Sec 4.5-5.1

Lay 4.5 #18: True/False: The number of variables in Ax = 0 equals nullity A.

False. The number of variables is n = rank A + nullity A. Nullity counts only the free variables.

Lay 4.5 #20: True/False: The dimension of P4 is 4.

False. dim P_n = n + 1, so dim P4 = 5 (basis 1, t, t^2, t^3, t^4).

Lay 4.5 #22: True/False: The row space and column space of A have the same dimension, even if A is not square.

True. Both dimensions equal rank A.

Lay 4.5 #24: True/False: The nullity of A is the number of columns of A that are not pivot columns.

True. Non-pivot columns correspond to free variables, and nullity = number of free variables.

Lay 4.5 #44: True/False: If a linearly dependent set {v1, …, vp} exists in V, then dim V ≤ p.

False. Counterexample: {0} in R^3 is dependent with p = 1, but dim V = 3. A dependent set puts no upper bound on dim V.

Lay 4.5 #46: True/False: If dim V = p, then V has a spanning set of p + 1 vectors.

True. Take a basis (p vectors) and add any other vector from V. The set still spans.

Lay 4.5 #48: True/False: If p ≥ 2 and dim V = p, then every set of p - 1 nonzero vectors is linearly independent.

False. In R^3, {e1, 2e1} is two nonzero vectors but dependent. Independence is not guaranteed by size.

Lay 5.1 #22: True/False: If Ax = λx for some scalar λ, then x is an eigenvector of A.

False. An eigenvector must be nonzero. x = 0 satisfies Ax = λx for every λ.

Lay 5.1 #24: True/False: A number c is an eigenvalue of A iff (A - cI)x = 0 has a nontrivial solution.

True. This is the definition rewritten: Ax = cx with x ≠ 0.

Lay 5.1 #26: True/False: To find the eigenvalues of A, reduce A to echelon form.

False. Row operations change eigenvalues. Solve det(A - λI) = 0 instead.

Lay 5.1 #30: True/False: An eigenspace of A is a null space of a certain matrix.

True. The eigenspace for λ is Nul(A - λI).

Lay 5.1 #33: Show that if λ is an eigenvalue of invertible A, then 1/λ is an eigenvalue of A^-1.

Let Ax = λx with x ≠ 0. λ ≠ 0, since otherwise Ax = 0 with x ≠ 0 contradicts invertibility. Multiply both sides by A^-1: x = λA^-1 x. Divide by λ: A^-1 x = (1/λ)x. So 1/λ is an eigenvalue of A^-1.

Lay 5.1 #35: Show that λ is an eigenvalue of A iff λ is an eigenvalue of A^T.

(A - λI)^T = A^T - λI. A matrix is singular iff its transpose is singular (equal determinants). So A - λI is singular iff A^T - λI is singular, i.e. λ is an eigenvalue of A iff it is one of A^T.