一个4×4的矩阵，有什么快速的方法求它的最大特征值 ，或者是它的模？
一个4×4的矩阵，有什么快速的方法求它的最大特征值 ，或者是它的模？

[解决办法]
这是我以前做AHP的时候用过的一个程序片断，由于需要所以把特征向量规一化了：

C# code

        public class solution        {            private double v;            private double [] weight;            public solution(double v, double [] weight)            {                this.v = v;                this.weight = weight;            }            public double V            {                get                {                    return this.v;                }                set                {                    this.v = value;                }            }            public double [] Weight            {                get                {                    return this.weight;                }                set                {                    this.weight = value;                }            }        }        /// <summary>        /// 幂法求矩阵最大特征根和特征向量实现        /// </summary>        /// <param name="matrix"></param>        /// <returns></returns>        public static solution GetSolution(double [][] matrix)        {            double [] DL = new double[matrix.Length];            double [] tmp = new double[matrix.Length];            double [] tmp1 = new double[matrix.Length];            for (int i = 0; i < matrix.Length; i++)                tmp[i] = 1;            do            {                DL = tmp;                double max = Max(tmp);                for (int i = 0; i < tmp.Length; i++)                    tmp1[i] = tmp[i] / max;                tmp = mMultiply(matrix, tmp1);            }while(Math.Abs(Max(tmp) - Max(DL)) > 0.000001);            DL = tmp;            double maxVL = Max(DL);            double sum = 0.0;            foreach (double d in DL)                sum += d;            for (int i = 0; i < DL.Length; i++)                DL[i] /= sum;            return new solution(maxVL, DL);        }