Practice Solutions and Feedback

Use after an independent attempt. Correct witnesses, parameterizations and proofs receive credit even when expressed differently.

P1: x=(3,2). Swap the two rows before pivoting; determinant is −1, so the matrix is invertible. A zero candidate pivot is not an invariant of the map.

P2: Rank one, particular (1,0,0), nullspace basis (−2,1,0),(−3,0,1). All solutions are their affine combinations with the particular vector. For the second b, y=(−2,1) annihilates every column but gives y^T b=1. Multiplying a hypothetical equation Ax=b by y^T is the contradiction.

P3: The homogeneous plane contains zero and is closed under linear combinations. The shifted plane excludes zero. The empty set excludes zero even though pairwise closure is vacuous. The union contains (1,0,0),(0,1,0) but not their sum. Distinguish union from the subspace sum.

P4: Column space is R², row space is span{(1,0,1),(0,1,1)}, nullspace span{(−1,−1,1)}, and left nullspace {0}. Dimensions are 2,2,1,0. A vector is annihilated by A exactly when orthogonal to each row; this proves the input-side orthogonal-complement equality, not merely containment.

P5: RREF=[[1,2],[0,0]]. Original pivot column (1,2) and reduced (1,0) span different lines. Invertible row operations preserve row space, nullspace, rank and the solution set when applied to the entire augmented system; they transform the output coordinates and generally change column space.

P6: Coordinates (3,2); the third coordinate must equal the sum of the first two. Hence (0,0,1) is outside the subspace. B is 3-by-2. Uniqueness of coordinates within its image follows from independent columns, without a square inverse.

P7: The new matrix is [[3/2,−1/2],[1/2,1/2]]. Derive it by expressing T(1,1)=(2,1) and T(1,−1)=(0,−1) in the new basis. With unrelated bases B,C, the expression is C^-1 A B; similarity uses the same basis change on both sides.

P8: Q^TQ=I₂ and QQ^T=diag(1,1,0). The latter drops the last coordinate, equals its transpose and squares to itself. It cannot be I₃ because Q reaches only a two-dimensional subspace.

P9: (c,m)=(7/6,1/2), fitted=(7/6,5/3,13/6), residual=(−1/6,1/3,−1/6), squared error 1/6. Both sum(r_i) and sum(t_i*r_i) vanish. For any h, the squared error at x+h is ||r||²+||Ah||², proving minimality. Column independence gives uniqueness.

P10: Row two is twice row one, so determinant zero. Multiplication sends (7,−5,1) to zero but sends (2,−1,0) to (0,0,−1). Row-dependence coefficients do not automatically describe a column dependence. epsilon*I has tiny determinant and condition number one when epsilon is nonzero.

P11: Rotation has characteristic equation lambda²+1=0, with no real root. For diag(−3,1), a start (1,1) has a dominant component whose sign alternates; normalized directions approach the first coordinate line. A start (0,1) remains an exact eigenvector for eigenvalue one, with zero residual and no dominance claim.

P12: Multiplying the proposed formula by J adds one to its upper-right entry, proving the induction. Powers are unbounded on e₂. For the decaying example, A^k has diagonal 0.5^k and upper-right 10k0.5^(k−1) for k>=1. This term eventually tends to zero but can initially amplify an input.

P13: Leading minors 2 and 3 are positive. L=[[sqrt(2),0], [−1/sqrt(2),sqrt(3/2)]]. For the other matrix, x=(1,−1) gives x^T A x=−2. Nonnegative leading minors are insufficient for PSD: diag(0,−1) has leading determinants zero and zero yet a negative direction.

P14: Approximation diag(4,0,0), operator error 2, Frobenius error sqrt(5). For diag(2,2), diag(2,0) and diag(0,2) are distinct optimal rank-one choices. Factors also admit simultaneous sign changes; compare their products and invariant properties instead of exact vector entries.

P15: Centered rows are (−1,−1),(0,0),(1,1); singular values are 2 and 0. Sample covariance is [[1,1],[1,1]] with eigenvalues 2 and 0. Raw largest singular value is sqrt(28). A constant feature becomes zero on centering; independence of raw columns alone does not establish positive-definite sample covariance.

P16: Solutions (2,0) and (1,1), relative input change 1/20000 and relative output change 1/2. The exact infinity-norm condition number is 40004.0001. Both exact solves have zero residual for their own data. Neither establishes which measured right-hand side is trustworthy.

P17: Exact inverse sends (0,10^-8) to (0,1); cutoff pseudoinverse sends it to zero. Exact rank is two; numerical rank under the declared cutoff is one. The discarded output amplitude may encode a coefficient that matters. State noise, units and task loss before calling a small direction unimportant.

P18: Write A=(1/4)I+(3/4)11^T; eigenvalues are 5/2 and 1/4 twice. Jacobi's iteration matrix has zero diagonal and off-diagonal −3/4, with eigenvalue −3/2 along the all-ones vector. Iterates are 1,−1/2,7/4 times that vector; errors from the exact 2/5 times that vector grow by factor 3/2 in magnitude.

P19: Weighted constant=(11+33)/4=5/2; unweighted is 2. In the separate duplicate-feature model, all minimizers satisfy x₁+x₂=1. Minimum norm is (1/2,1/2). At lambda=1 the normal matrix is [[6,5],[5,6]], right side (5,5), giving (5/11,5/11). The shrinkage changes the fitted values and the objective.

P20: Cholesky for a resolved SPD system; QR for a full-column-rank fit; a rank-revealing method/SVD for rank deficiency with an explicit cutoff and minimum-norm policy. The Laplacian has null vector (1,1), and b has sum zero, so it is compatible but not uniquely solvable. Fix x₂=0 to obtain x₁=1, or choose zero-mean x=(1/2,−1/2). Ordinary SPD CG assumptions do not apply to the unconstrained singular matrix. Equivalent justified approaches receive credit.