Original public practice · L-1 · October 2026

Connect a calculation to its proof.

Twelve multipart questions connect systems and structure, geometric computation, and operators over different fields. The form has 120 practice points.

This is supported practice across introductory and proof-oriented undergraduate linear algebra. It is not an official university paper, a calibrated grade prediction or a complete examination of every topic. Choose a section directly and return to unfamiliar prerequisites.

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State the scalar field, dimensions and hypotheses before using a theorem. Verify numerical factors and solutions in the original equations. Exact reasoning is expected; a numerical experiment or theorem name cannot replace a requested proof. Linked results may be cited unless a question asks you to prove them. Alternative valid arguments can meet the same criteria.

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1. Systems and structure

40 practice points.

Question 1: A changing rank, complete fibers and obstruction certificates

10 points

Work over R. Let A(t) be the 3×4 matrix with rows (1,0,1,2), (0,1,−1,1), (1,2,t,4), where t is real. A right-hand side b=(b₁,b₂,b₃) is arbitrary. A complete answer must cover every parameter and distinguish the four ambient spaces. Suggested time: 20 minutes.

(a) · 3 points

Reduce A(t)x=b by reversible row operations. Classify consistency for every t and b, and give every solution in each consistent case with all free parameters explicit.

(b) · 3 points

For t≠−1 and t=−1 separately, give bases and dimensions for col A, row A, ker A and ker Aᵀ. Use original matrix columns when giving a column-space basis. State the ambient space for each.

(c) · 4 points

At t=−1, give a numerical inconsistent b and a left-null-vector certificate. Then prove for an arbitrary real m×n matrix M that Mx=b is consistent iff zᵀb=0 for every z with Mᵀz=0. Finally prove that every nonempty solution set is a translate of ker M and has affine dimension n−rank M.

Review question 1: worked argument and scoring criteria

(a)

Replace row 3 by row 3 minus row 1 minus twice row 2. It becomes (0,0,t+1,0), with right side d=b₃−b₁−2b₂. If t≠−1, put q=d/(t+1). Every solution is (b₁−q−2s,b₂+q−s,q,s), s∈R, and every b is consistent. If t=−1, consistency is exactly d=0. When it holds, every solution is (b₁−r−2s,b₂+r−s,r,s), r,s∈R. When d≠0 the final equation 0=d is impossible. The first two pivot equations determine x₁,x₂ once the displayed free variables are chosen, proving exhaustiveness, not just producing examples.

1 point: Obtains the final row and right-side expression by a reversible operation.
1 point: Classifies all parameter/right-side cases including inconsistency.
1 point: Gives complete affine families with correct free variables and verifies exhaustiveness.

(b)

For t≠−1 the rank is 3. A column-space basis in R³ is (1,0,1),(0,1,2),(1,−1,t); their determinant is t+1≠0. A row-space basis in R⁴ is (1,0,1,2),(0,1,−1,1),(0,0,1,0). The kernel in R⁴ has basis (−2,−1,0,1), dimension 1. The left kernel in R³ is the zero space, with empty basis and dimension 0. At t=−1 the rank is 2. The first two original columns above form a basis for col A in R³. The first two displayed rows form a basis for row A in R⁴. A kernel basis is (−1,1,1,0),(−2,−1,0,1), dimension 2. The left kernel has basis (−1,−2,1), dimension 1. Row reduction preserves row space, while the independent original columns span the column space; kernel vectors follow from the homogeneous families in part a. Their independent free coordinates and the rank counts establish that the lists are full bases.

1 point: Gives all four correct bases and dimensions for t≠−1.
1 point: Gives all four correct bases and dimensions at t=−1.
1 point: States the four ambient spaces and justifies spanning/independence, including the empty basis.

(c)

Take b=(0,0,1) and z=(−1,−2,1). Then A(−1)ᵀz=0 but zᵀb=1, so an assumed solution would give 1=zᵀAx=0. Generally, (col M)⊥=ker Mᵀ: the dot product with every column vanishes exactly when Mᵀz=0. In a finite-dimensional real inner-product space, taking the orthogonal complement twice returns the original subspace, as follows by extending an orthonormal basis of that subspace. Therefore b orthogonal to ker Mᵀ lies in col M, proving sufficiency; necessity follows immediately by multiplying a solution by zᵀ. If x₀ is one solution, Mx=b iff M(x−x₀)=0, so the entire fiber is x₀+ker M. Rank-nullity gives its affine dimension n−rank M. No such dimension statement is assigned to an empty inconsistent fiber.

1 point: Checks a numerical inconsistent right side with an explicit nonzero left-null certificate.
1 point: Proves the orthogonal-complement identity for general M.
1 point: Uses justified finite-dimensional double complementation to prove sufficiency.
1 point: Proves the complete affine-fiber identity and its dimension with the nonempty qualification.

Targeted lesson returns:

Question 2: Turn observations into quotient coordinates and dual functionals

10 points

Let V be the real polynomials of degree at most 3, and define T:V→R³ by T(p)=(p(0),p(1),p′(0)). Here p′(0) is just the coefficient of t in p, so no analytic differentiation theorem is needed. Put W=ker T. Suggested time: 20 minutes.

(a) · 3 points

Find W, prove T is onto, and construct the induced isomorphism T̄:V/W→R³. Prove well-definedness and injectivity instead of just quoting a quotient theorem.

(b) · 3 points

Prove that B=(1+W,t+W,t²+W) is a basis of V/W. In the codomain use C=((1,0,0),(1,1,0),(0,1,1)). Find the matrix of T̄ from B-coordinates to C-coordinates. Use it to find the C-coordinates of T(2−t+3t²+4t³), checking against the actual vector.

