题目
Problem
A = ( a a 1 − a 4 0 4 a 5 ) A=\begin{pmatrix}
a&a&1\\
-a&4&0\\
4&a&5
\end{pmatrix} A = a − a 4 a 4 a 1 0 5 where a a a
Given that 2 2 2 A A A
(a) determine
(i) the value of a a a
(ii) the other two eigenvalues of A A A
A normalised eigenvector for the eigenvalue 2 2 2
( 1 6 1 6 − 2 6 ) \begin{pmatrix}
\frac1{\sqrt6}\\
\frac1{\sqrt6}\\
-\frac2{\sqrt6}
\end{pmatrix} 6 1 6 1 − 6 2 (b) Determine a normalised eigenvector for each of the other eigenvalues of A A A
(c) Hence determine a matrix P P P D D D P T A P = D P^TAP=D P T A P = D
(12)
题目中文翻译
A = ( a a 1 − a 4 0 4 a 5 ) A=\begin{pmatrix}
a&a&1\\
-a&4&0\\
4&a&5
\end{pmatrix} A = a − a 4 a 4 a 1 0 5 其中 a a a
已知 2 2 2 A A A
(a) 求
(i) a a a
(ii) A A A
已知特征值 2 2 2
( 1 6 1 6 − 2 6 ) \begin{pmatrix}
\frac1{\sqrt6}\\
\frac1{\sqrt6}\\
-\frac2{\sqrt6}
\end{pmatrix} 6 1 6 1 − 6 2 (b) 求另外两个特征值对应的单位特征向量
(c) 因此求矩阵 P P P D D D P T A P = D P^TAP=D P T A P = D
解答
(a)(i)
解法一：使用特征行列式
思路
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因为 2 是特征值，所以 det ( A − 2 I ) = 0 \det(A-2I)=0 det ( A − 2 I ) = 0 a a a a > 0 a>0 a > 0
答题过程
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Since 2 2 2
det ( A − 2 I ) = 0. \det(A-2I)=0. det ( A − 2 I ) = 0. Now
A − 2 I = ( a − 2 a 1 − a 2 0 4 a 3 ) . A-2I=
\begin{pmatrix}
a-2&a&1\\
-a&2&0\\
4&a&3
\end{pmatrix}. A − 2 I = a − 2 − a 4 a 2 a 1 0 3 . Expanding along the first row,
0 = 6 ( a − 2 ) + 3 a 2 − a 2 − 8 , 0 = 2 a 2 + 6 a − 20 , 0 = ( a − 2 ) ( a + 5 ) . \begin{align*}
0
=&\,6(a-2)+3a^2-a^2-8, \\
0
=&\,2a^2+6a-20, \\
0
=&\,(a-2)(a+5).
\end{align*} 0 = 0 = 0 = 6 ( a − 2 ) + 3 a 2 − a 2 − 8 , 2 a 2 + 6 a − 20 , ( a − 2 ) ( a + 5 ) . As a > 0 a>0 a > 0
a = 2 . \boxed{a=2}. a = 2 .
解法二：使用 A x = 2 x A\mathbf x=2\mathbf x A x = 2 x
思路
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令非零特征向量为 ( x , y , z ) T (x,y,z)^T ( x , y , z ) T A x = 2 x A\mathbf x=2\mathbf x A x = 2 x y , z y,z y , z a a a x ≠ 0 x\ne0 x = 0 x x x
答题过程
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Let
x = ( x y z ) \mathbf x=
\begin{pmatrix}
x\\y\\z
\end{pmatrix} x = x y z be a non-zero eigenvector for the eigenvalue 2 2 2 A x = 2 x A\mathbf x=2\mathbf x A x = 2 x
a x + a y + z = 2 x , − a x + 4 y = 2 y , 4 x + a y + 5 z = 2 z . \begin{align*}
ax+ay+z
=&\,2x, \\
-ax+4y
=&\,2y, \\
4x+ay+5z
=&\,2z.
\end{align*} a x + a y + z = − a x + 4 y = 4 x + a y + 5 z = 2 x , 2 y , 2 z . The second equation gives
y = a x 2 , y=\frac{ax}{2}, y = 2 a x , and the third gives
z = − 4 x + a y 3 = − ( 8 + a 2 ) x 6 . z=-\frac{4x+ay}{3}
=-\frac{(8+a^2)x}{6}. z = − 3 4 x + a y = − 6 ( 8 + a 2 ) x . If x = 0 x=0 x = 0 y = z = 0 y=z=0 y = z = 0 x ≠ 0 x\ne0 x = 0
Substituting for y y y z z z x x x
a + a 2 2 − 8 + a 2 6 = 2 , a 2 + 3 a − 10 = 0 , ( a − 2 ) ( a + 5 ) = 0. \begin{align*}
a+\frac{a^2}{2}-\frac{8+a^2}{6}
=&\,2, \\
a^2+3a-10
=&\,0, \\
(a-2)(a+5)
=&\,0.
\end{align*} a + 2 a 2 − 6 8 + a 2 = a 2 + 3 a − 10 = ( a − 2 ) ( a + 5 ) = 2 , 0 , 0. Since a > 0 a>0 a > 0
a = 2 . \boxed{a=2}. a = 2 .
