Discussion
Consider the linear system Mx = b, where \(M = \begin{bmatrix} 2 & -1 \\ -4 & 3 \end{bmatrix}\) and \(b = \begin{bmatrix} -2 \\ 5 \end{bmatrix}\).
Suppose M = LU, where L and U are lower triangular and upper triangular square matrices, respectively. Consider the following statements:
P: If each element of the main diagonal of L is 1, then trace(U) = 3.
Q: For any choice of the initial vector \(x^{(0)}\), the Jacobi iterates \(x^{(k)}\), k = 1, 2, 3, ... converge to the unique solution of the linear system Mx = b.
Then
Answer: A
Both P and Q are TRUE.
P: Doolittle factorisation (unit diagonal in L):
\[ L = \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix},\ U = \begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix} \]
since \(l_{21} = -4/2 = -2\) and \(u_{22} = 3 - (-2)(-1) = 1\). So trace(U) = 2 + 1 = 3.
Q: the Jacobi iteration matrix is \(-D^{-1}(L+U)\) built from the off-diagonal entries:
\[ J = \begin{bmatrix} 0 & 1/2 \\ 4/3 & 0 \end{bmatrix},\quad \lambda^{2} = \frac{2}{3} \]
Its spectral radius is \(\sqrt{2/3} \approx 0.82 \lt 1\), so the Jacobi iterates converge from every starting vector. (M is not diagonally dominant, so that sufficient condition would not settle it; the spectral radius does.)
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