(c) · 4 points

Let W° be the annihilator of W in V*, namely all linear functionals λ with λ(w)=0 for every w∈W. Prove the three functionals p↦p(0), p↦p(1), p↦p′(0) form a basis of W°. Generalize: for any surjective linear map S:E→F between finite-dimensional real spaces, prove that φ↦φ∘S is an isomorphism F*→(ker S)°, including the construction and uniqueness of φ.

Review question 2: worked argument and scoring criteria

(a)

For p=a+bt+ct²+dt³, T(p)=(a,a+b+c+d,b). Its kernel requires a=b=0 and c+d=0, so W=span(t³−t²). Given (r,s,u), the polynomial r+ut+(s−r−u)t² has that image, proving surjectivity. Define T̄(p+W)=T(p). If p+W=q+W then p−q∈W and T(p)=T(q), proving well-definedness. The induced map is linear, onto by the explicit preimage, and has zero kernel because T(p)=0 implies p∈W and hence p+W is the zero coset. Thus it is an isomorphism.

1 point: Computes the exact kernel from polynomial coefficients.
1 point: Constructs a preimage for every three-coordinate target.
1 point: Proves the induced map is well-defined, linear and injective, as well as onto.

(b)

Because t³−t²∈W, every coset has representative a+bt+(c+d)t², so B spans. If a+bt+ct² belongs to W, comparison with a multiple of t³−t² forces that multiple and then a,b,c to be zero, proving independence. For a target (r,s,u), C-coordinates are (r−s+u,s−u,u), obtained by solving (α+β,β+γ,γ)=(r,s,u). The images of the three B elements are (1,1,0),(0,1,1),(0,1,0), with C-coordinates (0,1,0),(0,0,1),(−1,1,0). Thus the matrix has rows (0,0,−1),(1,0,1),(0,1,0). The given polynomial has B-coordinates (2,−1,7), which map to (−7,9,−1). Its actual image is (2,8,−1), and −7(1,0,0)+9(1,1,0)−(0,1,1)=(2,8,−1), checking the convention.

1 point: Proves the stated quotient basis using both spanning and independence.
1 point: Computes the coordinate matrix from images of basis elements in the declared order.
1 point: Calculates and independently verifies the changed polynomial coordinates.

(c)

Each displayed functional vanishes on W. If a linear combination of them is zero on V, surjectivity of T lets the three output coordinates vary arbitrarily, forcing all three coefficients to be zero. For spanning, the general factorization argument applies. Given λ∈(ker S)°, define φ(y)=λ(x) for any x with Sx=y. Such an x exists by surjectivity; different choices differ by ker S, so φ is well-defined. Taking preimages of y₁,y₂ proves linearity of φ, and by definition λ=φ∘S. Uniqueness follows because S is onto. Conversely φ∘S always vanishes on ker S, and if φ∘S=0, onto implies φ=0. Thus the linear pullback is a bijection. Applying it to T, the standard coordinate functionals on R³ pull back to precisely the three given functionals, proving they span W°. The construction does not select a preferred polynomial representative or assume a quotient inner product.

1 point: Checks annihilation and independence of the three concrete functionals.
1 point: Constructs φ through preimages and proves choice independence using the kernel.
1 point: Proves linearity, equality λ=φ∘S and uniqueness.
1 point: Establishes the pullback isomorphism and uses it to prove the concrete spanning claim.

Targeted lesson returns:

Question 3: Preserve factors through a late exchange and justify existence

10 points

Let A have real rows (3,1,2),(6,2,5),(−3,1,1). Use unit lower triangular L and upper triangular U. A permutation P acts on the left, so the pivoted identity is PA=LU. Suggested time: 20 minutes.

(a) · 3 points

Begin elimination with the top-left pivot 3 and eliminate both entries below it. Show why a later row exchange is needed, compute det A from these operations, and prove that this nonsingular A has no factorization A=LU without exchanges.

(b) · 3 points

Give P,L,U for the pivoted identity, explicitly explaining where the earlier multipliers go after the exchange. Reuse these factors to solve Ax=(5,12,−3) and Ax=(−2,−4,4). Show both forward right-hand sides and forward solutions, and check the final vectors in the original equations.

(c) · 4 points

Prove the general theorem: a nonsingular real n×n matrix has unit-lower LU without exchanges iff all its leading principal minors Δ₁,…,Δₙ are nonzero. Supply a Schur-complement induction for sufficiency and derive uₖₖ=Δₖ/Δₖ₋₁, with Δ₀=1. Do not infer sufficiency merely from the numerical example.

Review question 3: worked argument and scoring criteria

(a)

The first multipliers are 2 and −1. Subtracting 2 row 1 from row 2 and adding row 1 to row 3 gives rows (3,1,2),(0,0,1),(0,2,3). Swap rows 2 and 3 to obtain U with rows (3,1,2),(0,2,3),(0,0,1), whose determinant is 6. Row additions preserve determinant and the single exchange reverses its sign, so det A=−6. The leading 2×2 minor of A is 3·2−6·1=0. If nonsingular A were unit-lower LU, its leading 2×2 determinant would be u₁₁u₂₂, and all three U diagonal entries would be nonzero because det A≠0. That is impossible. A selected zero pivot here reflects ordering, not singularity.

1 point: Computes both first-step multipliers and the late exchange accurately.
1 point: Tracks determinant preservation and the exchange sign to obtain −6.
1 point: Proves no unpivoted LU using the leading block and nonsingularity, not just algorithm failure.

(b)

P reorders rows as 1,3,2. The earlier multipliers must follow those rows, so L has rows (1,0,0),(−1,1,0),(2,0,1), and U is as in part a. Multiplication gives rows (3,1,2),(−3,1,1),(6,2,5)=PA. For the first request Pb=(5,−3,12); Ly=Pb gives y=(5,2,2). Back substitution gives x=(1,−2,2), whose original outputs are (5,12,−3). For the second Pb=(−2,4,−4); Ly=Pb gives y=(−2,2,0), and back substitution gives x=(−1,1,0), whose original outputs are (−2,−4,4). Swapping only the active U rows while leaving previous multipliers in their old order would break PA=LU. The same factors can be reused because A has not changed; only b has.

1 point: Gives correct P,L,U and verifies the relocation of stored multipliers through PA=LU.
1 point: Shows the permuted right side, forward solution and final solution for the first request.
1 point: Does the same for the second request and verifies both original-system outputs.

(c)

If A=LU, the leading k-block equals LₖUₖ, since terms with summation index greater than k vanish through the lower factor. Thus Δₖ is the product of the first k U diagonal entries. Nonsingularity forces all of them nonzero, proving necessity and the successive-product ratio. For sufficiency, n=1 is immediate. Write A=(a,w;r,C) with a=Δ₁≠0 and S=C−rw/a. Eliminating the first column on the leading (j+1)-block preserves determinant and gives a times the leading j-minor of S. Therefore all those minors equal Δⱼ₊₁/a and are nonzero. By induction S=LₛUₛ. Then L=(1,0;r/a,Lₛ) and U=(a,w;0,Uₛ) multiply to A and have the required shapes and normalization. Every division is by a proved nonzero pivot. The hypothesis nonsingular matters for the necessity statement as formulated: singular matrices such as diag(1,0) may have LU with a zero final pivot.

1 point: Proves the leading-block determinant identity and necessity.
1 point: Derives the Schur leading-minor relation through determinant-preserving operations.
1 point: Completes the induction with explicit block factors and the base case.
1 point: Derives the pivot ratios and correctly qualifies the singular boundary.

Targeted lesson returns:

Question 4: Prove rank bounds by locating lost directions

10 points

All vector spaces below are finite-dimensional over R. General proofs must identify the relevant subspaces and restrictions; a list of matrix examples alone is insufficient. Suggested time: 20 minutes.

(a) · 3 points

For T:U→V and S:V→W, prove rank(ST)=rank T−dim(im T∩ker S). Deduce rank(ST)≥rank S+rank T−dim V, and characterize equality in this lower bound by a subspace containment.

(b) · 3 points

Fix S:R⁴→R² by S(v)=(v₁+v₃,v₂+v₄). Give two injective maps T₁,T₂:R³→R⁴ such that rank(ST₁)=1 and rank(ST₂)=2. Provide their three column vectors and verify all claimed ranks. Explain how one attains the lower bound and the other does not.

(c) · 4 points

Let A,B:V→W have im A∩im B={0}. Prove ker(A+B)=ker A∩ker B. Prove that rank(A+B)=rank A+rank B holds iff ker A+ker B=V. Give explicit 2×2 real matrices showing that disjoint images alone do not force rank additivity.

Review question 4: worked argument and scoring criteria

(a)

Restrict S to im T. Its image is im(ST), while its kernel is im T∩ker S. Rank-nullity for this restricted map gives the exact identity. Since dim(im T∩ker S)≤dim ker S=dim V−rank S, subtraction gives the stated lower bound. Equality holds exactly when the intersection has the full dimension of ker S. As it is a subspace of ker S and dimensions are finite, this is equivalent to ker S⊆im T. This condition describes all directions that S loses being supplied by T, not independence of the two maps.

1 point: Identifies the correct restricted map, image and kernel.
1 point: Uses rank-nullity to derive the exact identity and lower bound.
1 point: Proves the necessary and sufficient containment condition for equality.

(b)

S has rank 2 and kernel spanned by u=(−1,0,1,0), v=(0,−1,0,1). Let T₁ have columns u,v,e₁. They are independent: coordinates 3 and 4 first force the coefficients of u,v to vanish, then coordinate 1 forces the last coefficient to vanish. Their S-images are (0,0),(0,0),(1,0), so the composition has rank 1. This is 2+3−4, and ker S lies in im T₁. Let T₂ have columns e₁,e₂,e₃. These are independent, and their S-images are (1,0),(0,1),(1,0), so the composition has rank 2. Its image subspace intersects ker S only in span(u), not the whole two-dimensional kernel, explaining the strict lower-bound inequality.

1 point: Computes rank S and its full kernel.
1 point: Supplies and verifies an injective T₁ with composition rank 1.
1 point: Supplies and verifies T₂ with composition rank 2 and relates the difference to the subspace condition.

(c)

If (A+B)x=0, then Ax=−Bx lies in both images; disjointness forces Ax=Bx=0. The converse is immediate, proving the kernel identity. Write n=dim V, K=ker A and H=ker B. Rank-nullity gives rank(A+B)=n−dim(K∩H) and rank A+rank B=2n−dim K−dim H. These are equal exactly when dim K+dim H−dim(K∩H)=n. The dimension formula for a sum identifies the left side with dim(K+H); because K+H⊆V, equality is equivalent to K+H=V. For a counterexample take A with rows (1,0),(0,0), and B with rows (0,0),(1,0). Their images are the distinct coordinate lines, intersecting only at zero. Each rank is 1, but A+B has columns (1,1),(0,0) and rank 1, not 2. Both kernels are span(e₂), so their sum is not V. The missing independent input directions explain the failure even though the output directions are disjoint.

1 point: Proves both inclusions in the kernel identity using disjoint images.
1 point: Derives the exact dimension equality from rank-nullity.
1 point: Uses the sum/intersection dimension formula to prove the iff condition.
1 point: Checks explicit matrices with disjoint images but nonadditive ranks and relates their kernels to the theorem.

Targeted lesson returns:

2. Geometry and computation

40 practice points.

Question 5: A rectangular fit and a nearly dependent redesign

10 points

Work over R with the Euclidean norm. Let u=(1,1,0)ᵀ, w=(1,−1,2)ᵀ, and let B have columns u and u+w. The observed vector is y=(1,2,2)ᵀ. Use exact arithmetic and prove the geometric claims. Suggested time: 20 minutes.

