Assessment Solutions and Scoring

Open after the independent checkpoint. Use the gate rubric. Alternate correct proofs, bases and witnesses receive full credit.

A1/C1: Swap rows so the first pivot is two. The first original equation gives x₂=3 and the second gives x₁=1. For the inconsistent system, y=(−2,1) satisfies y^T A=0 but y^T b=1; this contradicts any hypothetical solution.

A2/C2: Rank two; original columns one and three give basis (1,0),(0,1) for R². Nullspace is span{(−2,1,0)} because x₁=−2x₂ and x₃=0. Nullity one plus rank two equals three input dimensions. Empty sets vacuously satisfy a pairwise universal closure statement but fail zero membership.

A3/C3: B^-1 A B=[[2,1],[0,1]], sending (1,2) to (4,2) in B coordinates. The original vector is B(1,2)=(3,2); its image (6,2)=B(4,2). A nonzero translation sends zero to a nonzero vector, contradicting a necessary linearity condition.

A4/C4: Coefficient one, fitted (1,1,0), residual (1,−1,1), squared norm three. The residual is perpendicular to a; Pythagoras proves no other point on the line is closer. With columns a,2a, all minimizers satisfy x₁+2x₂=1. Minimum norm is (1/5,2/5), parallel to (1,2) and perpendicular to the null direction (−2,1). Decomposing any other minimizer along that null direction adds a nonnegative squared norm.

A5/C5: First determinant 10^-12 and condition one. Second determinant one and condition 10^12. Magnitude of determinant does not bound distortion by itself. A negative real determinant reverses orientation; it is not a sensitivity metric.

A6/C6: The start stays an eigenvector for two, giving zero eigenpair residual while missing magnitude four. J=[[1,1],[0,1]] has J^k=[[1,k],[0,1]], so spectral radius one allows polynomial growth. A 90-degree rotation has eigenvalues ±i, so it lacks a real eigenbasis; over C it has distinct eigenvalues and diagonalizes.

A7/C7: Leading minors are three and eight; eigenvalues four and two. L=[[sqrt(3),0],[1/sqrt(3),sqrt(8/3)]] reconstructs A. Positive diagonal is insufficient for [[1,2],[2,1]]: vector (1,−1) gives −2. diag(0,−1) has nonnegative leading determinants but is not PSD. A floating tiny-pivot rejection must be distinguished from these exact negative-direction proofs.

A8/C8: Rank two, pseudoinverse diag(1/5,1/2,0), approximation diag(5,0,0), operator error two and Frobenius error two. Here the full and reduced SVD both have 3-by-3 factors; the compact nonzero form has U and V of shape 3-by-2 and a 2-by-2 diagonal. A threshold that removes the positive value two produces a different effective model, not the exact original pseudoinverse.

A9/C9: Residual (0,1/1000), relative residual 1/1000, backward error 1/2000 because ||A||∞=||x||∞=||b||∞=1. Condition number 1000; forward-error bound one. Exact solution (1,1), so actual relative infinity-norm error is also one. The small residual alone hid a full-size error in a sensitive direction.

A10/C10: Weighted constant=(1+2*4)/3=3, compared with unweighted 5/2 for that data. In the separate ridge problem, (2+2)c=6 gives c=3/2; ordinary least squares gives three. Weighted least squares can be transformed by sqrt(W) and solved with QR. Ridge can be represented as an augmented least-squares system; Cholesky of its positive-definite normal matrix is another option with conditioning costs to assess. Rank-revealing/SVD methods need a declared cutoff. Accurate computation of a changed objective is not evidence that the change answers the original question.