(a)(ii)
解法一
思路
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把 a = 2 a=2 a = 2
答题过程
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With a = 2 a=2 a = 2
A = ( 2 2 1 − 2 4 0 4 2 5 ) . A=
\begin{pmatrix}
2&2&1\\
-2&4&0\\
4&2&5
\end{pmatrix}. A = 2 − 2 4 2 4 2 1 0 5 . The characteristic equation is
0 = det ( A − λ I ) = ( 2 − λ ) ( 4 − λ ) ( 5 − λ ) + 4 ( 5 − λ ) − 4 − 4 ( 4 − λ ) = ( 2 − λ ) ( 4 − λ ) ( 5 − λ ) . \begin{align*}
0
=&\,\det(A-\lambda I) \\
=&\,(2-\lambda)(4-\lambda)(5-\lambda) \\
&\,+4(5-\lambda)-4-4(4-\lambda) \\
=&\,(2-\lambda)(4-\lambda)(5-\lambda).
\end{align*} 0 = = = det ( A − λ I ) ( 2 − λ ) ( 4 − λ ) ( 5 − λ ) + 4 ( 5 − λ ) − 4 − 4 ( 4 − λ ) ( 2 − λ ) ( 4 − λ ) ( 5 − λ ) . Therefore, the other two eigenvalues are
4 and 5 . \boxed{4\text{ and }5}. 4 and 5 .
(b)
解法一
思路
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对每个新特征值分别解 ( A − λ I ) x = 0 (A-\lambda I)\mathbf x=\mathbf0 ( A − λ I ) x = 0
答题过程
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For λ = 4 \lambda=4 λ = 4
A − 4 I = ( − 2 2 1 − 2 0 0 4 2 1 ) . A-4I=
\begin{pmatrix}
-2&2&1\\
-2&0&0\\
4&2&1
\end{pmatrix}. A − 4 I = − 2 − 2 4 2 0 2 1 0 1 . Solving ( A − 4 I ) x = 0 (A-4I)\mathbf x=\mathbf0 ( A − 4 I ) x = 0
x = 0 , z = − 2 y . x=0,
\qquad
z=-2y. x = 0 , z = − 2 y . Taking y = − 1 y=-1 y = − 1
( 0 − 1 2 ) , \begin{pmatrix}
0\\-1\\2
\end{pmatrix}, 0 − 1 2 , whose magnitude is 5 \sqrt5 5
1 5 ( 0 − 1 2 ) . \boxed{
\frac1{\sqrt5}
\begin{pmatrix}
0\\-1\\2
\end{pmatrix}
}. 5 1 0 − 1 2 . For λ = 5 \lambda=5 λ = 5
A − 5 I = ( − 3 2 1 − 2 − 1 0 4 2 0 ) . A-5I=
\begin{pmatrix}
-3&2&1\\
-2&-1&0\\
4&2&0
\end{pmatrix}. A − 5 I = − 3 − 2 4 2 − 1 2 1 0 0 . Solving ( A − 5 I ) x = 0 (A-5I)\mathbf x=\mathbf0 ( A − 5 I ) x = 0
y = − 2 x , z = 7 x . y=-2x,
\qquad
z=7x. y = − 2 x , z = 7 x . Taking x = 1 x=1 x = 1
( 1 − 2 7 ) , \begin{pmatrix}
1\\-2\\7
\end{pmatrix}, 1 − 2 7 , whose magnitude is 54 \sqrt{54} 54
1 54 ( 1 − 2 7 ) . \boxed{
\frac1{\sqrt{54}}
\begin{pmatrix}
1\\-2\\7
\end{pmatrix}
}. 54 1 1 − 2 7 .
(c)
解法一：官方评分资料的预期构造
思路
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按照题目的预期，把三个单位特征向量依次作为 P P P D D D P P P P − 1 A P = D P^{-1}AP=D P − 1 A P = D P T A P = D P^TAP=D P T A P = D
答题过程
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Following the construction credited in the official mark scheme, take
P = ( 1 6 0 1 54 1 6 − 1 5 − 2 54 − 2 6 2 5 7 54 ) , P=
\begin{pmatrix}
\dfrac1{\sqrt6}&0&\dfrac1{\sqrt{54}}\\[2mm]
\dfrac1{\sqrt6}&-\dfrac1{\sqrt5}&-\dfrac2{\sqrt{54}}\\[2mm]
-\dfrac2{\sqrt6}&\dfrac2{\sqrt5}&\dfrac7{\sqrt{54}}
\end{pmatrix}, P = 6 1 6 1 − 6 2 0 − 5 1 5 2 54 1 − 54 2 54 7 , with
D = ( 2 0 0 0 4 0 0 0 5 ) . D=
\begin{pmatrix}
2&0&0\\
0&4&0\\
0&0&5
\end{pmatrix}. D = 2 0 0 0 4 0 0 0 5 . The columns of P P P D D D
A P = P D . AP=PD. A P = P D . However, the first two columns are not orthogonal, since
1 6 ( 1 1 − 2 ) ⋅ 1 5 ( 0 − 1 2 ) = − 5 30 ≠ 0. \begin{align*}
&\,\frac1{\sqrt6}
\begin{pmatrix}
1\\1\\-2
\end{pmatrix}
\mathbin{\boldsymbol{\cdot}}
\frac1{\sqrt5}
\begin{pmatrix}
0\\-1\\2
\end{pmatrix}
\\
=&\,-\frac5{\sqrt{30}}\ne0.
\end{align*} = 6 1 1 1 − 2 ⋅ 5 1 0 − 1 2 − 30 5 = 0. Thus P T P ≠ I P^TP\ne I P T P = I P T ≠ P − 1 P^T\ne P^{-1} P T = P − 1
P − 1 A P = D , \boxed{P^{-1}AP=D}, P − 1 A P = D , but they do not satisfy P T A P = D P^TAP=D P T A P = D
Indeed, no invertible P P P P T A P P^TAP P T A P
P T ( A − A T ) P = 0. P^T(A-A^T)P=0. P T ( A − A T ) P = 0. Since P P P A = A T A=A^T A = A T A A A P T P^T P T P − 1 P^{-1} P − 1
