0 · Exam Blueprintread first

PROVE, don't just compute. This is the Advanced unit (≈4 h/wk vs 3): nearly every definition is stated rigorously and almost every theorem is proved. The 60% exam rewards a correct ε–δ argument, a clean induction, a fully-justified IVT/MVT, and linear-algebra reasoning — not the final number alone.

Assessment: exam 60% · mid-sem quiz 13% · 10 online quizzes 10% · 2 assignments 15% · tutorials 2%. Both streams (Calculus + Linear Algebra) on one paper.

Where marks are won: ε–δ limit proofs · sup/inf completeness · induction (weak & strong) · the Rolle→MVT→Taylor→FTC chain · rank–nullity & eigenvalue reasoning. These are exactly what mainstream 1061 can't do.

Two streams, one paper: Calculus (Cirstea, Daners notes) runs complex numbers → limits → continuity → differentiation → Taylor → integration; Linear Algebra (Brownlowe) runs vectors → systems → matrices → subspaces → transformations → eigenvalues. Reproduce named proofs verbatim where you can — they recur.

1 · Proof Toolkitname the method

Quantifiers & negation — get these exact or lose partial credit. ¬(∀x P(x)) ≡ ∃x ¬P(x); ¬(∃x P(x)) ≡ ∀x ¬P(x). Prove ∃x:P(x) by exhibiting one witness; prove ∀x:P(x) by a generalised argument on arbitrary x; disprove ∀ by a single counterexample.

Three generalised proofs

• Direct — assume P, deduce Q. e.g. m=j², n=k² ⇒ mn=(jk)².
• Contrapositive — to prove P⇒Q, prove ¬Q⇒¬P. e.g. n² even ⇒ n even (assume n=2k+1 ⇒ n² odd).
• Contradiction — assume P ∧ ¬Q, derive absurdity. Classic: √2 irrational.
• Cases — for (P or Q)⇒R, prove each branch.

⚠️ A counterexample is a complete proof of a "false/disprove" claim.

Notation marks: ∈ "element of"; := "defined to be" (vs = a deduced equality); distinguish ⇒, ⇐, ⇔. Number systems ℕ⊂ℤ⊂ℚ⊂ℝ⊂ℂ. Set-builder {x∈ℤ : 1≤x≤5}. Sloppy quantifier order or arrows lose marks even when the idea is right.

1b · Induction Templateexamined every sem

Weak induction. P(n) for all n∈ℕ if: (Basis) P(0) (or P(1)) holds; (Step) assume P(k) [the inductive hypothesis], prove P(k+1).

Skeleton1. State P(n) precisely.
2. Basis: verify P(0).
3. Step: "Assume P(k). Then …" ⇒ P(k+1).
4. By induction P(n) ∀n. ∎

Canonical: 3 | 4ⁿ−1, via 4^{k+1}−1 = 4(4^k−1)+3.

Strong (complete) induction. Basis P(0),P(1); step: if P(i) holds for all 0≤i≤k then P(k+1). Use strong induction when P(k+1) needs several earlier cases — e.g. order-≥2 recurrences (Fibonacci-type f(m+2)=f(m+1)+f(m)).

2 · The ε–δ Limitcentral definition

f defined on an open interval around a (not necessarily at a).

lim(x→a) f(x) = L∀ε>0 ∃δ=δ(ε)>0 :
0 < |x−a| < δ ⇒ |f(x)−L| < ε

Variants (all examinable):

lim(x→a) f = +∞ : ∀M ∃δ>0, 0<|x−a|<δ ⇒ f(x)>M
lim(x→∞) f = L : ∀ε>0 ∃N, x>N ⇒ |f(x)−L|<ε

plus one-sided (x→a⁺, a⁻) and ±∞ point/value combinations. The quantifier ORDER is the marked part.

2b · ε–δ Proof Skeletonprove a limit

Method1. Fix ε>0 (arbitrary).
2. Bound |f(x)−L| ≤ (expr in |x−a|).
3. Choose δ so that bound < ε.
4. Verify 0<|x−a|<δ ⇒ |f(x)−L|<ε. ∎

Worked: lim(x→a) √x = √a (a>0). |√x−√a| = |x−a|/(√x+√a) ≤ |x−a|/√a. So take δ = ε√a: then |x−a|<δ ⇒ |√x−√a| ≤ δ/√a = ε. ∎

Squeeze law: f ≤ h ≤ g near a and lim f = lim g = L ⇒ lim h = L. Kills x·cos(1/x)→0. Fundamental limit: lim(x→0) sin x / x = 1.

DNE by two sequences ⚠️: if xₙ→a, x̃ₙ→a give lim f(xₙ) ≠ lim f(x̃ₙ), the limit does not exist. You must exhibit both sequences — don't just assert oscillation.

3 · Continuitylimit = value

f is continuous at a if lim(x→a) f(x) exists and equals f(a). One-sided: right-cts at a if lim(x→a⁺) f = f(a). Continuous on A = cts at each point.

⚠️ A limit can exist yet f be discontinuous because f(a) ≠ lim. Piecewise "find a,b so f continuous": match one-sided limits to the value.

