MAST10007 Cheatsheet · Linear Algebra

0 · Exam mapcourse facts

“The only permitted material for the final exam is one A4 sheet of notes, with writing back and front.” (MAST10007 subject materials, 2026)

1 · Row reductionCh 1–2 · solve/classify

Fix variable order; encode all coefficients and constants in [A|b]. Legal reversible moves: Rᵢ↔Rⱼ; Rᵢ←cRᵢ with c≠0; Rᵢ←Rᵢ+cRⱼ. Apply each move across the augmented bar.

exam routineleftmost useful pivot → swap if zero → clear below → continue → back-substitute; use exact arithmetic → substitute into original equations.

1b · What the rows meanCh 1–2 · interpret

2 · Matrix algebraCh 3 · structure first

productAₘₓₙBₙₓₚ exists → AB is m×p; (AB)ᵢⱼ = rowᵢ(A)·colⱼ(B). Usually AB≠BA; one order may not exist.

transpose / inverse(AB)ᵀ=BᵀAᵀ · (AB)⁻¹=B⁻¹A⁻¹ · solve AX=I or reduce [A|I]. Never “divide by A”.

2×2 determinant and inverse

A=[[a,b],[c,d]] · det A=ad−bc.
det A≠0 ⇒ A⁻¹=(1/(ad−bc))[[d,−b],[−c,a]].

Determinant row effects: swap rows → sign flips; scale a row by c → determinant scales by c; row replacement → determinant unchanged. For triangular A, det A is the product of diagonal entries.

3 · Vector geometryCh 4 · choose output

u·v=Σuᵢvᵢ=||u||||v||cosθ · u⊥v ⇔ u·v=0
projᵤ(v)=((v·u)/(u·u))u, u≠0
r=v−projᵤ(v) ⇒ r·u=0.

3b · Determinant geometryCh 3–4 · scale

4 · Subspace or affine?Ch 5 · proof

Subspace test: for W⊆V, show 0∈W and for arbitrary u,v∈W, scalars a,b, au+bv∈W. A homogeneous linear constraint has a subspace solution set; a translated constraint may be affine.

5 · Span · independenceCh 6–7 · pivot logic

two different systemsb∈span{v₁,…,vₖ}? Solve [v₁ … vₖ]c=b; consistency gives coefficients.
Independent? Solve [v₁ … vₖ]c=0; only c=0.

A nontrivial relation identifies redundancy: isolate any vector with nonzero coefficient; removing it preserves the span. The zero vector makes a list dependent. More vectors than ambient dimension forces dependence.

Basis = independent + spanning; coordinates are unique only relative to a declared ordered basis B. If v=Σcᵢbᵢ, then [v]ᴮ=(c₁,…,cₙ).

Fundamental-space bookkeeping

5b · Redundancy pictureCh 6 · explain relation

6 · Linear mapsCh 8–9 · build/diagnose

linearityT(u+v)=T(u)+T(v) and T(cu)=cT(u). Hence T(0)=0 and T(au+bv)=aT(u)+bT(v).

For T:Rⁿ→Rᵐ, the standard matrix is A=[T(e₁) … T(eₙ)] of size m×n. A nonzero translation fails T(0)=0. If S acts first, then T, the composite matrix is A_T A_S.

Kernel = solutions of Ax=0, living in the domain. Image = Col(A), living in the codomain. A nonzero kernel vector records a lost input direction.

7 · Change of basisCh 7, 9 · direction

If Pᴮ=[b₁ … bₙ], then v=Pᴮ[v]ᴮ. Thus [v]ᶜ=(Pᶜ)⁻¹Pᴮ[v]ᴮ. Label source and destination; the reverse conversion is the inverse.

safest checksource coordinates → reconstruct standard v → solve in destination basis → reconstruct the same standard v.

7b · Coordinate diagramCh 9 · commute

Composition audit → Send a basis vector through the maps in order; it must match the corresponding column of the product. The first map is rightmost. Domain and codomain belong to T.

8 · EigenstructureCh 10 · diagonalise

methoddet(A−λI)=0 → for each λ solve (A−λI)v=0 → collect independent eigenvectors → P=[v₁ … vₙ], D=diag(λ₁,…,λₙ) in matching order.

An eigenvector is nonzero and obeys Av=λv. Algebraic multiplicity counts roots; geometric multiplicity is dim eigenspace. Always 1≤geom≤alg.

A diagonalizable ⇔ n independent eigenvectors.
Distinct eigenvalues ⇒ independent eigenvectors.
A=PDP⁻¹ ⇒ Aᵏ=PDᵏP⁻¹.

