Question

我正在尝试求解类似Ax = 0的齐次线性方程组。这是一个为简化起见已简化的示例矩阵：

1 2 | 0
3 6 | 0

 The solution I am hoping to get to is at least [ 2, -1 ].
 But the fundamental solution is folow: [2C; -1C].
 You can see than Det(A)=0 and Rank(A) =1;
 Of course you know that such systems have trivial solution [0,0].

我正在尝试：

Matrix<double> A = Matrix<double>.Build.DenseOfArray(new double[,] {
{ 1, 2 }, 
{ 3, 6 }, 
});
Vector<double> B = Vector<double>.Build.Dense(new double[] { 0, 0 });
var result = A.Solve(B); //result = Nan, Nan

请帮助我通过MathNet查找此类任务的解决方案，因为在实际任务中，A的尺寸大于20。

此问题的解决方案：Math.Net system of linear equations with a 0 value in solution对我的情况不起作用（B = 0，Det（A）= 0）。

谢谢

Answer 1

要求解线性方程，可以尝试Matrix<T>.SolveIterative(Vector<T> input, IIterativeSolver<T> solver, Iterator<T> iterator = null, IPreconditioner<T> preconditioner = null)。

要获取可用的求解器：

var solvers = Assembly.GetAssembly(typeof(Matrix<double>))
    .GetTypes()
    .Where(t => t.GetInterfaces().Contains(typeof(IIterativeSolver<double>)) &&
                t.GetConstructors().Any(ctor => ctor.GetParameters().Count() == 0))
    .Select(t => Activator.CreateInstance(t))
    .Cast<IIterativeSolver<double>>();

它给出：

MathNet.Numerics.LinearAlgebra.Double.Solvers.BiCgStab
MathNet.Numerics.LinearAlgebra.Double.Solvers.GpBiCg
MathNet.Numerics.LinearAlgebra.Double.Solvers.MlkBiCgStab
MathNet.Numerics.LinearAlgebra.Double.Solvers.TFQMR

使用数据尝试所有操作：

Matrix<double> A = Matrix<double>.Build.DenseOfArray(new double[,]
{
    { 1, 2 },
    { 3, 6 }
});
Vector<double> B = Vector<double>.Build.Dense(new double[] { 0, 0 });

赞：

foreach (var solver in solvers)
{
    try
    {
        Console.WriteLine(solver);
        Console.WriteLine(A.SolveIterative(B, solver));
    }
    catch (Exception e)
    {
        Console.WriteLine(e.Message);
    }
}

在您的情况下，这不是运气。结果是：

MathNet.Numerics.LinearAlgebra.Double.Solvers.BiCgStab
Algorithm experience a numerical break down

MathNet.Numerics.LinearAlgebra.Double.Solvers.GpBiCg
DenseVector 2-Double
NaN
NaN

MathNet.Numerics.LinearAlgebra.Double.Solvers.MlkBiCgStab
Algorithm experience a numerical break down

MathNet.Numerics.LinearAlgebra.Double.Solvers.TFQMR
DenseVector 2-Double
0
0

尽管它可以像here一样工作。

无论如何，MathNet是如此出色，以至于人们可以轻松构建解决方案。

使用大学的线性代数和库功能：

static class MatrixExtension
{
    public static Vector<double>[] SolveDegenerate(this Matrix<double> matrix, Vector<double> input)
    {
        var augmentedMatrix =
            Matrix<double>.Build.DenseOfColumnVectors(matrix.EnumerateColumns().Append(input));

        if (augmentedMatrix.Rank() != matrix.Rank())
            throw new ArgumentException("Augmented matrix rank does not match coefficient matrix rank.");

        return augmentedMatrix.SolveAugmented();
    }

    private static Vector<double>[] SolveAugmented(this Matrix<double> matrix)
    {
        var rank = matrix.Rank();
        var cut = matrix.CutByRank(rank);
        // [A|R]x[X] = [B]            
        var A = Matrix<double>.Build.DenseOfColumnVectors(cut.EnumerateColumns().Take(rank));
        var R = Matrix<double>.Build.DenseOfColumnVectors(cut.EnumerateColumns().Skip(rank).Take(cut.ColumnCount - rank - 1));
        var B = cut.EnumerateColumns().Last();

        var vectors = Matrix<double>.Build.DenseDiagonal(R.ColumnCount, 1)
            .EnumerateColumns().ToArray();

        return vectors.Select(v => A.Solve(B - R * v))
            .Zip(vectors, (x, v) => Vector<double>.Build.DenseOfEnumerable(x.Concat(v)))
            .ToArray();
    }

    private static Matrix<double> CutByRank(this Matrix<double> matrix, int rank)
    {
        var result = Matrix<double>.Build.DenseOfMatrix(matrix);
        while (result.Rank() < result.RowCount)
            result = result.EnumerateRows()
                           .Select((r, index) => result.RemoveRow(index))
                           .FirstOrDefault(m => m.Rank() == rank);
        return result;
    }
}

现在：

Console.WriteLine(A.SolveDegenerate(B).First());

礼物：

DenseVector 2-Double
-2
 1

另一个例子：

Matrix<double> A = Matrix<double>.Build.DenseOfArray(new double[,]
{
    { 1, 2, 1 },
    { 3, 6, 3 },
    { 4, 8, 4 }
});
Vector<double> B = Vector<double>.Build.Dense(new double[3]);
foreach (var x in A.SolveDegenerate(B))
    Console.WriteLine(x);

礼物：

DenseVector 3-Double
-2
 1
 0

DenseVector 3-Double
-1
 0
 1

齐次线性方程组的基本解：Ax = 0（Det（A）= 0）MathNet

1 个答案: