Question

我有以下矩阵，它是对称且真实的（它是一个汉密尔顿算子）：（Matlab友好）

[63.000000，-1.732051,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000; -1.732051,61.000000，-2.000000,0.000000，-1.000000,0.000000,0.000000,0.000000,0.000000,0.000000; 0.000000，-2.000000,61.000000，-1.732051,0.000000，-1.414214,0.000000,0.000000,0.000000,0.000000; 0.000000,0.000000，-1.732051,63.000000,0.000000,0.000000，-1.732051,0.000000,0.000000,0.000000; 0.000000，-1.000000,0.000000,0.000000,6.00.000000，-1.414214,0.000000,0.000000,0.000000,0.000000; 0.000000,0.000000，-1.414214,0.000000，-1.414214,60.000000，-1.414214，-1.414214,0.000000,0.000000; 0.000000,0.000000,0.000000，-1.732051,0.000000，-1.414214,61.000000,0.000000，-2.000000,0.000000; 0.000000,0.000000,0.000000,0.000000,0.000000，-1.414214,0.000000,6.00.000000，-1.000000,0.000000; 0.000000,0.000000,0.000000,0.000000,0.000000,0.000000，-2.000000，-1.000000,61.000000，-1.732051; 0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000, -1.732051,63.000000]

如果我用这些值制作JAMA矩阵，并执行特征值分解，则V * D *转置（V）不等于汉密尔顿。你们中的任何人都知道出了什么问题吗？特征值与MATLAB一致，但特征向量不是。

这是一个测试它的类

public class TestJama {


public static void main(String[] args) {

    double[][] m = new double[][] {
            {63.000000, -1.732051, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000},
            { -1.732051, 61.000000, -2.000000, 0.000000, -1.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000},
            { 0.000000, -2.000000, 61.000000, -1.732051, 0.000000, -1.414214, 0.000000, 0.000000, 0.000000, 0.000000},
            { 0.000000, 0.000000, -1.732051, 63.000000, 0.000000, 0.000000, -1.732051, 0.000000, 0.000000, 0.000000},
            { 0.000000, -1.000000, 0.000000, 0.000000, 61.000000, -1.414214, 0.000000, 0.000000, 0.000000, 0.000000},
            { 0.000000, 0.000000, -1.414214, 0.000000, -1.414214, 60.000000, -1.414214, -1.414214, 0.000000, 0.000000},
            { 0.000000, 0.000000, 0.000000, -1.732051, 0.000000, -1.414214, 61.000000, 0.000000, -2.000000, 0.000000},
            { 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, -1.414214, 0.000000, 61.000000, -1.000000, 0.000000},
            { 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, -2.000000, -1.000000, 61.000000, -1.732051},
            { 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, -1.732051, 63.000000}
    };


    Matrix hamilton = new Matrix(m);
    System.out.println(jamaToString(hamilton));

    EigenvalueDecomposition e = hamilton.eig();
    System.out.println(jamaToString(e.getD()));
    System.out.println(jamaToString(e.getV()));

    Matrix recomb = e.getV().times(e.getD()).times(e.getV().transpose());
    System.out.println(jamaToString(recomb));

    System.out.println(hamilton.equals(recomb));

}

private static String jamaToString(Matrix m) {
    StringBuilder b = new StringBuilder();
    b.append("[");
    for(int i=0; i<m.getRowDimension(); i++) {

        for(int j=0; j<m.getColumnDimension(); j++) {
            b.append(m.get(i, j));
            if(j<m.getColumnDimension() - 1) b.append(",");
        }
        if(i<m.getRowDimension() - 1) b.append(";");
    }
    b.append("]");
    return b.toString();
}
}

编辑：结果（V * D *转置（V））产生

63.1093 -0.6379 0.3400 -0.6576 0.0938 -0.0437 -0.6056 -0.5066 0.3463 0.5039 -0.6379 61.3082 -0.2368 -1.7813 -0.0851 0.7949 -0.1337 0.7668 -0.0422 -2.4329 0.3400 -0.2368 60.1481 1.3323 -0.4099 -1.8834 -0.5780 0.7516 0.0946 0.1705 -0.6576 -1.7813 1.3323 61.2455 0.0972 -0.8075 -0.9004 0.0242 0.3963 -1.2527 0.0938 -0.0851 -0.4099 0.0972 60.3086 -0.1899 0.0394 0.1987 -0.0484 -0.1495 -0.0437 0.7949 -1.8834 -0.8075 -0.1899 61.7941 0.3741 0.8237 0.7772 0.7557 -0.6056 -0.1337 -0.5780 -0.9004 0.0394 0.3741 60.6415 0.6351 0.7099 0.3349 -0.5066 0.7668 0.7516 0.0242 0.1987 0.8237 0.6351 62.8108 1.3507 1.3002 0.3463 -0.0422 0.0946 0.3963 -0.0484 0.7772 0.7099 1.3507 63.3270 0.1244 0.5039 -2.4329 0.1705 -1.2527 -0.1495 0.7557 0.3349 1.3002 0.1244 60.3069

Answer 1

如果它只是矩阵的值：它是正确的。

编辑：上面的陈述是在之前写的，结果矩阵被插入到原始问题中。当然，这个矩阵不正确。根据评论，似乎这个错误的矩阵是由旧的JAMA版本中的一些错误引起的。有了JAMA 1.0.3，它就可以了。尽管如此，这个答案的其余部分仍然有效且与问题相关：

这里有两个问题。首先，不会覆盖JAMA equals类的Matrix方法来比较矩阵的内容。 equals的默认实现比较了引用对象的身份。所以即使是像这样的微不足道的比较

Matrix A = new Matrix(new double[][]{{1.0}}); Matrix B = new Matrix(new double[][]{{1.0}}); System.out.println(A.equals(B));

会打印false。

第二个问题相当简单（并且很常见）：具有double值的计算不是无限精确的。您可以在StackOverflow上找到许多有关此问题的问题，但可能需要查看Wikipedia about Floating Point Accuracy Problems。（有些人会推荐关于What Every Computer Scientist Should Know About Floating-Point Arithmetic的文章，但它涉及到......）。

使用这个小例子可以轻松复制基本问题：

double x = 0.1; double y = 0; for (int i=0; i<10; i++) { y += x; } System.out.println(y+" == 1.0: "+(y==1.0));

此计算的结果是不 y==1.0，而是y==0.99999999999999...。

所以即使如果覆盖了equals方法来执行逐元素比较，由于浮点错误，比较仍会产生false。

缓解这个问题的一种方法是检查某些值是否等于小ε＆＃34;。这可能仍然很棘手，因为要比较的值的大小在这里会有所不同，但一种解决方案可能是使用以下方法来比较epsilon-equality的矩阵：

private static boolean epsilonEqual(Matrix a, Matrix b) { int ra = a.getRowDimension(); int rb = b.getRowDimension(); if (ra != rb) { return false; } int ca = a.getColumnDimension(); int cb = b.getColumnDimension(); if (ca != cb) { return false; } for (int c=0; c<ca; c++) { for (int r=0; r<ra; r++) { double ea = a.get(r, c); double eb = b.get(r, c); if (!epsilonEqual(ea, eb)) { return false; } } } return true; } private static boolean epsilonEqual(double x, double y) { final double epsilon = 1e-8; return Math.abs(x - y) <= epsilon; }

JAMA没有计算正确的特征向量

1 个答案: