《统计学习方法》第15章习题答案（持续更新）
15.1
A = [ 1 2 0 2 0 2 ] A=\begin{bmatrix} 1 & 2 & 0\\2 & 0 & 2 \end {bmatrix} A=[122002]
A T A = [ 1 2 2 0 0 2 ] [ 1 2 0 2 0 2 ] = [ 5 2 4 2 4 0 4 0 4 ] A^TA= \begin{bmatrix} 1 & 2\\2 & 0\\ 0 & 2 \end {bmatrix}\begin{bmatrix} 1 & 2 & 0\\2 & 0 & 2 \end {bmatrix} = \begin{bmatrix} 5 & 2 & 4\\2 & 4& 0\\ 4& 0 & 4 \end {bmatrix} ATA=⎣⎡120202⎦⎤[122002]=⎣⎡524240404⎦⎤
∣ A T A − λ I ∣ = ∣ 5 − λ 2 4 2 4 − λ 0 4 0 4 − λ ∣ = ( 5 − λ ) ( 4 − λ ) 2 − 20 ( 4 − λ ) − 4 ( 4 − λ ) \begin{vmatrix} A^TA - \lambda I\end{vmatrix} = \begin{vmatrix} 5-\lambda & 2 & 4 \\ 2 & 4-\lambda & 0 \\ 4 & 0 & 4 - \lambda \end{vmatrix} = (5-\lambda)(4-\lambda)^2 - 20(4 - \lambda) - 4(4 - \lambda) ∣∣ATA−λI∣∣=∣∣∣∣∣∣5−λ2424−λ0404−λ∣∣∣∣∣∣=(5−λ)(4−λ)2−20(4−λ)−4(4−λ)
= ( 4 − λ ) ( ( 5 − λ ) ( 4 − λ ) − 20 ) = ( 4 − λ ) ( λ 2 − 9 λ ) = 0 = (4 - \lambda)((5 - \lambda)(4 - \lambda) - 20) = (4 - \lambda)(\lambda^2 - 9\lambda)=0 =(4−λ)((5−λ)(4−λ)−20)=(4−λ)(λ2−9λ)=0
λ 1 = 9 , λ 2 = 4 , λ 0 = 0 \lambda_1 = 9,\lambda_2 = 4,\lambda_0 = 0 λ1=9,λ2=4,λ0=0
σ 1 = 3 , σ 2 = 2 , σ 0 = 0 \sigma_1 = 3,\sigma_2 = 2,\sigma_0 = 0 σ1=3,σ2=2,σ0=0
( A − λ 1 I ) x = [ − 4 2 4 2 − 5 0 4 0 − 5 ] x = 0 (A - \lambda_1 I)x =\begin{bmatrix}-4 & 2 & 4\\2 & -5& 0\\ 4& 0 & -5 \end {bmatrix} x = 0 (A−λ1I)x=⎣⎡−4242−5040−5⎦⎤x=0
v 1 = [ 5 3 2 5 15 4 5 15 ] , v 2 = [ 0 2 5 5 − 5 5 ] v_1 = \begin{bmatrix} \frac{\sqrt5 }{3} \\ \frac{2\sqrt5 }{15} \\ \frac{4\sqrt5}{15}\end{bmatrix},v_2 = \begin{bmatrix} 0 \\ \frac{2\sqrt5 }{5} \\ -\frac{\sqrt5}{5}\end{bmatrix} v1=⎣⎢⎡3515251545⎦⎥⎤,v2=⎣⎢⎡0525−55⎦⎥⎤
N ( A ) N(A) N(A)的正交基是 v 3 = [ − 2 3 1 3 2 3 ] v_3 = \begin{bmatrix} -\frac{2}{3} \\ \frac{1 }{3} \\ \frac{2}{3}\end{bmatrix} v3=⎣⎡−323132⎦⎤
u 1 = 1 σ 1 A v 1 = 1 3 [ 7 5 15 14 5 15 ] = [ 7 5 45 14 5 45 ] u_1 = \frac{1}{\sigma_1}Av_1= \frac{1}{3}\begin{bmatrix} \frac{7\sqrt5 }{15} \\ \frac{14\sqrt5 }{15} \end{bmatrix}=\begin{bmatrix} \frac{7\sqrt5 }{45} \\ \frac{14\sqrt5 }{45} \end{bmatrix} u1=σ11Av1=31[157515145]=[457545145]