(a) · 3 points

Construct a thin QR factorization B=QR with orthonormal columns of Q and positive diagonal of R. Use it to find the unique least-squares coefficient vector x⋆. Show the triangular system explicitly.

(b) · 3 points

Compute the residual r=y−Bx⋆ and its squared norm. Prove, for every h∈R², the exact least-squares comparison identity and use it to establish unique minimality, rather than merely quoting the normal equations.

(c) · 4 points

Now replace B by Bε with columns u and u+εw, where ε>0. Starting from zero data, perturb the data by δε=εw. Find the exact new least-squares coefficients and show that arbitrarily small data perturbations can produce a fixed coefficient change. Compute BεᵀBε and its determinant. Explain what this example does and does not prove about QR versus normal equations in floating-point arithmetic.

Review question 5: worked argument and scoring criteria

(a)

Since uᵀw=0, ‖u‖=√2 and ‖w‖=√6, take Q with columns u/√2 and w/√6. Then R has rows (√2,√2),(0,√6). The coefficient system is Rx=Qᵀy=(3/√2,3/√6). Its second equation gives x₂=1/2, and the first gives x₁+x₂=3/2, hence x⋆=(1,1/2). Both diagonal entries are nonzero, so the coefficient solution is unique.

1 point: Constructs Q and verifies its orthonormal columns.
1 point: Obtains and checks the positive-diagonal R and Qᵀy.
1 point: Solves the displayed triangular equations to obtain (1,1/2).

(b)

Bx⋆=(2,1,1), so r=(−1,1,1) and ‖r‖²=3. Dotting r with u and w gives zero, hence Bᵀr=0. Therefore ‖y−B(x⋆+h)‖²=‖r−Bh‖²=3+‖Bh‖². The columns are independent since u,w are nonzero orthogonal vectors, so Bh=0 implies h=0. The identity proves that x⋆ is the unique global minimizer and the minimum squared residual is 3.

1 point: Computes the residual and squared norm correctly.
1 point: Verifies orthogonality and derives the exact cross-term cancellation.
1 point: Uses column independence to prove strict improvement for every nonzero change.

(c)

The perturbation is exactly Bε(−1,1), and Bε has independent columns for every ε>0, so its unique fit is (−1,1), with residual zero. The data perturbation norm is ε√6→0 but the coefficient change norm is √2. Thus no uniform coefficient-sensitivity bound exists as ε approaches zero. The Gram matrix has rows (2,2),(2,2+6ε²), determinant 12ε². At ε=0 rank is lost; for positive ε the columns approach dependence. This proves sensitivity of the mathematical coefficient problem, regardless of solver. Forming normal equations additionally squares the spectral condition number under the singular-value theorem and may worsen numerical behavior. QR avoids explicitly forming that Gram matrix, but cannot remove the intrinsic sensitivity. No measured error bound for either floating implementation follows from these exact calculations.

1 point: Obtains the exact coefficient change and zero residual using column independence.
1 point: Computes both norms and proves the absence of a uniform sensitivity bound.
1 point: Computes the Gram matrix and determinant 12ε² with the ε=0 boundary.
1 point: Separates intrinsic conditioning from numerical algorithm behavior and avoids an unsupported accuracy guarantee.

Targeted lesson returns:

Question 6: A changing last pivot and the fate of a minimum

10 points

Let A(t) be the real symmetric matrix with rows (4,2,0),(2,2,1),(0,1,t). For each b define E_b(x)=xᵀA(t)x/2−bᵀx. Suggested time: 20 minutes.

(a) · 3 points

Derive an LDLᵀ factorization with unit lower L. Use it to classify every t as positive definite, positive semidefinite singular or indefinite. Give a kernel vector at the boundary and both a positive and a negative quadratic witness in the indefinite range.

(b) · 3 points

At t=2 construct the positive-diagonal Cholesky factor C. Solve A(2)x=(0,−1,0) by Cy=b and Cᵀx=y. Give the minimum of E_b and prove it is unique.

(c) · 4 points

At t=1 compare b₁=(0,−1,0) and b₂=(2,1,0). For each, determine solvability of Ax=b and boundedness/minimizers of E_b. Give complete solution sets where appropriate. Explain why existence of a semidefinite factor CCᵀ does not settle these questions without examining b.

Review question 6: worked argument and scoring criteria

(a)

L has rows (1,0,0),(1/2,1,0),(0,1,1) and D=diag(4,1,t−1). Multiplication gives A(t); equivalently xᵀAx=4(x₁+x₂/2)²+(x₂+x₃)²+(t−1)x₃². Thus t>1 is positive definite. At t=1 the form is semidefinite with kernel span of z=(1,−2,2), so it is singular. For t<1, z gives 4(t−1)<0 while e₁ gives 4>0, proving indefiniteness.

1 point: Derives and verifies L,D or the equivalent complete square identity.
1 point: Classifies strict positivity and the singular boundary with the full kernel.
1 point: Supplies positive and negative witnesses for every t<1.

(b)

C has rows (2,0,0),(1,1,0),(0,1,1). Forward substitution gives y=(0,−1,1), and backward substitution gives x⋆=(1,−2,1). Multiplication checks b. The energy at the solution is −bᵀx⋆/2=−1. For any h, expansion gives E_b(x⋆+h)−E_b(x⋆)=hᵀA(2)h/2=‖Cᵀh‖²/2. Since C is invertible, this is positive for h≠0, proving the unique minimum.

1 point: Constructs and checks C with positive diagonal.
1 point: Shows both triangular solutions and the original-system check.
1 point: Computes minimum −1 and proves uniqueness by the exact energy difference.