Check → Verify AP=PD before computing P⁻¹. Repeated λ needs an eigenspace-dimension check.

9 · Powers · MarkovCh 11 · convention

Recursion xₖ₊₁=Axₖ gives xₖ=Aᵏx₀. For column states xₖ₊₁=Pxₖ, columns sum to 1; row-state convention uses row sums.

stationary statePx=x, xᵢ≥0, Σxᵢ=1. Solve eigenvalue 1, then normalise.

9b · State motionCh 11 · interpret

10 · OrthogonalityCh 12 · project/GS

||v||=√⟨v,v⟩ · u⊥v ⇔ ⟨u,v⟩=0
projᵤ(v)=(⟨v,u⟩/⟨u,u⟩)u, u≠0.
Orthonormal {qᵢ}: proj_W(v)=Σ⟨v,qᵢ⟩qᵢ.

Gram–Schmidt: u₁=v₁; uⱼ=vⱼ−Σᵢ<ⱼ projᵤᵢ(vⱼ); then qⱼ=uⱼ/||uⱼ||. Subtract every earlier projection; the resulting nonzero list preserves the original span.

11 · Least squares · SVDCh 13–14 · fit/factor

least squaresx̂ minimises ||b−Ax||; r=b−Ax̂ ⟂ Col(A).
AᵀA x̂=Aᵀb. Unique if columns of A are independent.

Orthogonal Q: QᵀQ=I, Q⁻¹=Qᵀ; preserves norms, dot products and angles. Real symmetric A has real eigenvalues and an orthonormal eigenbasis: A=QDQᵀ.

SVDA=UΣVᵀ: Vᵀ selects orthonormal input directions → Σ scales by nonnegative singular values → U places orthonormal output directions.

Complex: A*=Āᵀ; (AB)*=B*A*. Hermitian H*=H. Unitary U*U=I, so U⁻¹=U*. For real entries these become symmetric and orthogonal.

Trap → Normal equations express closest fit, not exact solvability; singular values are nonnegative and are not generally eigenvalues of A.

11b · SVD geometryCh 13 · all real A

12 · Worked 1 · eliminationCh 1 · 4 steps

Solve x+2y−z=3; 2x+5y+z=9; −x−y+2z=0.

[1 2 −1 | 3]
[2 5  1 | 9]
[−1 −1 2 | 0]

1. R₂←R₂−2R₁ and R₃←R₃+R₁ give new rows (0,1,3|3) and (0,1,1|3).
2. R₃←R₃−R₂ gives (0,0,−2|0), so z=0.
3. Back-substitute: y=3; then x=−3.
4. Answer (−3,3,0). Substitution into all three originals returns 3, 9 and 0.

13 · Parametric answerCh 2 · whole family

For x+2y−z=4 and 2x+4y−2z=8, the second row is twice the first. No contradiction; y,z are free.

y=s, z=t ⇒ x=4−2s+t
(x,y,z)=(4,0,0)+s(−2,1,0)+t(1,0,1), s,t∈R.

Check: each direction solves the homogeneous equation; the particular vector supplies 4. Never report only one convenient solution.

13b · Classify at a glanceCh 2 · geometry

Stop rule → Echelon form is already enough for pivots, zero or contradiction rows and back-substitution. Reduce to RREF only if easier to read; extra operations add arithmetic-error risk.

14 · Worked 2 · change basisCh 9 · 4 steps

Given B={(1,0),(1,1)}, C={(1,1),(0,1)}, [v]ᴮ=(2,3). Find [v]ᶜ.

1. Reconstruct: v=2(1,0)+3(1,1)=(5,3).
2. Write (5,3)=a(1,1)+b(0,1)=(a,a+b).
3. a=5; a+b=3 ⇒ b=−2.
4. [v]ᶜ=(5,−2). Check: 5(1,1)−2(0,1)=(5,3).

15 · Basis/space trapsCh 6–9 · recall

Mini-check: B={(1,1),(1,−1)}, x=(5,1). Solve a+b=5, a−b=1 ⇒ a=3,b=2, so [x]ᴮ=(3,2); reconstruct 3(1,1)+2(1,−1)=(5,1).

Language → Kernel vectors are inputs; image vectors are outputs. Codomain is declared; image is what is actually reached.

15b · Four spaces snapshotCh 7 · dimensions

Direction audit → Label each coordinate vector with its ordered basis. A basis-column matrix maps basis coordinates to standard; its inverse returns standard to basis. Reconstruct from both coefficient lists: the same standard vector must result. This catches arithmetic and a reversed matrix.

