java求矩阵的特征值和特征向量(AHP层次分析法计算权重)(附源代码) 这几天做一个项目，需要用到求矩阵的特征值特征向量。
我c++学的不好，所以就去网站找了很多java的源代码，来实现这个功能。
很多都不完善，甚至是不准确。
所以自己参考写了一个。
这个用于我一个朋友的毕业设计。
结果肯定正确。
话不多说，贴源代码！import java.math.BigDecimal;import java.util.Arrays;/*** AHP层次分析法计算权重** @since jdk1.6* @author 刘兴* @version 1.0* @date 2012.05.25**/public class AHPComputeWeight {/*** @param args*/public static void main(String[] args) {/** a为N*N矩阵*///double[][] a= {{1,1,1},{1,1,1},{1,1,1}};double[][] a ={{1,3,5},{2,3,1,},{4,7,3}};//double[][] a = {{1 ,1/5, 1/3},{5, 1, 1},{3,1,1}};//double[][] a ={{1, 1/2, 2, 1},{2, 1, 3, 4},{1/2 ,1/3, 1, 1},{1 ,1/4, 1, 1}};//double[][] a = {{1 ,0.5, 0.5},{2 ,1, 1},{2 ,1, 1}};//double[][] a = {{1, 1/4, 1/3, 1},{4, 1 ,3 ,5},{3, 1/3, 1, 4},{1, 1/5, 1/4, 1}};// double[][] a= {{1,2,3,5},{0.5,1,2,3},{0.33,0.5,1,2},{0.2,0.33,0.5,1}};int N = a[0].length;double[] weight = new double[N];AHPComputeWeight instance = AHPComputeWeight.getInstance();instance.weight(a, weight, N);System.out.println(Arrays.toString(weight));}// 单例private static final AHPComputeWeight acw = new AHPComputeWeight();// 平均随机一致性指针private double[] RI = { 0.00, 0.00, 0.58, 0.90, 1.12, 1.21, 1.32, 1.41,1.45, 1.49 };// 随机一致性比率private double CR = 0.0;// 最大特征值private double lamta = 0.0;/*** 私有构造*/private AHPComputeWeight() {}/*** 返回单例** @return*/public static AHPComputeWeight getInstance() { return acw;}/*** 计算权重** @param a* @param weight* @param N*/public void weight(double[][] a, double[] weight, int N) { // 初始向量Wkdouble[] w0 = new double[N];for (int i = 0; i < N; i++) {w0[i] = 1.0 / N;}// 一般向量W（k+1）double[] w1 = new double[N];// W（k+1）的归一化向量double[] w2 = new double[N];double sum = 1.0;double d = 1.0;// 误差double delt = 0.00001;while (d > delt) {d = 0.0;sum = 0;// 获取向量int index = 0;for (int j = 0; j < N; j++) {double t = 0.0;for (int l = 0; l < N; l++)t += a[j][l] * w0[l];// w1[j] = a[j][0] * w0[0] + a[j][1] * w0[1] + a[j][2] * w0[2];w1[j] = t;sum += w1[j];}// 向量归一化for (int k = 0; k < N; k++) {w2[k] = w1[k] / sum;// 最大差值d = Math.max(Math.abs(w2[k] - w0[k]), d);// 用于下次迭代使用w0[k] = w2[k];}}// 计算矩阵最大特征值lamta，CI，RIlamta = 0.0;for (int k = 0; k < N; k++) {lamta += w1[k] / (N * w0[k]);}double CI = (lamta - N) / (N - 1);if (RI[N - 1] != 0) {CR = CI / RI[N - 1];}// 四舍五入处理lamta = round(lamta, 3);CI = Math.abs(round(CI, 3));CR = Math.abs(round(CR, 3));for (int i = 0; i < N; i++) {w0[i] = round(w0[i], 4);w1[i] = round(w1[i], 4);w2[i] = round(w2[i], 4);}// 控制台打印输出System.out.println("lamta=" + lamta);System.out.println("CI=" + CI);System.out.println("CR=" + CR);// 控制台打印权重System.out.println("w0[]=");for (int i = 0; i < N; i++) {System.out.print(w0[i] + " ");}System.out.println("");System.out.println("w1[]=");for (int i = 0; i < N; i++) {System.out.print(w1[i] + " ");}System.out.println("");System.out.println("w2[]=");for (int i = 0; i < N; i++) {weight[i] = w2[i];System.out.print(w2[i] + " ");}System.out.println("");}/*** 四舍五入** @param v* @param scale* @return*/public double round(double v, int scale) {if (scale < 0) {throw new IllegalArgumentException("The scale must be a positive integer or zero");}BigDecimal b = new BigDecimal(Double.toString(v));BigDecimal one = new BigDecimal("1");return b.divide(one, scale, BigDecimal.ROUND_HALF_UP).doubleV alue();}/*** 返回随机一致性比率** @return*/public double getCR() {return CR;}}。
