层次分析法在实际应用时,可以用成对比较阵A的列向量的平均值近似代替特征向量,称为和法,其步骤是:先将A的每一列向量归一化,按行求和后再归一化,得到 w = ( a 1 , a 2 , . . . , a n ) T w = (a_1,a_2,...,a_n)^T w=(a1?,a2?,...,an?)T 即为近似特征向量,并将 1 n ∑ i = 1 n ( A w ) i a i \frac{1}{n}\sum_{i=1}^n\frac{(Aw)_i}{a_i} n1?∑i=1n?ai?(Aw)i??(可以看例子理解)作为近似最大特征根。
设 A = [ 1 1 / 2 1 / 3 2 1 1 3 1 1 ] A = \left [ \begin{matrix} 1 & 1/2 & 1/3 \\ 2 & 1 & 1 \\ 3 & 1 & 1 \end {matrix} \right ] A= ?123?1/211?1/311? ?,用和法计算近似特征向量和近似最大特征根。
解:
【1】求特征向量:
A = [ 1 1 / 2 1 / 3 2 1 1 3 1 1 ] A = \left [ \begin{matrix} 1 & 1/2 & 1/3 \\ 2 & 1 & 1 \\ 3 & 1 & 1 \end {matrix} \right ] A= ?123?1/211?1/311? ?
各列归一化 → [ 1 / 6 1 / 5 1 / 7 1 / 3 2 / 5 3 / 7 1 / 2 2 / 5 3 / 7 ] \underrightarrow{各列归一化} \left [ \begin{matrix} 1/6 & 1/5 & 1/7 \\ 1/3 & 2/5 & 3/7 \\ 1/2 & 2/5 & 3/7 \end{matrix}\right ] 各列归一化? ?1/61/31/2?1/52/52/5?1/73/73/7? ?
各行求和 → [ 107 / 210 122 / 105 93 / 70 ] \underrightarrow{各行求和}\left [ \begin{matrix} 107/210 \\ 122/105 \\ 93/70 \end{matrix}\right ] 各行求和? ?107/210122/10593/70? ?
再归一化 → [ 107 / 630 122 / 315 31 / 70 ] ≈ [ 0.1698 0.3873 0.4429 ] = w \underrightarrow{再归一化}\left [ \begin{matrix} 107/630 \\ 122/315 \\ 31/70 \end{matrix}\right ] \approx \left [ \begin{matrix} 0.1698 \\ 0.3873 \\ 0.4429 \end{matrix}\right ] = w 再归一化? ?107/630122/31531/70? ?≈ ?0.16980.38730.4429? ?=w
【2】求最大特征根:
λ = 1 n ∑ i = 1 n ( A w ) i a i \lambda = \frac{1}{n}\sum_{i=1}^{n}\frac{(Aw)_i}{a_i} λ=n1?∑i=1n?ai?(Aw)i??
其中 A w = [ 1 1 / 2 1 / 3 2 1 1 3 1 1 ] [ 0.1698 0.3873 0.4429 ] = [ 0.5111 1.1698 1.3396 ] Aw = \left [ \begin{matrix} 1 & 1/2 & 1/3 \\ 2 & 1 & 1 \\ 3 & 1 & 1 \end{matrix}\right ] \left [ \begin{matrix} 0.1698 \\ 0.3873 \\ 0.4429 \end{matrix}\right ] = \left [ \begin{matrix} 0.5111 \\ 1.1698 \\ 1.3396 \end{matrix}\right ] Aw= ?123?1/211?1/311? ? ?0.16980.38730.4429? ?= ?0.51111.16981.3396? ?
则 λ = 1 3 ( 0.5111 0.1698 + 1.1698 0.3873 + 1.3396 0.4429 ) = 3.0183 \lambda = \frac{1}{3}(\frac{0.5111}{0.1698}+\frac{1.1698}{0.3873}+\frac{1.3396}{0.4429}) = 3.0183 λ=31?(0.16980.5111?+0.38731.1698?+0.44291.3396?)=3.0183