u 2 = 1 σ 2 A v 1 = 2 = 1 2 [ 4 5 5 − 2 5 5 ] = [ 2 5 5 − 5 5 ] u_2 = \frac{1}{\sigma_2}Av_1=2= \frac{1}{2}\begin{bmatrix} \frac{4\sqrt5 }{5} \\ -\frac{2\sqrt5 }{5} \end{bmatrix}=\begin{bmatrix} \frac{2\sqrt5 }{5} \\ -\frac{\sqrt5 }{5} \end{bmatrix} u2=σ21Av1=2=21[545−525]=[525−55]
V = [ 5 3 0 − 2 3 2 5 15 2 5 5 1 3 4 5 15 − 5 5 2 3 ] V = \begin{bmatrix} \frac{\sqrt5 }{3} & 0 & -\frac{2}{3}\\ \frac{2\sqrt5 }{15} & \frac{2\sqrt5 }{5} & \frac{1}{3}\\ \frac{4\sqrt5}{15} & -\frac{\sqrt5 }{5} &\frac{2}{3}\end{bmatrix} V=⎣⎢⎡35152515450525−55−323132⎦⎥⎤
U = [ 2 5 5 − 2 5 5 − 5 5 − 5 5 ] U=\begin{bmatrix} \frac{2\sqrt5 }{5}&-\frac{2\sqrt5 }{5} \\ -\frac{\sqrt5 }{5} & -\frac{\sqrt5 }{5} \end{bmatrix} U=[525−55−525−55]
Σ = [ 3 0 0 2 0 0 ] \Sigma =\begin{bmatrix} 3&0 \\ 0 & 2 \\0&0\end{bmatrix} Σ=⎣⎡300020⎦⎤
15.2
A T A = [ 2 4 1 3 0 0 0 0 ] [ 2 1 0 0 4 3 0 0 ] = [ 5 11 11 25 ] A^TA= \begin{bmatrix} 2 & 4\\1 & 3\\ 0 & 0\\ 0 & 0 \end {bmatrix}\begin{bmatrix} 2 &1 & 0 & 0\\4 & 3&0 & 0 \end {bmatrix} = \begin{bmatrix} 5 & 11\\11 & 25 \end {bmatrix} ATA=⎣⎢⎢⎡21004300⎦⎥⎥⎤[24130000]=[5111125]
∣ A T A − λ I ∣ = ∣ 5 − λ 11 11 25 − λ ∣ = ( 5 − λ ) ( 25 − λ ) − 121 = λ 2 − 30 λ + 4 = 0 \begin{vmatrix} A^TA - \lambda I\end{vmatrix} = \begin{vmatrix} 5-\lambda & 11 \\ 11 & 25-\lambda \end{vmatrix} = (5-\lambda)(25-\lambda) - 121 = \lambda^2-30 \lambda + 4 = 0 ∣∣ATA−λI∣∣=∣∣∣∣5−λ111125−λ∣∣∣∣=(5−λ)(25−λ)−121=λ2−30λ+4=0
15.4
r a n k ( A ) = 1 rank(A) = 1 rank(A)=1
则 A A A的列向量组的秩是1，因此各列向量是某一列向量 α \alpha α的倍数，设这些倍数构成向量 β \beta β，则 A = α β T A=\alpha \beta^T A=αβT
实例：
A = [ 1 2 2 4 ] = [ 1 2 ] [ 1 2 ] A=\begin{bmatrix} 1 & 2 \\2 & 4 \end{bmatrix} = \begin{bmatrix} 1 \\2 \end{bmatrix} \begin{bmatrix} 1 & 2 \end{bmatrix} A=[1224]=[12][12]