16 · Worked 3 · diagonaliseCh 10 · 4 steps

A=[[4,1],[0,2]]. Diagonalise and state Aᵏ.

1. Triangular entries give λ₁=4, λ₂=2.
2. λ=4: v₁=(1,0). λ=2: 2x+y=0, choose v₂=(1,−2).
3. P=[[1,1],[0,−2]], D=diag(4,2). Independent columns license A=PDP⁻¹.
4. Aᵏ=PDᵏP⁻¹, Dᵏ=diag(4ᵏ,2ᵏ). Verify AP=PD.

17 · Markov drillCh 11 · fixed point

P=[[0.8,0.3],[0.2,0.7]] acts on column states. Columns sum to one. Solve Px=x with x₁+x₂=1:

0.8x₁+0.3x₂=x₁ ⇒ 0.2x₁=0.3x₂ ⇒ x₁:x₂=3:2 ⇒ x=(0.6,0.4).

Check P(0.6,0.4)=(0.60,0.40), entries nonnegative, total 1. This proves stationarity only; do not overclaim convergence from every initial state.

17b · Diagonalisation auditCh 10–11

Failure check → A repeated root gives algebraic multiplicity, not the number of independent eigenvectors. Solve (A−λI)v=0 and count the eigenspace dimension before claiming diagonalisation. P needs n independent columns, each aligned with its λ in D. Test AP=PD: it checks all eigenpairs at once before powers are computed.

18 · Worked 4 · Gram–SchmidtCh 12 · 4 steps

v₁=(1,1,0), v₂=(1,0,1).

1. u₁=v₁=(1,1,0); u₁·u₁=2.
2. projᵤ₁(v₂)=((v₂·u₁)/(u₁·u₁))u₁=(1/2)(1,1,0).
3. u₂=v₂−proj=(1/2,−1/2,1); u₁·u₂=0.
4. If orthonormal requested: e₁=u₁/√2; e₂=u₂/√(3/2). Otherwise {u₁,u₂} is already an orthogonal basis for the same span.

19 · Projection trapsCh 4, 12 · diagnose

Cauchy–Schwarz: |⟨u,v⟩|≤||u||||v||; equality iff u,v dependent. An orthogonal list of nonzero vectors is automatically independent.

19b · Projection decisionCh 12 · basis type

Projection audit → Projection lies in the target subspace; its residual is orthogonal to every basis direction. In Gram–Schmidt test each new residual against every earlier uᵢ; normalise after checking the orthogonal list.

20 · Worked 5 · least squaresCh 13 · 4 steps

Fit y≈a+bx to (0,1),(1,2),(2,2).

1. A has rows (1,0),(1,1),(1,2); b=(1,2,2). AᵀA=[[3,3],[3,5]], Aᵀb=(5,6).
2. Solve 3a+3b=5; 3a+5b=6. Subtract: 2b=1 ⇒ b=1/2.
3. 3a+3/2=5 ⇒ a=7/6. Fit: y=7/6+x/2.
4. Fitted values (7/6,5/3,13/6); residual b−A(a,b)=(-1/6,1/3,−1/6). Its dot products with (1,1,1) and (0,1,2) are both 0.

21 · Orthogonal · symmetric · SVDCh 13 · compare

AᵀA vᵢ=σᵢ²vᵢ; σᵢ≥0. Right singular vectors vᵢ are input directions; left singular vectors uᵢ are output directions.

Check → Orthogonal diagonalisation is for real symmetric square matrices; SVD works for every real matrix.

21b · Fit versus exact solveCh 13 · residual

22 · SVD picture · complex checkCh 13–14

complex adjointA*=Āᵀ · (AB)*=B*A*
Hermitian: H*=H. Unitary: U*U=I ⇒ U⁻¹=U*.

Source check: U=(1/√2)[[1,i],[i,1]]. Then U*=(1/√2)[[1,−i],[−i,1]]. In U*U, diagonal (1+(−i)i)/2=1 and off-diagonal (i−i)/2=0, hence U*U=I.

Trap → Ordinary transpose is the right adjoint only when entries are real. Conjugation makes ⟨v,v⟩ real and nonnegative.

23 · Method choosertrigger → move → proof

24 · Complex geometryCh 14 · conjugation

MATLAB audit → Use supported arithmetic or RREF to check a relation; then state the definition tested and conclusion justified. Output checks arithmetic, not proof; transpose replaces the adjoint only for real entries.