(c)

The kernel is span z with z=(1,−2,2). For b₁, zᵀb₁=2, so Ax=b₁ is impossible since zᵀA=0. Along x=sz, E_b₁(sz)=−2s→−∞ as s→∞. For b₂, x₀=(1/2,0,0) is a particular solution. Every solution is x₀+sz because the kernel is exactly one-dimensional. Its energy is −b₂ᵀx₀/2=−1/2. For arbitrary h, the energy difference from x₀ is hᵀA(1)h/2≥0, and equality holds exactly for h in that kernel. Thus the same affine line is the entire set of minimizers. A singular real square-root factor exists but has a zero diagonal; it supplies neither invertibility nor compatibility with every right-hand side. The null-direction linear term decides whether the energy is bounded below here.

1 point: Uses the left-kernel obstruction to prove incompatibility for b₁.
1 point: Proves unboundedness by an explicit null-direction path.
1 point: Gives the entire compatible solution line and minimum −1/2.
1 point: Proves the complete minimizer classification and states the semidefinite-factor limitation.

Targeted lesson returns:

Question 7: A rectangular spectral factor and an optimal truncation

10 points

Let A=(1/√2) times the real matrix with rows (3,3),(4,4),(2,−2). You may use the definition of singular values and basic Euclidean orthogonality, but prove the approximation bound requested below. Suggested time: 20 minutes.

(a) · 3 points

Construct a thin singular-value decomposition A=UΣVᵀ with positive singular values in decreasing order. Verify both orthonormal-column conditions and the product, and identify rank and kernel.

(b) · 3 points

Use the factors to write the pseudoinverse and solve the least-squares problem with y=(1,2,3). Give the coefficient vector in the v₁,v₂ basis, the fitted vector and squared residual. Justify uniqueness without assuming y is in the image.

(c) · 4 points

Let A₁=5u₁v₁ᵀ. Compute ‖A−A₁‖₂ and ‖A−A₁‖F. Prove no real 3×2 matrix of rank at most one has a smaller error in either norm. Your proof should exhibit the lower-bound direction rather than merely cite a low-rank approximation theorem.

Review question 7: worked argument and scoring criteria

(a)

Take u₁=(3/5,4/5,0), u₂=(0,0,1), v₁=(1,1)/√2 and v₂=(1,−1)/√2. Both pairs are orthonormal. With Σ=diag(5,2), the sum 5u₁v₁ᵀ+2u₂v₂ᵀ has precisely the stated rows, proving A=UΣVᵀ. The two positive diagonal values and orthonormal factors show rank 2 and zero kernel. They are its singular values since AᵀA=V diag(25,4)Vᵀ.

1 point: Constructs and verifies the two orthonormal pairs.
1 point: Checks the full product with singular values 5 and 2.
1 point: Justifies the singular-value identification, rank and kernel.

(b)

A⁺=V diag(1/5,1/2)Uᵀ. Since Uᵀy=(11/5,3), the coefficients are x⋆=(11/25)v₁+(3/2)v₂. The fitted vector is U Uᵀy=(33/25,44/25,3). The residual y−Ax⋆=(−8/25,6/25,0) is orthogonal to both u columns and has squared norm (64+36)/625=4/25. For any h, squared residual becomes 4/25+‖Ah‖²; rank 2 makes the added term positive for nonzero h. Thus this is the unique least-squares coefficient vector, even though the nonzero residual shows y is not in the image.

1 point: Writes the pseudoinverse and computes the two spectral coefficients.
1 point: Computes the fitted vector, orthogonal residual and squared norm 4/25.
1 point: Proves uniqueness using the residual identity and injectivity.

(c)

The error is 2u₂v₂ᵀ. Its spectral norm and Frobenius norm are both 2 because u₂,v₂ are unit vectors. For any rank-at-most-one B:R²→R³, rank-nullity gives a unit w∈ker B. Write w=av₁+bv₂ with a²+b²=1. Then ‖(A−B)w‖²=‖Aw‖²=25a²+4b²≥4. Hence ‖A−B‖₂≥2. For every matrix M the Frobenius norm bounds its spectral norm: for any unit w, rowwise Cauchy–Schwarz gives ‖Mw‖²≤Σ over rows of their squared norms=‖M‖F². Therefore ‖A−B‖F≥2 as well. The displayed A₁ attains both lower bounds, proving optimality in both norms. This establishes an optimal rank-one approximation, not a claim that every rank-one candidate or every norm has the same optimum.

1 point: Computes both error norms of the rank-one truncation.
1 point: Finds a unit kernel direction for any competing rank-at-most-one map.
1 point: Uses the singular directions to prove the spectral lower bound 2.
1 point: Proves the Frobenius domination and concludes attainment in both norms.

Targeted lesson returns:

Question 8: Lost cyclic modes and the cost of avoiding wraparound

10 points

Use ωₙ=exp(−2πi/n), X(k)=Σⱼx(j)ωₙ^(jk), and inverse normalization 1/n. Circular convolution is (x∗h)(j)=Σℓ h(ℓ)x(j−ℓ modulo n). Complex arithmetic is exact unless a work estimate explicitly states otherwise. Suggested time: 20 minutes.

(a) · 3 points

At length 6 take h=(1,0,0,−1,0,0). Determine all its Fourier multipliers and prove complete kernel and output-compatibility descriptions in ordinary coordinates. Compute the output for x=(1,0,2,−1,3,0), and determine whether the impulse (1,0,0,0,0,0) can be an output.

(b) · 3 points

For ordinary linear convolution use a=(1,0,−1,2) and b=(1,2,1). Compute the output and the result if both inputs are instead padded to circular length 4. Prove a sufficient padding bound for arbitrary lengths m and ℓ, then choose the smallest power-of-two padding length guaranteed to work for these inputs.

(c) · 4 points

Use two forward radix-two FFTs, pointwise multiplication and one inverse FFT at the padding length from (b). Derive the FFT addition and multiplication recurrences, counting trivial twiddle products too. Give the totals for this three-transform convolution, with inverse scaling listed separately. Explain what this calculation proves about correctness and what it does not prove about floating-point error or measured speed.

