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IAL 2021 Jan FP3 Q5

A Level / Edexcel / FP3

IAL 2021 Jan Paper · Question 5

题目

Problem

The matrix MM is given by

M=(6−2−1−26−1−1−15).M=\begin{pmatrix} 6&-2&-1\\ -2&6&-1\\ -1&-1&5 \end{pmatrix}.

Given that 8 is an eigenvalue of MM,

(a) determine an eigenvector corresponding to the eigenvalue 8.

(b) Determine the other two eigenvalues of MM.

(c) Hence find an orthogonal matrix PP and a diagonal matrix DD such that PTMP=DP^{\mathsf T}MP=D.

(9)
题目中文翻译

矩阵 MM 定义为

M=(6−2−1−26−1−1−15).M=\begin{pmatrix} 6&-2&-1\\ -2&6&-1\\ -1&-1&5 \end{pmatrix}.

已知 8 是 MM 的一个特征值，

(a) 求一个与特征值 8 对应的特征向量。

(b) 求 MM 的另外两个特征值。

(c) 因此，求一个正交矩阵 PP 和一个对角矩阵 DD，使得 PTMP=DP^{\mathsf T}MP=D。

解答

(a)

解法一

思路

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与特征值 8 对应的特征向量满足 (M−8I)v=0(M-8I)\mathbf{v}=\mathbf0。联立所得方程，取一个最简单的非零解即可。

答题过程

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Let

v=(xyz).\mathbf{v}=\begin{pmatrix}x\\y\\z\end{pmatrix}.

For the eigenvalue 8,

(M−8I)v=0,(M-8I)\mathbf{v}=\mathbf0,

so

(−2−2−1−2−2−1−1−1−3)(xyz)=(000).\begin{pmatrix} -2&-2&-1\\ -2&-2&-1\\ -1&-1&-3 \end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} =\begin{pmatrix}0\\0\\0\end{pmatrix}.

The independent equations are

2x+2y+z=02x+2y+z=0

and

x+y+3z=0.x+y+3z=0.

Writing u=x+yu=x+y, these become 2u+z=02u+z=0 and u+3z=0u+3z=0. Hence z=0z=0 and u=0u=0, so y=−xy=-x.

Taking x=1x=1, an eigenvector is

(1−10).\boxed{\begin{pmatrix}1\\-1\\0\end{pmatrix}}.

(b)

解法一

思路

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由 det⁡(M−λI)=0\det(M-\lambda I)=0 建立特征方程。题目已给出 8 是特征值，因此三次多项式应含因子 λ−8\lambda-8；除去此因子即可求另外两个特征值。

答题过程

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The characteristic equation is

det⁡(M−λI)=0.\det(M-\lambda I)=0.

Expanding the determinant,

0= ∣6−λ−2−1−26−λ−1−1−15−λ∣= (6−λ)((6−λ)(5−λ)−1) +2(−2(5−λ)−1) −(2+6−λ).\begin{align*} 0=&\, \begin{vmatrix} 6-\lambda&-2&-1\\ -2&6-\lambda&-1\\ -1&-1&5-\lambda \end{vmatrix}\\[4mm] =&\,(6-\lambda) \bigl((6-\lambda)(5-\lambda)-1\bigr)\\[4mm] &\,\hspace{2pt}+2\bigl(-2(5-\lambda)-1\bigr)\\[4mm] &\,\hspace{4pt}-\bigl(2+6-\lambda\bigr). \end{align*}

After expanding and multiplying by −1-1,

λ3−17λ2+90λ−144=0.\lambda^3-17\lambda^2+90\lambda-144=0.

Since 8 is already known to be an eigenvalue,

 λ3−17λ2+90λ−144= (λ−8)(λ2−9λ+18)= (λ−8)(λ−3)(λ−6).\begin{align*} &\,\lambda^3-17\lambda^2+90\lambda-144\\[4mm] =&\,(\lambda-8)(\lambda^2-9\lambda+18)\\[4mm] =&\,(\lambda-8)(\lambda-3)(\lambda-6). \end{align*}

Therefore the other two eigenvalues are

3and6.\boxed{3\quad\text{and}\quad 6}.

(c)

解法一

思路

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实对称矩阵可由正交矩阵对角化。分别求特征值 3、6 的特征向量，并与 (a) 中特征值 8 的向量一起单位化；把这些单位特征向量按列组成 PP，同时按相同顺序把特征值放入 DD。

答题过程

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For λ=3\lambda=3, solving (M−3I)v=0(M-3I)\mathbf{v}=\mathbf0 gives an eigenvector

v3=(111).\mathbf{v}_3=\begin{pmatrix}1\\1\\1\end{pmatrix}.

For λ=6\lambda=6, solving (M−6I)v=0(M-6I)\mathbf{v}=\mathbf0 gives an eigenvector

v6=(11−2).\mathbf{v}_6=\begin{pmatrix}1\\1\\-2\end{pmatrix}.

Together with the eigenvector from part (a), take

v8=(1−10).\mathbf{v}_8=\begin{pmatrix}1\\-1\\0\end{pmatrix}.

Their scalar products are all zero:

v8⋅v3=0,v8⋅v6=0,v3⋅v6=0.\mathbf{v}_8\cdot\mathbf{v}_3=0, \qquad \mathbf{v}_8\cdot\mathbf{v}_6=0, \qquad \mathbf{v}_3\cdot\mathbf{v}_6=0.

Their corresponding unit eigenvectors are

u8=12(1−10),\mathbf{u}_8=\frac1{\sqrt2} \begin{pmatrix}1\\-1\\0\end{pmatrix}, u3=13(111),\mathbf{u}_3=\frac1{\sqrt3} \begin{pmatrix}1\\1\\1\end{pmatrix},

and

u6=16(11−2).\mathbf{u}_6=\frac1{\sqrt6} \begin{pmatrix}1\\1\\-2\end{pmatrix}.

Using the order 8,3,68,3,6, we may therefore take

P=(121316−121316013−26)\boxed{ P=\begin{pmatrix} \dfrac1{\sqrt2}&\dfrac1{\sqrt3}&\dfrac1{\sqrt6}\\[3mm] -\dfrac1{\sqrt2}&\dfrac1{\sqrt3}&\dfrac1{\sqrt6}\\[3mm] 0&\dfrac1{\sqrt3}&-\dfrac2{\sqrt6} \end{pmatrix}}

and

D=(800030006).\boxed{ D=\begin{pmatrix} 8&0&0\\ 0&3&0\\ 0&0&6 \end{pmatrix}}.

The columns of PP are orthonormal, so PP is orthogonal, and the matching column order gives PTMP=DP^{\mathsf T}MP=D.