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京都大学 情報学研究科 知能情報学専攻 2023年2月実施 基礎科目 F1-1

Author

祭音Myyura (co-authored with GPT 5.6 SOL)

Description

Let

A=(1a0b30112),a,bR.A=\begin{pmatrix} 1&a&0\\ b&3&0\\ -1&1&2 \end{pmatrix},\qquad a,b\in\mathbb R.
  1. Find the condition on a,ba,b under which all eigenvalues of AA are distinct real numbers.
  2. Let λ1\lambda_1 and λ2\lambda_2 be the largest and smallest eigenvalues, and let v1,v2v_1,v_2 be corresponding eigenvectors. For a=6,b=4a=6,b=4:
    1. find λ1,λ2,v1,v2\lambda_1,\lambda_2,v_1,v_2;
    2. find an orthonormal basis of W=span{v1,v2}W=\operatorname{span}\{v_1,v_2\};
    3. find the orthogonal projection of r=(1,4,6)r=(1,4,-6)^\top onto WW.

题目描述

A=(1a0b30112),a,bR.A=\begin{pmatrix} 1&a&0\\ b&3&0\\ -1&1&2 \end{pmatrix},\qquad a,b\in\mathbb R.
  1. 求使 AA 的特征值均为两两不同实数的 a,ba,b 条件。
  2. 记最大、最小特征值分别为 λ1,λ2\lambda_1,\lambda_2,对应特征向量为 v1,v2v_1,v_2。当 a=6,b=4a=6,b=4 时:
    1. λ1,λ2,v1,v2\lambda_1,\lambda_2,v_1,v_2
    2. W=span{v1,v2}W=\operatorname{span}\{v_1,v_2\} 的一组标准正交基;
    3. r=(1,4,6)r=(1,4,-6)^\topWW 上的正交投影。

Kai

Q.1

The characteristic polynomial factors as

det(λIA)=(λ2)((λ1)(λ3)ab)=(λ2)((λ2)2(1+ab)).\det(\lambda I-A) =(\lambda-2)\left((\lambda-1)(\lambda-3)-ab\right) =(\lambda-2)\left((\lambda-2)^2-(1+ab)\right).

Hence the eigenvalues are

2,2+1+ab,21+ab.2,\qquad 2+\sqrt{1+ab},\qquad 2-\sqrt{1+ab}.

They are real and pairwise distinct if and only if

ab>1.\boxed{ab>-1}.

Q.2

For a=6,b=4a=6,b=4, the eigenvalues are 7,2,37,2,-3. Therefore

λ1=7,λ2=3.\lambda_1=7,\qquad \lambda_2=-3.

(1)

Solving (A7I)v1=0(A-7I)v_1=0 and (A+3I)v2=0(A+3I)v_2=0, we may take

v1=(110),v2=(321).\boxed{v_1=\begin{pmatrix}1\\1\\0\end{pmatrix}}, \qquad \boxed{v_2=\begin{pmatrix}-3\\2\\-1\end{pmatrix}}.

(2)

Normalize v1v_1 and apply Gram-Schmidt to v2v_2:

e1=12(110),e_1=\frac1{\sqrt2}\begin{pmatrix}1\\1\\0\end{pmatrix},
v2(v2e1)e1=12(552).v_2-(v_2^\top e_1)e_1 =\frac12\begin{pmatrix}-5\\5\\-2\end{pmatrix}.

Thus an orthonormal basis is

e1=12(110),e2=136(552).\boxed{ e_1=\frac1{\sqrt2}\begin{pmatrix}1\\1\\0\end{pmatrix}, \qquad e_2=\frac1{3\sqrt6}\begin{pmatrix}-5\\5\\-2\end{pmatrix} }.

(3)

Using this basis,

projWr=(re1)e1+(re2)e2=(5/25/20)+(5/25/21)=(051).\begin{aligned} \operatorname{proj}_W r &=(r^\top e_1)e_1+(r^\top e_2)e_2\\ &=\begin{pmatrix}5/2\\5/2\\0\end{pmatrix} +\begin{pmatrix}-5/2\\5/2\\-1\end{pmatrix} =\boxed{\begin{pmatrix}0\\5\\-1\end{pmatrix}}. \end{aligned}