(y = Mx\) with some \(x \in Z^*\).
  • B) No \(y \in Z^*\) can be expressed as \(y = Mx\) with some \(x \in Z^*\).
  • C) At least one \(y \in Z^*\) can be expressed as \(y = (M - \lambda_1 I)x\) with some \(x \in Z^*\).

设 \(Z^* = \left\{ \begin{pmatrix} z_1 \\ z_2 \end{pmatrix} : \text{复数}, |z_1|^2 + |z_2|^2 \neq 0 \right\}\) 为非零复二维向量的集合。设 \(M = \begin{pmatrix} a & b \\ b & d \end{pmatrix}\) 为一个 2×2 的实对称矩阵，\(I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) 为单位矩阵。

(1) 找出 \(M\) 的所有特征值 \(\lambda_1, \lambda_2\)。

(2) 在假设 \(\lambda_1 \neq \lambda_2\) 的条件下，回答 i) 和 ii)。

  • i) 设 \(U = (v_1, v_2)\) 为一个矩阵，其第一列和第二列分别由特征值 \(\lambda_1\) 和 \(\lambda_2\) 的特征向量 \(v_1\) 和 \(v_2\) 组成。证明 \(U\) 是可逆的，并且满足 \(M = U \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} U^{-1}\)。
  • ii) 证明集合 \(\{ U^{-1} x | x \in Z^* \}\) 和 \(Z^*\) 是相等的。

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

  • A) 每个 \(y \in Z^*\) 都可以表示为 \(y = Mx\)，其中 \(x \in Z^*\)。
  • B) 没有 \(y \in Z^*\) 可以表示为 \(y = Mx\)，其中 \(x \in Z^*\)。
  • C) 至少有一个 \(y \in Z^*\) 可以表示为 \(y = (M - \lambda_1 I)x\)，其中 \(x \in Z^*\)。

Kai

(1)

To find the eigenvalues of the matrix \(M = \begin{pmatrix} a & b \\ b & d \end{pmatrix}\), we solve the characteristic equation:

\[ \det(M - \lambda I) = 0 \]

The characteristic polynomial of \(M\) is:

\[ \det \begin{pmatrix} a - \lambda & b \\ b & d - \lambda \end{pmatrix} = (a - \lambda)(d - \lambda) - b^2 = 0 \]

This simplifies to:

\[ \lambda^2 - (a + d)\lambda + (ad - b^2) = 0 \]

The eigenvalues \(\lambda_1\) and \(\lambda_2\) are the roots of this quadratic equation:

\[ \lambda_{1,2} = \frac{(a + d) \pm \sqrt{(a + d)^2 - 4(ad - b^2)}}{2} \]

(2)

i) Showing \(U\) is Invertible

Let \(v_1\) and \(v_2\) be the eigenvectors corresponding to \(\lambda_1\) and \(\lambda_2\), respectively. Define the matrix \(U = (v_1, v_2)\). Since \(\lambda_1 \neq \lambda_2\), the eigenvectors \(v_1\) and \(v_2\) are linearly independent, and thus \(U\) is invertible.

To show that \(M = U \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} U^{-1}\), consider the action of \(M\) on the eigenvectors:

\[ Mv_1 = \lambda_1 v_1 \quad \text{and} \quad Mv_2 = \lambda_2 v_2 \]

Therefore,

\[ M(v_1, v_2) = (Mv_1, Mv_2) = (\lambda_1 v_1, \lambda_2 v_2) = (v_1, v_2) \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} \]

Thus, we have:

\[ M = U \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} U^{-1} \]

ii) Proving Set Equality

To prove that the set \(\left\{ U^{-1} x | x \in Z^* \right\}\) and \(Z^*\) are equal, consider any \(x \in Z^*\). Then \(U^{-1}x \in Z^*\) if and only if \(|z_1|^2 + |z_2|^2 \neq 0\). Since \(U\) is invertible and \(Z^*\) consists of all non-zero complex vectors, applying \(U^{-1}\) to any vector in \(Z^*\) yields another non-zero complex vector, ensuring the sets are equal.

(3)

A) For every \(y \in Z^*\) to be expressible as \(y = Mx\) for some \(x \in Z^*\), \(M\) must be invertible. This requires \(\lambda_1 \neq 0\) and \(\lambda_2 \neq 0\), ensuring \(a \neq 0\), \(d \neq 0\), and \(ad - b^2 \neq 0\).

B) No \(y \in Z^*\) can be expressed as \(y = Mx\) for some \(x \in Z^*\) if \(M\) is singular and its image does not cover \(Z^*\). This happens when \(M\) has a zero eigenvalue, i.e., \(ad - b^2 = 0\) and one of the eigenvalues is zero.

C) At least one \(y \in Z^*\) can be expressed as \(y = (M - \lambda_1 I)x\) for some \(x \in Z^*\) if \(a=d\) and \(b=0\) do not both hold true. This requires \(M - \lambda_1 I\) to be invertible or have a non-trivial image, which is true if \(\lambda_1\) is not an eigenvalue of \(M\), ensuring \(\lambda_2 \neq \lambda_1\).

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