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1 , 设a=[10−2]\begin{bmatrix} 1 \\ 0 \\ - 2 \end{bmatrix} , b=[−423]\begin{bmatrix} - 4 \\ 2 \\ 3 \end{bmatrix} , c与a正交 , 且b=λa+c , 求λ和c

解：

步骤一：由 b = λa + c 得到c关于 λ 的表达式

因为 b = λa + c，移项可得 c = b - λa。

已知 a = [10−2]\begin{bmatrix} 1 \\ 0 \\ - 2 \end{bmatrix}，b = [−423]\begin{bmatrix} - 4 \\ 2 \\ 3 \end{bmatrix}，将其代入上式可得：

c = [−423]\begin{bmatrix} - 4 \\ 2 \\ 3 \end{bmatrix} - λ [10−2]\begin{bmatrix} 1 \\ 0 \\ - 2 \end{bmatrix} = [−4−λ23+2λ]\begin{bmatrix} - 4 - \lambda \\ 2 \\ 3 + 2\lambda \end{bmatrix}

步骤二：利用向量正交的性质列方程求λ

由于c与a正交，则 aᵀc = 0,

a = [10−2]\begin{bmatrix} 1 \\ 0 \\ - 2 \end{bmatrix}, c = [−4−λ23+2λ]\begin{bmatrix} - 4 - \lambda \\ 2 \\ 3 + 2\lambda \end{bmatrix}，

则：aᵀc = [1 0 -2][−4−λ23+2λ]\begin{bmatrix} - 4 - \lambda \\ 2 \\ 3 + 2\lambda \end{bmatrix}

= 1×(-4 - λ) + 0×2 + (-2)×(3 + 2λ)

= -4 - λ - 6 - 4λ

= -10 - 5λ

令 aᵀc = 0，即 -10 - 5λ = 0， 解得 λ = -2。

步骤三：将λ的值代入c的表达式，求出c

把λ=-2代c=[−4−λ23+2λ]\begin{bmatrix} - 4 - \lambda \\ 2 \\ 3 + 2\lambda \end{bmatrix} 可得c=[−4−(−2)23+2(−2)]=[−22−1]\begin{bmatrix} - 4 - ( - 2) \\ 2 \\ 3 + 2( - 2) \end{bmatrix} = \begin{bmatrix} - 2 \\ 2 \\ - 1 \end{bmatrix}

综上，λ=-2 ，c=[−22−1]\begin{bmatrix} - 2 \\ 2 \\ - 1 \end{bmatrix}

2.试把下列向量组施密特正交化 , 然后再单位化

(1) (a1 , a2 , a3)=[111124139]\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 3 & 9 \end{bmatrix} , (2) (a1 , a2 , a3)=[10−111−101−1110]\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} - 1 \\ 1 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ - 1 \end{matrix} \\ \begin{matrix} 0 \\ 1 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} - 1 \\ 1 \end{matrix} \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{array} \right\rbrack

解: (1)b1=a1=[111]\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} , 单位化得p1=b1∥b1∥\frac{b_{1}}{\left\| b_{1} \right\|}=13[111]\frac{1}{\sqrt{3}}\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}

b2=a2- [b1,a2][b1,b1]\frac{\lbrack b_{1}\ ,\ a_{2}\rbrack}{\lbrack b_{1}\ ,\ b_{1}\rbrack}b1=[123]\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}-63[111]\frac{6}{3}\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}=[−101]\begin{bmatrix} - 1 \\ 0 \\ 1 \end{bmatrix} , 单位化得p2=b2∥b2∥\frac{b_{2}}{\left\| b_{2} \right\|}=12[−101]\frac{1}{\sqrt{2}}\begin{bmatrix} - 1 \\ 0 \\ 1 \end{bmatrix}

b3=a3- [b1,a3][b1,b1]\frac{\lbrack b_{1}\ ,\ a_{3}\rbrack}{\lbrack b_{1}\ ,\ b_{1}\rbrack}b1- [b2,a3][b2,b2]\frac{\lbrack b_{2}\ ,\ a_{3}\rbrack}{\lbrack b_{2}\ ,\ b_{2}\rbrack}b2=[149]\begin{bmatrix} 1 \\ 4 \\ 9 \end{bmatrix}-143[111]\frac{14}{3}\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}-82[−101]\frac{8}{2}\begin{bmatrix} - 1 \\ 0 \\ 1 \end{bmatrix}=13[1−21]\frac{1}{3}\begin{bmatrix} 1 \\ - 2 \\ 1 \end{bmatrix}

单位化得p3=b3∥b3∥\frac{b_{3}}{\left\| b_{3} \right\|}=13[1−21]∥13[1−21]∥\frac{\frac{1}{3}\begin{bmatrix} 1 \\ - 2 \\ 1 \end{bmatrix}}{\left\| \frac{1}{3}\begin{bmatrix} 1 \\ - 2 \\ 1 \end{bmatrix} \right\|}=13[1−21]13∥[1−21]∥=[1−21]∥[1−21]∥=16[1−21]\frac{\frac{1}{3}\begin{bmatrix} 1 \\ - 2 \\ 1 \end{bmatrix}}{\frac{1}{3}\left\| \begin{bmatrix} 1 \\ - 2 \\ 1 \end{bmatrix} \right\|} = \frac{\begin{bmatrix} 1 \\ - 2 \\ 1 \end{bmatrix}}{\left\| \begin{bmatrix} 1 \\ - 2 \\ 1 \end{bmatrix} \right\|} = \frac{1}{\sqrt{6}}\begin{bmatrix} 1 \\ - 2 \\ 1 \end{bmatrix}

