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IAL 2024 Jan FP3 Q4

A Level / Edexcel / FP3

IAL 2024 Jan Paper · Question 4

题目

Problem

M=(0−13−14−13−10)M=\begin{pmatrix} 0&-1&3\\ -1&4&-1\\ 3&-1&0 \end{pmatrix}

Given that (1−21)\begin{pmatrix}1\\-2\\1\end{pmatrix} is an eigenvector of MM

(a) determine its corresponding eigenvalue.

Given that −3-3 is an eigenvalue of MM

(b) determine a corresponding eigenvector.

Hence, given that (111)\begin{pmatrix}1\\1\\1\end{pmatrix} is also an eigenvector of MM

(c) determine a diagonal matrix DD and an orthogonal matrix PP such that D=PTMPD=P^TMP

(8)
题目中文翻译 M=(0−13−14−13−10)M=\begin{pmatrix} 0&-1&3\\ -1&4&-1\\ 3&-1&0 \end{pmatrix}

已知 (1−21)\begin{pmatrix}1\\-2\\1\end{pmatrix} 是 MM 的特征向量

(a) 求其对应的特征值

已知 −3-3 是 MM 的特征值

(b) 求一个对应的特征向量

因此，已知 (111)\begin{pmatrix}1\\1\\1\end{pmatrix} 也是 MM 的特征向量

(c) 求对角矩阵 DD 和正交矩阵 PP，使得 D=PTMPD=P^TMP

解答

(a)

解法一

思路

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把给定特征向量左乘矩阵 MM。所得向量若是原向量的 λ\lambda 倍，则这个倍数就是对应的特征值。

答题过程

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Let

v=(1−21).\mathbf v= \begin{pmatrix}1\\-2\\1\end{pmatrix}.

Then

Mv= (0−13−14−13−10)(1−21)= (5−105)= 5(1−21).\begin{align*} M\mathbf v =&\, \begin{pmatrix} 0&-1&3\\ -1&4&-1\\ 3&-1&0 \end{pmatrix} \begin{pmatrix}1\\-2\\1\end{pmatrix} \\ =&\, \begin{pmatrix}5\\-10\\5\end{pmatrix} \\ =&\,5 \begin{pmatrix}1\\-2\\1\end{pmatrix}. \end{align*}

Therefore, the corresponding eigenvalue is

5.\boxed{5}.

(b)

解法一

思路

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对特征值 −3-3，解齐次方程 (M+3I)v=0(M+3I)\mathbf v=\mathbf0。两个独立方程会给出 z=−xz=-x、y=0y=0，再任选一个非零的 xx 即可得到特征向量。

答题过程

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For the eigenvalue −3-3, let

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

Then (M+3I)v=0(M+3I)\mathbf v=\mathbf0, so

(3−13−17−13−13)(xyz)=(000).\begin{pmatrix} 3&-1&3\\ -1&7&-1\\ 3&-1&3 \end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix}.

The independent equations are

3x−y+3z=0,−x+7y−z=0.3x-y+3z=0, \qquad -x+7y-z=0.

From the first equation, y=3x+3zy=3x+3z. Substituting this into the second equation,

−x+7(3x+3z)−z= 0,20x+20z= 0,\begin{align*} -x+7(3x+3z)-z=&\,0, \\ 20x+20z=&\,0, \end{align*}

so z=−xz=-x and hence y=0y=0. Taking x=1x=1, a corresponding eigenvector is

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

(c)

解法一

思路

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先求第三个已知特征向量 (1,1,1)T(1,1,1)^T 的特征值。由于 MM 是实对称矩阵，三个不同特征值对应的特征向量彼此正交；把它们分别单位化后按列组成 PP，并按相同顺序把特征值放入 DD 的主对角线。

答题过程

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For

w=(111),\mathbf w= \begin{pmatrix}1\\1\\1\end{pmatrix},

we have

Mw= (222)= 2(111).\begin{align*} M\mathbf w =&\, \begin{pmatrix}2\\2\\2\end{pmatrix} \\ =&\,2 \begin{pmatrix}1\\1\\1\end{pmatrix}. \end{align*}

Thus the third eigenvalue is 22.

Choose the eigenvectors in the order corresponding to the eigenvalues −3-3, 22 and 55:

v−3=(10−1),v2=(111),v5=(1−21).\mathbf v_{-3}= \begin{pmatrix}1\\0\\-1\end{pmatrix}, \qquad \mathbf v_2= \begin{pmatrix}1\\1\\1\end{pmatrix}, \qquad \mathbf v_5= \begin{pmatrix}1\\-2\\1\end{pmatrix}.

Their norms are 2\sqrt2, 3\sqrt3 and 6\sqrt6 respectively. Therefore, the corresponding unit eigenvectors are

u−3=12(10−1),\mathbf u_{-3}= \frac1{\sqrt2} \begin{pmatrix}1\\0\\-1\end{pmatrix}, u2=13(111),\mathbf u_2= \frac1{\sqrt3} \begin{pmatrix}1\\1\\1\end{pmatrix},

and

u5=16(1−21).\mathbf u_5= \frac1{\sqrt6} \begin{pmatrix}1\\-2\\1\end{pmatrix}.

The vectors are mutually orthogonal, so placing them as the columns of PP makes PP orthogonal. Thus one valid pair is

D=(−300020005)\boxed{ D= \begin{pmatrix} -3&0&0\\ 0&2&0\\ 0&0&5 \end{pmatrix} }

and

P=(121316013−26−121316).\boxed{ P= \begin{pmatrix} \frac1{\sqrt2}&\frac1{\sqrt3}&\frac1{\sqrt6}\\ 0&\frac1{\sqrt3}&-\frac2{\sqrt6}\\ -\frac1{\sqrt2}&\frac1{\sqrt3}&\frac1{\sqrt6} \end{pmatrix} }.

With the columns and diagonal entries in this matching order,

D=PTMP.\boxed{D=P^TMP}.