Review question 8: worked argument and scoring criteria

(a)

H(k)=1−ω₆^(3k)=1−(−1)^k, so even modes have multiplier zero and odd modes multiplier 2. Directly y(j)=x(j)−x(j−3). Thus the kernel is exactly x(j+3)=x(j), a three-dimensional space, and every output satisfies y(j+3)=−y(j). Conversely any y satisfying that condition is attained by x=y/2, since x(j)−x(j−3)=y(j). These are the complete descriptions, equivalent to killing the even Fourier modes. The stated input gives (2,−3,2,−2,3,−2). The impulse violates the opposite-half condition and therefore is unattainable.

1 point: Computes every spectral multiplier and the correct cyclic difference.
1 point: Proves both directions of the kernel and output descriptions.
1 point: Computes the changed output and proves impulse incompatibility.

(b)

The polynomial product (1−z²+2z³)(1+2z+z²) is 1+2z+3z⁴+2z⁵, giving linear output (1,2,0,0,3,2). Modulo z⁴−1 the last two coefficients fold into the first two positions, giving (4,4,0,0). For lengths m and ℓ the product degree is at most m+ℓ−2; choosing n≥m+ℓ−1 prevents any coefficient from wrapping under zⁿ=1. Here the sufficient bound is 6, and the smallest power of two at least 6 is 8. Length 6 is a valid DFT length but is not supported by the stated radix-two recurrence.

1 point: Computes the linear coefficients and the length-four wraparound output.
1 point: Proves the padding bound from polynomial degree.
1 point: Chooses length 8 and distinguishes DFT existence from radix-two support.

(c)

A length-n butterfly stage has n/2 pairs, each with one twiddle product and two additions/subtractions. Thus A(n)=2A(n/2)+n and M(n)=2M(n/2)+n/2, with A(1)=M(1)=0. Hence A(n)=n log₂n and M(n)=(n/2)log₂n. Each length-eight FFT uses 24 additions and 12 multiplications. Three transforms use 72 additions and 36 multiplications; the eight pointwise products give total 44 complex multiplications. The inverse, implemented by conjugation around a forward FFT, also requires eight complex-value scalings by 1/8 (sixteen real-component divisions if performed componentwise) and conjugations, listed separately. Root generation, memory operations and indexing are excluded. Correctness follows from the proven transform inverse, convolution theorem and sufficient padding; these work counts alone prove neither correctness nor numerical stability. The arithmetic counts are not elapsed times, and exact identities do not bound roundoff. Independent symbolic cases and reconstruction checks are useful diagnostics, not an all-input floating-point theorem.

1 point: Derives both butterfly recurrences with explicit counting conventions.
1 point: Solves them and obtains 72 additions and 44 complex multiplications for this workflow.
1 point: Separately records inverse scaling/conjugation and omitted work.
1 point: Identifies the mathematical correctness dependencies and limits of counts and numerical checks.

Targeted lesson returns:

3. Operators and fields

40 practice points.

Question 9: Recover chains from partial observations

10 points

Let V be a ten-dimensional complex vector space and T:V→V linear. Put N=T−2I. Suppose dim ker N=4, dim ker N²=7, dim ker N³=9 and dim ker N⁴=10. You may use existence of Jordan form over C, but must justify how the observations determine its blocks. Suggested time: 20 minutes.

(a) · 3 points

For a nilpotent operator with Jordan block sizes s₁,…,s_r, prove dim ker N^j=Σ min(j,s_i). Deduce a general interpretation of the consecutive nullity differences.

(b) · 3 points

Determine all Jordan block sizes of T, its characteristic and minimal polynomials, and whether T is diagonalizable. Prove that the minimal polynomial degree cannot be reduced.

(c) · 4 points

A second report for a nilpotent ten-dimensional operator lists nullities 3,7,9,10. Prove it is impossible. Then explain the field hypothesis: does the valid nullity profile still determine the same block sizes for T−2I over R or F₃? Why is it unsafe to infer Jordan blocks over R for an arbitrary real operator without a splitting assumption?

Review question 9: worked argument and scoring criteria

(a)

On a block with chain e₁,…,e_s and Ne₁=0, Ne_i=e_(i−1), N^j kills exactly the span of the first min(j,s) vectors. The direct sum of blocks makes the kernel dimensions add. Writing d₀=0 and d_j=dim ker N^j, the difference d_j−d_(j−1) counts precisely those blocks with s_i≥j: each such block contributes one new kernel dimension. Thus these differences are nonnegative and nonincreasing until they become zero.

1 point: Computes the kernel on one chain, including j≥s.
1 point: Uses direct sums to derive the general sum formula.
1 point: Interprets consecutive differences as numbers of blocks of length at least j.

(b)

N⁴=0 because its kernel is V. The differences are 4,3,2,1, then 0. Taking their successive differences gives exactly one block of each size 1,2,3,4. Hence T has blocks J₁(2),J₂(2),J₃(2),J₄(2), characteristic polynomial (z−2)^10 and minimal polynomial (z−2)^4. On the size-four block, e₄,Ne₄,N²e₄,N³e₄ are independent. Any polynomial of degree at most three in N annihilating that vector therefore has all coefficients zero. Division of a polynomial in T by (z−2)^4 reduces to such a remainder, proving minimality. A block of size greater than one prevents diagonalizability.

1 point: Recovers exactly one block of each size 1,2,3,4.
1 point: States both polynomials and proves minimality using a chain or an equivalent argument.
1 point: Correctly decides diagonalizability from the nontrivial blocks.

