東京大学 情報理工学系研究科 コンピュータ科学専攻 2015年2月実施 問題1

标签：

• Tokyo-University
• Eigenvalues-and-Eigenvectors
• Inner-Product-and-Orthogonality
• Matrix-Determinant

Author

kainoj

Description

(1) Suppose that $A$ is a real symmetric matrix. Let $\boldsymbol{x}$ and $\lambda$ be an eigenvector and the corresponding eigenvalue of $A$, respectively. Prove that $\lambda$ is a real number.

(2) Suppose that $A$ is a real symmetric matrix. Let $\boldsymbol{x}_1$ and $\boldsymbol{x}_2$ be eigenvectors of $A$; and $\lambda_1$ and $\lambda_2$ be the eigenvalues of $A$ that correspond to $\boldsymbol{x}_1$ and $\boldsymbol{x}_2$, respectively. Furthermore, assume that $\lambda_1 \neq \lambda_2$.

Prove that $\boldsymbol{x}_1 \cdot \boldsymbol{x}_2 = 0$. Here $\boldsymbol{x}_1 \cdot \boldsymbol{x}_2$ denotes the inner product of the two vectors $\boldsymbol{x}_1$ and $\boldsymbol{x}_2$.

(3) Let $A$ be a matrix

$$A = \begin{pmatrix} x & a_1 & b_1 & c_1 \\ y & a_2 & b_2 & c_2 \\ z & a_3 & b_3 & c_3 \\ 1 & 1 & 1 & 1 \end{pmatrix}$$

where $a_i$, $b_i$, and $c_i$ ( $i \in \{1, 2, 3\}$) are real numbers such that the two vectors $(b_1 - a_1, b_2 - a_2, b_3 - a_3)$ and $(c_1 - a_1, c_2 - a_2, c_3 - a_3)$ are linearly independent. Let us define a function $f: \mathbb{R}^3 \to \mathbb{R}$ by

$$f(x, y, z) = \text{det}(A)$$

Prove that the equation $f(x, y, z) = 0$ determines a plane in the $xyz$ space.

(4) In the setting of Question (3), give three points that lie on the plane determined by $f(x, y, z) = 0$.

题目描述

（1）设 $A$ 为实对称矩阵，$\boldsymbol{x}$ 是 $A$ 的特征向量，对应特征值为 $\lambda$。证明 $\lambda$ 是实数。

（2）设 $A$ 为实对称矩阵，$\boldsymbol{x}_1,\boldsymbol{x}_2$ 分别是对应特征值 $\lambda_1,\lambda_2$ 的特征向量，且 $\lambda_1\ne\lambda_2$。证明

$$\boldsymbol{x}_1\cdot\boldsymbol{x}_2=0,$$

其中“$\cdot$”表示向量内积。

（3）令

$$A=\begin{pmatrix} x&a_1&b_1&c_1\\ y&a_2&b_2&c_2\\ z&a_3&b_3&c_3\\ 1&1&1&1 \end{pmatrix},$$

其中 $a_i,b_i,c_i\ (i\in\{1,2,3\})$ 均为实数，并且向量 $(b_1-a_1,b_2-a_2,b_3-a_3)$ 与 $(c_1-a_1,c_2-a_2,c_3-a_3)$ 线性无关。定义 $f:\mathbb{R}^3\to\mathbb{R}$ 为

$$f(x,y,z)=\det(A).$$

证明方程 $f(x,y,z)=0$ 在 $xyz$ 空间中确定一个平面。

（4）在第（3）问的设定下，给出位于该平面上的三个点。

考点

• 实对称矩阵的谱性质：证明特征值为实数，并证明不同特征值对应的特征向量正交。
• 行列式与线性相关：把 $\det(A)=0$ 解释为矩阵列向量线性相关，并据此刻画几何对象。
• 仿射平面：利用给定的线性无关条件证明零点集合确为平面，并识别平面上的具体点。

Kai

(1)

Let $A\in \mathbb{R}^{n\times n}$ such that $A = A^T$ and $Ax = \lambda x$ for some eigenvector $x$ and corresponding eigenvalue $\lambda$. Suppose that $\lambda \in \mathbb{C}$, i.e. $\lambda = a + ib$ for some $a,b\in\mathbb{R}$. Let's start with $Ax = \lambda x$:

$$\begin{aligned} Ax &= \lambda x && \text{take conjugate transpose} \\ x^H A^H &= \lambda^H x^H && \text{$A$ is real, so $A=A^H$} \\ x^H A &= \lambda^H x^H && \text{multiply by $x$} \\ x^H A x &= \lambda^H x^H x && \text{now, $Ax = \lambda x$} \\ \lambda x^H x &= \lambda^H x^H x \\ \lambda &= \lambda^H \\ a + bi &= a - bi \end{aligned}$$

Hence, $\text{Im}(\lambda)$ = 0, which means $\lambda \in \mathbb{R}$.

(2)

Start with the first eigenpair:

$$\begin{aligned} Ax_1 &= \lambda_1 x_1 && \text{take transpose. $A = A^T$} \\ x_1^T A &= \lambda_1 x_1^T && \text{multiply by $x_2$} \\ x_1^T A x_2 &= \lambda_1 x_1^T x_2 && \\ \end{aligned}$$

Now fiddle with the second eigenpair:

$$\begin{aligned} Ax_2 &= \lambda_2 x_2 && \text{left-multiply by $x_1^T$} \\ x_1^T Ax_2 &= \lambda_2 x_1^T x_2 \end{aligned}$$

If we subtract both result, we get:

$$\begin{aligned} 0 = (\lambda_1 - \lambda_2) x_1^T x_2 \end{aligned}$$

Since $\lambda_1 \neq \lambda_2$, then $x_1^T x_2 = 0$. In other words, $x_1 \cdot x_2 =0$.

(3) - send help

$$\begin{aligned} A = \begin{pmatrix} x & a_1 & b_1 & c_1 \\ x & a_2 & b_2 & c_2 \\ x & a_3 & b_3 & c_3 \\ 1 & 1 & 1 & 1 \end{pmatrix} \end{aligned}$$

$$\begin{aligned} f(x,y,z) = \text{det}(A) \end{aligned}$$

This seems to be difficult. I succeeded in computing determinant using Laplace expansion wrt first column, but it gets hardcore later. Any ideas?

(4)

Give 3 points that lie on plane $f(x,y,z) = 0$. When is determinant $=0$ ? For example, when some columns are dependent. Let's make some columns dependent. Zum Beispiel, let's take:

$$\begin{aligned} x = a_1 && y = a_2 && z = a_3 \end{aligned}$$

Now first and second column are dependent, so $\text{det}(A) = f(x,y,z) = 0$. To get two others solutions, take $x = b_1, y = b_2, z = b_3$ and $x = c_1, y = c_2, z = c_3$ or simply multiply the first solution by some constants.