(2)b1=a1=[10−11]\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} - 1 \\ 1 \end{matrix} \end{array} \right\rbrack , 单位化向量为p1=b1∥b1∥=13[10−11]\frac{b_{1}}{\left\| b_{1} \right\|} = \frac{1}{\sqrt{3}}\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} - 1 \\ 1 \end{matrix} \end{array} \right\rbrack

b2=a2- [b1,a2][b1,b1]\frac{\lbrack b_{1}\ ,\ a_{2}\rbrack}{\lbrack b_{1}\ ,\ b_{1}\rbrack}b1=[1−101]\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ - 1 \end{matrix} \\ \begin{matrix} 0 \\ 1 \end{matrix} \end{array} \right\rbrack- 23[10−11]\frac{2}{3}\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} - 1 \\ 1 \end{matrix} \end{array} \right\rbrack=13[1−321]\frac{1}{3}\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ - 3 \end{matrix} \\ \begin{matrix} 2 \\ 1 \end{matrix} \end{array} \right\rbrack , 单位化得p2=b2∥b2∥\frac{b_{2}}{\left\| b_{2} \right\|}=115[1−321]\frac{1}{\sqrt{15}}\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ - 3 \end{matrix} \\ \begin{matrix} 2 \\ 1 \end{matrix} \end{array} \right\rbrack

b3=a3- [b1,a3][b1,b1]\frac{\lbrack b_{1}\ ,\ a_{3}\rbrack}{\lbrack b_{1}\ ,\ b_{1}\rbrack}b1- [b2,a3][b2,b2]\frac{\lbrack b_{2}\ ,\ a_{3}\rbrack}{\lbrack b_{2}\ ,\ b_{2}\rbrack}b2=[−1110]+23[10−11]+215[1−321]\left\lbrack \begin{array}{r} \begin{matrix} - 1 \\ 1 \end{matrix} \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{array} \right\rbrack + \frac{2}{3}\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} - 1 \\ 1 \end{matrix} \end{array} \right\rbrack + \frac{2}{15}\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ - 3 \end{matrix} \\ \begin{matrix} 2 \\ 1 \end{matrix} \end{array} \right\rbrack=15[−1334]\frac{1}{5}\left\lbrack \begin{array}{r} \begin{matrix} - 1 \\ 3 \end{matrix} \\ \begin{matrix} 3 \\ 4 \end{matrix} \end{array} \right\rbrack,

单位化得p3=b3∥b3∥\frac{b_{3}}{\left\| b_{3} \right\|}=135[−1334]\frac{1}{\sqrt{35}}\left\lbrack \begin{array}{r} \begin{matrix} - 1 \\ 3 \end{matrix} \\ \begin{matrix} 3 \\ 4 \end{matrix} \end{array} \right\rbrack

3 , 下列矩阵是不是正交矩阵? 并说明理由

(1)[1−1213−121121312−1]\begin{bmatrix} 1 & \frac{- 1}{2} & \frac{1}{3} \\ \frac{- 1}{2} & 1 & \frac{1}{2} \\ \frac{1}{3} & \frac{1}{2} & - 1 \end{bmatrix} (2)[19−89−49−8919−49−49−4979]\begin{bmatrix} \frac{1}{9} & \frac{- 8}{9} & \frac{- 4}{9} \\ \frac{- 8}{9} & \frac{1}{9} & \frac{- 4}{9} \\ \frac{- 4}{9} & \frac{- 4}{9} & \frac{7}{9} \end{bmatrix}

解:

正交矩阵的定义：

当且仅当ATA=IA^{T}A = I时，矩阵 AA 是正交矩阵，

(1)ATA=[1−1213−121121312−1][1−1213−121121312−1]A^{T}A = \begin{bmatrix} 1 & \frac{- 1}{2} & \frac{1}{3} \\ \frac{- 1}{2} & 1 & \frac{1}{2} \\ \frac{1}{3} & \frac{1}{2} & - 1 \end{bmatrix}\begin{bmatrix} 1 & \frac{- 1}{2} & \frac{1}{3} \\ \frac{- 1}{2} & 1 & \frac{1}{2} \\ \frac{1}{3} & \frac{1}{2} & - 1 \end{bmatrix}

