/// <summary>
/// Solve x for Ax = b, where A is a invertible matrix
///
/// The method works as follows:
/// 1. QR factorization A = Q * R
/// 2. Let: c = Q.transpose * b, we have R * x = c, where R is a triangular matrix if A is n x n
/// 3. Solve x using backward substitution
/// </summary>
/// <param name="A">An m x n matrix with linearly independent columns, where m >= n</param>
/// <param name="b">An m x 1 column vector </param>
/// <returns>An n x 1 column vector, x, such that Ax = b when m = n and Ax ~ b when m > n </returns>
public static IVector Solve(IMatrix A, IVector b)
{
// Q is a m x m matrix, R is a m x n matrix
// Q = [Q1 Q2] Q1 is a m x n matrix, Q2 is a m x (m-n) matrix
// R = [R1; 0] R1 is a n x n upper triangular matrix matrix
// A = Q * R = Q1 * R1
IMatrix Q, R;
QR.Factorize(A, out Q, out R);
IVector c = Q.Transpose().Multiply(b);
IVector x = BackwardSubstitution.Solve(R, c);
return(x);
}
/// <summary>
/// This is used for data fitting / regression
/// A is a m x n matrix, where m >= n
/// b is a m x 1 column vector
/// The method solves for x, which is a n x 1 column vectors such that A * x is closest to b
///
/// The method works as follows:
/// 1. Let C = A.transpose * A, we have A.transpose * A * x = C * x = A.transpose * b
/// 2. Decompose C : C = Q * R, we have Q * R * x = A.transpose * b
/// 3. Multiply both side by Q.transpose = Q.inverse, we have Q.transpose * Q * R * x = Q.transpose * A.transpose * b
/// 4. Since Q.tranpose * Q = I, we have R * x = Q.transpose * A.transpose * b
/// 5. Solve x from R * x = Q.transpose * A.transpose * b using backward substitution
/// </summary>
/// <param name="A"></param>
/// <param name="b"></param>
/// <returns></returns>
public static IVector SolveLeastSquare(IMatrix A, IVector b)
{
IMatrix At = A.Transpose();
IMatrix C = At.Multiply(A); //C is a n x n matrix
IMatrix Q, R;
QR.Factorize(C, out Q, out R);
IMatrix Qt = Q.Transpose();
IVector d = Qt.Multiply(At).Multiply(b);
IVector x = BackwardSubstitution.Solve(R, d);
return(x);
}
/// <summary>
/// Given A which is a n x n symmetric matrix, we want to find U and T
/// such that:
/// 1. A = U * T * U.transpose
/// 2. U is a n x n matrix whose columns are eigen vectors of A (A * x = lambda * x, where lambda is the eigen value, and x is the eigen vector)
/// 3. T is a diagonal matrix whose diagonal entries are the eigen values
///
/// The method works in the following manner:
/// 1. intialze A_0 = A, U_0 = I
/// 2. iterate for k = 1, ... K, K is termination criteria
/// 3. In each iteration k:
/// 3.1 QR factorization to find Q_k and R_k such that A_{k-1}=Q_k * R_k
/// 3.2 Let A_k = R_k * Q_k
/// 3.3 Let U_k = U_{k-1} * Q_k
/// 4. Set T = A_K, U = U_K
///
/// Note that U is an orthogonal matrix if A is a symmetric matrix, in other words, if A.transpose = A, then U.inverse = U.transpose
/// </summary>
/// <param name="A">The matrix to be factorized</param>
/// <param name="K">maximum number of iterations</param>
/// <param name="T">T is a diagonal matrix whose diagonal entries are the eigen values</param>
/// <param name="U">U is a n x n matrix whose columns are eigen vectors of A</param>
public static void Factorize(IMatrix A, out IMatrix T, out IMatrix U, int K = 100, double epsilon = 1e-10)
{
Debug.Assert(A.RowCount == A.ColCount);
int n = A.RowCount;
IMatrix A_k = A.Clone();
IMatrix U_k = A.Identity(n);
IMatrix Q_k, R_k;
for (int k = 1; k <= K; ++k)
{
QR.Factorize(A_k, out Q_k, out R_k);
A_k = R_k.Multiply(Q_k);
U_k = U_k.Multiply(Q_k);
double sum = 0;
foreach (IVector rowVec in A_k.NonEmptyRows)
{
int rowId = rowVec.ID;
foreach (int key in rowVec.NonEmptyKeys)
{
if (key == rowId)
{
continue;
}
sum += System.Math.Abs(rowVec[key]);
}
}
if (sum <= epsilon)
{
break;
}
}
T = new SparseMatrix(n, n, A.DefaultValue);
for (int i = 0; i < n; ++i)
{
T[i, i] = A_k[i, i];
}
U = U_k;
}