(c)

The second report gives increments 3,4,2,1, contradicting the nonincreasing block counts. This obstruction does not require C: N induces an injection ker N^(j+1)/ker N^j → ker N^j/ker N^(j−1), since Nx in ker N^(j−1) implies x in ker N^j. Thus consecutive increments cannot increase over any field. For the valid profile, N⁴=0 makes its minimal polynomial a power of z, which splits over every field; nilpotent Jordan chains therefore exist over R and F₃ as well and have the same sizes. The scalar 2 is interpreted in the chosen field. In contrast the real matrix ((0,−1),(1,0)) has characteristic polynomial z²+1 and no real eigenvalue, so its complex Jordan description is not a Jordan eigenvector basis over R.

1 point: Identifies the impossible increasing nullity increment.
1 point: Proves the quotient injection or supplies another field-independent argument.
1 point: Explains why nilpotence supplies splitting over both specified fields.
1 point: Gives and explains a real nonsplitting counterexample.

Targeted lesson returns:

Question 10: Separate normality, symmetry and length preservation

10 points

All inner products in this problem are on finite-dimensional complex spaces; A* denotes the adjoint. You may use existence of a complex eigenvalue for a nonzero-dimensional operator. Do not invoke the normal spectral theorem as an unexplained premise. Suggested time: 20 minutes.

(a) · 4 points

Prove that a normal operator has an orthonormal eigenbasis. Establish the shifted norm identity, show an eigenvector is also an adjoint eigenvector, and justify the invariant orthogonal complement and induction.

(b) · 3 points

Use that result to prove: a normal A is self-adjoint exactly when all eigenvalues are real, and is unitary exactly when all eigenvalues have modulus one. Give a normal matrix that is neither, and a unitary matrix that is not self-adjoint.

(c) · 3 points

Let S=((1,2),(0,1)). Its eigenvalues all have modulus one. Compute S^m for integers m≥0, show that S is not normal, and exhibit unbounded growth from a unit input. Explain precisely which hypothesis makes the conclusion in part (b) inapplicable.

Review question 10: worked argument and scoring criteria

(a)

For normal B, ||Bx||²=||B*x||² since B*B=BB*. If A is normal, B=A−λI is normal by expansion. For Av=λv the norm identity implies (A*−conj(λ)I)v=0. The line spanned by v is invariant under both A and A*. Its orthogonal complement is also invariant under both, by the adjoint identity. Consequently the restrictions to that complement are mutual adjoints and remain normal. Normalize v and apply induction on dimension to the complement, with the zero-dimensional case empty. Existence of λ uses the complex field. Combining v with the complement basis gives the required orthonormal eigenbasis.

1 point: Proves the norm identity and checks normality after shifting by λ.
1 point: Deduces the adjoint eigenvector equation.
1 point: Proves the complement is reducing and its restriction normal.
1 point: Completes dimension induction, including the complex eigenvalue premise.

(b)

In the orthonormal eigenbasis A=UDU*, so A*=U D* U*. Equality A=A* is equivalent to λ_j=conj(λ_j) for every j. Likewise A*A=I is equivalent to |λ_j|²=1 for every j. The diagonal matrix diag(2,i) is normal but neither self-adjoint nor unitary. The diagonal matrix diag(i,−1) is unitary but not self-adjoint. Conversely every self-adjoint or unitary operator is normal directly from its defining identities, so the criteria are consistent with those classes.

1 point: Derives the self-adjoint criterion in an orthonormal basis.
1 point: Derives the unitary criterion and checks normality of these classes.
1 point: Provides and verifies both requested counterexamples.

(c)

Write S=I+K with K²=0. Induction or the finite binomial expansion gives S^m=I+mK=((1,2m),(0,1)). Here S*S=((1,2),(2,5)), whereas SS*=((5,2),(2,1)); they differ. For the unit vector e₂, ||S^m e₂||=sqrt(4m²+1), which is unbounded. The eigenvalue condition alone omits normality and does not control the nilpotent part of a Jordan block. The unitary equivalence in (b) required an orthonormal eigenbasis, which S does not have.

1 point: Derives the exact powers, including m=0.
1 point: Checks nonnormality and an explicit growing unit input.
1 point: Identifies the missing normality hypothesis rather than rejecting the correct spectral theorem.

Targeted lesson returns:

Question 11: Count solutions after changing the field

10 points

Use the integer matrix A=((1,1,0,1),(1,−1,2,3)) as a linear map F_q⁴→F_q², reducing its entries in the stated prime field. Distinguish vector counts, ordered basis counts and counts of matrices. Suggested time: 20 minutes.

(a) · 3 points

Over F₇ solve Ax=(1,3) completely, give a kernel basis and count the solutions. Justify that the map is onto.

(b) · 4 points

Prove for any rank-r linear map F_q^n→F_q^m that each nonempty fiber has q^(n−r) elements and that there are q^r nonempty fibers. Count the ordered bases of the kernel in part (a), explaining every factor.

(c) · 3 points

Now reduce the same matrix modulo two. Determine rank, kernel size and the full image; count solutions for right sides (1,1) and (1,0). Explain why reducing a real or F₇ rank calculation modulo two without rechecking pivots is invalid.

Review question 11: worked argument and scoring criteria

(a)

Let x₃=s,x₄=t. The equations give x₂=s+t−1 and x₁=2−s−2t. Thus every solution is (2,−1,0,0)+s(−1,1,1,0)+t(−2,1,0,1), with s,t∈F₇. The two displayed kernel vectors are independent because their third and fourth coordinates are the coordinate basis. The first two columns have determinant −2=5≠0 in F₇, so rank is two and the map is onto. There are exactly 7²=49 solutions.

1 point: Derives the complete affine solution with arithmetic in F₇.
1 point: Verifies the kernel basis and rank/onto claim.
1 point: Counts 49 distinct solutions using the free parameters.

(b)

If Ax₀=b, then Ax=b iff A(x−x₀)=0, so translation by x₀ is a bijection from ker A to the fiber. Rank-nullity makes that kernel dimension n−r. Choosing a basis identifies it with F_q^(n−r), hence gives q^(n−r) points. The nonempty fibers are indexed exactly by im A, a dimension-r space with q^r elements. This also partitions q^n inputs into q^r equal fibers. For the two-dimensional kernel in (a), its first ordered basis vector may be any of 49−1=48 nonzero vectors. Its span has seven elements; the second vector has 49−7=42 choices. The product is 2016 ordered bases, not merely 49 vectors or a count of all invertible four-by-four matrices.

1 point: Proves both directions of the affine-fiber translation.
1 point: Uses rank-nullity and basis coordinates to count each fiber.
1 point: Counts nonempty fibers via the image and checks the partition.
1 point: Derives the ordered kernel-basis count 48·42=2016.

(c)

Both rows become (1,1,0,1), which is nonzero, so rank is one. The image is exactly {(0,0),(1,1)} and the kernel has dimension three and 2³=8 points. The side (1,1) has eight solutions; (1,0) has none. The minor −2 that was invertible over F₇ becomes zero in F₂. Division by that pivot cannot be transported to F₂. The matrix entries are the same integers, but linear dependence and allowable inverses depend on the field.

1 point: Computes the reduced rank and complete image.
1 point: Counts the kernel and both requested fibers correctly.
1 point: Identifies the vanished pivot/minor and the field dependence.

Targeted lesson returns:

Question 12: Compute discrete and continuous redistribution exactly

10 points

A population is redistributed among three states by the row-stochastic matrix P=((13/24,1/24,5/12),(1/24,13/24,5/12),(5/12,5/12,1/6)). A row probability vector evolves as r_(m+1)=r_m P. Put u=(1,−1,0), v=(1,1,−2), π=(1/3,1/3,1/3). All quantities are exact. Suggested time: 20 minutes.

(a) · 3 points

Verify stochasticity and the three orthogonal eigenmodes 1,u,v. Derive P^m for all integers m≥0 using their orthogonal projections, not just an assertion about spectral radius.

(b) · 3 points

Start entirely in state three. Obtain the exact distribution at step m, prove convergence and calculate total variation distance to π. Determine the least integer m for which that distance is at most 1/100, and explain any oscillation.

(c) · 4 points

Consider instead the row-vector ODE r′(t)=r(t)(P−I), t≥0, with the same initial row. Derive its solution and distance to π. Prove that exp(t(P−I)) is stochastic, and compare this continuous process with simply putting m=t in the discrete formula.

Review question 12: worked argument and scoring criteria

(a)

All entries are nonnegative and each row sums to one. Symmetry also gives column sums one. Direct multiplication gives P1=1, Puᵀ=(1/2)uᵀ and Pvᵀ=−(1/4)vᵀ. The mutually orthogonal vectors have squared norms 3,2,6. Their projections E=11ᵀ/3, U=uᵀu/2 and V=vᵀv/6 sum to I and have pairwise zero products. Thus P=E+(1/2)U−(1/4)V and P^m=E+2^(−m)U+(−1/4)^m V, including m=0 when the sum is I.

1 point: Checks nonnegativity, row sums and all eigenvalues.
1 point: Constructs the correctly normalized orthogonal projections.
1 point: Proves the power formula and includes m=0.

(b)

For r₀=(0,0,1), the u coefficient is zero and r₀V=−v/3. Hence r_m=π−(1/3)(−1/4)^m v. The coordinate deviations are −a/3,−a/3,2a/3 with a=(−1/4)^m. Their half absolute sum is (2/3)4^(−m). It tends to zero, proving convergence explicitly. At m=3 the distance is 1/96>1/100; at m=4 it is 1/384<1/100, so four is the least step. The negative eigenvalue alternates the signs of deviations; positivity is nevertheless preserved by the stochastic transition matrix.

1 point: Obtains the exact distribution from the initial row vector.
1 point: Derives the total variation formula and least step four.
1 point: Explains convergence and the negative-mode oscillation without confusing signs with negative probabilities.

(c)

The generator Q=P−I has eigenvalues 0,−1/2,−5/4 on the three respective modes. Its exponential is E+exp(−t/2)U+exp(−5t/4)V. Therefore r(t)=π−exp(−5t/4)v/3 and its total variation distance is (2/3)exp(−5t/4). To prove stochasticity rather than infer it from eigenvalues, use commuting I and P: exp(t(P−I))=exp(−t)Σ_(j≥0) t^j P^j/j!. Every P^j is stochastic by induction, the nonnegative coefficients sum to one, and the convergent series gives nonnegative entries and row sums one. This is a mixture of integer-step transitions. It is not substitution of a real t into (−1/4)^m, which is not a real-valued mode for general t. Even at integer times the exponential transition is generally different from P^m.

1 point: Derives the exponential from the shifted eigenvalues.
1 point: Computes the solution and its exact total variation distance.
1 point: Proves stochasticity using the convergent nonnegative mixture of stochastic powers.
1 point: Distinguishes continuous-time evolution from noninteger discrete powers.

Targeted lesson returns:

Use errors to choose the next problem

Award points only for reasoning present in the response. Allow valid later work based on a carried-forward value unless the criterion requires the correct earlier result; do not award a correct-result point for an incorrect value. Record missing hypotheses, wrong ambient spaces, invalid inverses and scalar-field errors separately from arithmetic slips.

The total is out of 120, with no pass threshold or university-grade conversion. This public sample cannot establish full-course readiness. Preserve a first attempt, consult targeted lessons, explain each repair and solve a fresh case. A peer or teacher can review the complete argument; if no reviewer participated, label it self-review.

For further practice, use linear algebra foundations, operator structure, and their mixed assessments. Continue with the network calibration investigation and its separate changed case. Source-by-source course reconciliation remains ongoing.

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