The correct statement for solving a linear system $A X = b$ using the Conjugate Gradient Method is:

$X_j \in X_0 + \mathcal{K}_j(A, b - A X_0)$

This means that the solution at the $j$-th iteration, $X_j$, lies in the initial guess $X_0$ plus the Krylov subspace $\mathcal{K}_j(A, b - A X_0)$. The Krylov subspace is spanned by vectors generated by repeated application of $A$ on the residual vector $b - A X_0$.

Why this is true:

The Conjugate Gradient (CG) method works by iterating over a Krylov subspace formed from the residuals. The Krylov subspace at the $j$-th iteration is given by: $\mathcal{K}_j(A, b - A X_0) = \text{span} \{ r_0, A r_0, A^2 r_0, \dots, A^{j-1} r_0 \}$ where $r_0 = b - A X_0$ is the initial residual.

Explanations for the other options:

• $\langle b - A x_j, \mathcal{K}_j(A, b) \rangle \geq 1$: This inequality doesn't follow the structure of inner product relationships in the CG method.

• The error $e_j$ has minimum norm: While the CG method minimizes the A-norm of the error at each step, this is not the complete statement.

• $r_j$ is orthogonal to a different space $\mathcal{K}_j(A^T, b)$: The residuals $r_j$ are orthogonal to the Krylov subspace $\mathcal{K}_j(A, b)$, not $\mathcal{K}_j(A^T, b)$.

Would you like more details on how the Conjugate Gradient method works?

Here are 5 related questions:

1. What is the Krylov subspace in the Conjugate Gradient method?
2. How does the Conjugate Gradient method minimize the error?
3. Why is orthogonality of residuals important in the Conjugate Gradient method?
4. How does the CG method handle symmetric positive definite matrices?
5. Can the Conjugate Gradient method be applied to non-symmetric matrices?

Tip: In Conjugate Gradient, the residuals become orthogonal to the Krylov subspace, which ensures optimal convergence in finite steps.