=[1+14+19−12−12+1613−14−13−12−12+1614+1+14−16+12−1213−14−13−16+12−1219+14+1]=[4936−56−14−5632−16−14−164936]≠I= \begin{bmatrix} 1 + \frac{1}{4} + \frac{1}{9} & - \frac{1}{2} - \frac{1}{2} + \frac{1}{6} & \frac{1}{3} - \frac{1}{4} - \frac{1}{3} \\ - \frac{1}{2} - \frac{1}{2} + \frac{1}{6} & \frac{1}{4} + 1 + \frac{1}{4} & - \frac{1}{6} + \frac{1}{2} - \frac{1}{2} \\ \frac{1}{3} - \frac{1}{4} - \frac{1}{3} & - \frac{1}{6} + \frac{1}{2} - \frac{1}{2} & \frac{1}{9} + \frac{1}{4} + 1 \end{bmatrix} = \begin{bmatrix} \frac{49}{36} & - \frac{5}{6} & - \frac{1}{4} \\ - \frac{5}{6} & \frac{3}{2} & - \frac{1}{6} \\ - \frac{1}{4} & - \frac{1}{6} & \frac{49}{36} \end{bmatrix} \neq I，

因为 ATA≠IA^{T}A \neq I，所以矩阵不是正交矩阵。

(2)ATA=[19−89−49−8919−49−49−4979][19−89−49−8919−49−49−4979]A^{T}A = \begin{bmatrix} \frac{1}{9} & \frac{- 8}{9} & \frac{- 4}{9} \\ \frac{- 8}{9} & \frac{1}{9} & \frac{- 4}{9} \\ \frac{- 4}{9} & \frac{- 4}{9} & \frac{7}{9} \end{bmatrix}\begin{bmatrix} \frac{1}{9} & \frac{- 8}{9} & \frac{- 4}{9} \\ \frac{- 8}{9} & \frac{1}{9} & \frac{- 4}{9} \\ \frac{- 4}{9} & \frac{- 4}{9} & \frac{7}{9} \end{bmatrix}

=[181+6481+1681−881−881+1681−481+3281−2881−881−881+16816481+181+16813281−481−2881−481+3281−28813281−481−28811681+1681+4981]=[818100081810008181]=I= \begin{bmatrix} \frac{1}{81} + \frac{64}{81} + \frac{16}{81} & - \frac{8}{81} - \frac{8}{81} + \frac{16}{81} & - \frac{4}{81} + \frac{32}{81} - \frac{28}{81} \\ - \frac{8}{81} - \frac{8}{81} + \frac{16}{81} & \frac{64}{81} + \frac{1}{81} + \frac{16}{81} & \frac{32}{81} - \frac{4}{81} - \frac{28}{81} \\ - \frac{4}{81} + \frac{32}{81} - \frac{28}{81} & \frac{32}{81} - \frac{4}{81} - \frac{28}{81} & \frac{16}{81} + \frac{16}{81} + \frac{49}{81} \end{bmatrix} = \begin{bmatrix} \frac{81}{81} & 0 & 0 \\ 0 & \frac{81}{81} & 0 \\ 0 & 0 & \frac{81}{81} \end{bmatrix} = I，

因为 ATA=IA^{T}A = I，所以矩阵是正交矩阵。

4 , (1)设x为n维列向量 , xTx=1 , 令H=I-2xxT , 试证明H是对称的正交阵

(2)设A , B都是正交阵 , 试证明AB也是正交阵

证：(1)

1. 证明 HH 是对称矩阵

对称矩阵的定义：HT=HH^{T} = H。

H=I−2xxTH = I - 2xx^{T}

HT=IT−2(xxT)T=I−2(xT)TxT=I−2xxT=HH^{T} = I^{T} - 2(xx^{T})^{T} = I - 2(x^{T})^{T}x^{T} = I - 2xx^{T} = H

所以 HH 是对称矩阵。

2. 证明 HH 是正交矩阵

正交矩阵的定义：HTH=IH^{T}H = I 或 HHT=IHH^{T} = I，这里 HH 对称，所以只需证 H2=IH^{2} = I。

H2=(I−2xxT)(I−2xxT)H^{2} = (I - 2xx^{T})(I - 2xx^{T})

=I−2xxT−2xxT+4xxTxxT= I - 2xx^{T} - 2xx^{T} + 4xx^{T}xx^{T}

=I−4xxT+4x(xTx)xT= I - 4xx^{T} + 4x(x^{T}x)x^{T}

=I−4xxT+4x⋅1⋅xT= I - 4xx^{T} + 4x \cdot 1 \cdot x^{T}

=I−4xxT+4xxT=I= I - 4xx^{T} + 4xx^{T} = I

所以 HH 是正交矩阵。

3. 结论

HH 既对称又正交，得证。

(2)已知 AA、BB 都是 n×nn \times n 正交矩阵，即ATA=I,BTB=I.A^{T}A = I,\quad B^{T}B = I.

要证 ABAB 也是正交矩阵，即证(AB)T(AB)=I.(AB)^{T}(AB) = I.

证明：因为(AB)T(AB)=BTAT(AB)=BT(ATA)B=BTIB=BTB=I.(AB)^{T}(AB) = B^{T}A^{T}(AB) = B^{T}(A^{T}A)B = B^{T}IB = B^{T}B = I.

因此(AB)T(AB)=I,(AB)^{T}(AB) = I,即 ABAB 是正交矩阵。