Substitution (composition): f cts at m and lim(x→a) g = m ⇒ lim(x→a) f(g(x)) = f(m). Proof chains two ε–δ statements.

Worked (DNE): lim(x→0⁺) cos(1/√x) fails to exist. Take xₙ = 1/(4n²π²) → 0 with cos→1, and x̃ₙ = 1/(π/2+2nπ)² → 0 with cos→0. Two limits differ ⇒ no limit. ∎

3b · Limit Lawsuse after proving once

Once a few limits are proved from the definition, combine by the laws: lim(f±g)=lim f ± lim g; lim(fg)=lim f·lim g; lim(f/g)=lim f / lim g (denominator limit ≠ 0). Polynomials & rationals are continuous on their domain, so limits = substitution there.

⚠️ Laws presuppose each piece's limit exists — never split a limit you haven't shown exists (0·∞, ∞−∞ traps).

Asymptotes from limits: lim(x→a)f = ±∞ ⇒ vertical asymptote x=a; lim(x→±∞)f = L ⇒ horizontal asymptote y=L. e.g. arctan x has horizontals y=±π/2; 1/x has both.

Worked squeeze: show lim(x→0) x²sin(1/x) = 0. Since −x² ≤ x²sin(1/x) ≤ x² and both bounds → 0, the squeeze law gives 0. ∎ (note sin(1/x) alone has no limit at 0 — the x² factor is what forces it.) Likewise lim(x→∞) cos x/x = 0 by squeeze between ±1/x.

The sequential criterion (DNE via two sequences) is also the bridge to limits of sequences — the same ε–N machinery underlies both.

4 · Completeness · sup/infthe core of ℝ

A⊆ℝ nonempty is bounded above if ∃b: a≤b ∀a∈A (b an upper bound); dually bounded below.

c = sup A (least upper bound)(1) a ≤ c ∀a∈A [c is an upper bound]
(2) ∀ upper bound b : c ≤ b [least one]

β = inf A = greatest lower bound (dual). LUB Axiom ⚠️ (defining property of ℝ): every nonempty A⊆ℝ bounded above has a supremum in ℝ (dual GLB axiom for inf).

sup A need not lie in A: A={1−1/n} has sup A=1∉A. If sup A∈A then sup A = max A.

4b · How to PROVE c = sup Aboth clauses

Two-clause recipe(1) Show a ≤ c for every a∈A.
(2) Show no smaller bound: ∀ε>0 ∃a∈A with c−ε < a ≤ c.

Clause (2) says "c−ε is not an upper bound for any ε>0" — the standard way to pin the least bound. ⚠️ Verifying only clause (1) is the most common lost-mark.

Approximating sequence: if c=sup A then taking ε=1/n gives aₙ∈A with c−1/n < aₙ ≤ c, so aₙ→c. A monotone increasing sequence in A also reaches sup A.

5 · IVT & EVTproved via sup

Intermediate Value Thm ⚠️. f:[a,b]→ℝ continuous ⇒ for every ℓ strictly between f(a),f(b), ∃x₁∈(a,b): f(x₁)=ℓ.

Proof idea: WLOG f(a)<ℓ<f(b). Set A={x∈(a,b): f<ℓ on [a,x)}, let x₁=sup A (completeness!). Continuity squeezes f(x₁)=ℓ from both sides. ∎

Use: "show f has a root" → sign change at two points (or limits at ±∞), then IVT.

Extreme Value Thm ⚠️. f continuous on closed bounded [a,b] ⇒ ∃ x_m,x_M with f(x_m) ≤ f ≤ f(x_M) (min & max attained), so Range = [f(x_m),f(x_M)].

⚠️ EVT needs closed + bounded + continuous; drop any (open interval, jump) and max/min can fail. Surjectivity: cts f with lim_{−∞}=−∞, lim_{+∞}=+∞ ⇒ IVT gives every value.

5b · Worked · sup proofboth clauses

Claim: A = {1−1/n : n≥1} has sup A = 1.

(1) Upper bound: 1−1/n < 1 for all n≥1, so 1 is an upper bound.

(2) Least: take any ε>0. Pick n with 1/n < ε (Archimedean). Then 1−1/n > 1−ε, so 1−ε is not an upper bound. Hence no number < 1 bounds A. ⇒ sup A = 1 (and 1∉A, so A has no maximum). ∎

Note inf A = 0 (attained at n=1, so inf A = min A = 0).

⚠️ Conventions: A unbounded above ⇒ sup A = +∞; A = ∅ ⇒ sup A = −∞. These edge cases appear in "state the sup" questions.

The EVT also proves: a continuous function on a closed bounded interval is bounded — boundedness comes first, then the bounds are attained.

5c · Worked · IVT rootexam reflex

Show x⁵+x−1=0 has a real root. Let f(x)=x⁵+x−1, continuous. f(0)=−1<0, f(1)=1>0; by IVT ∃ c∈(0,1) with f(c)=0. f'(x)=5x⁴+1>0 ⇒ f strictly increasing ⇒ exactly one root. ∎ This "IVT for existence + monotonicity for uniqueness" pairing is the standard root-counting answer.

6 · Differentiationrigorous def

f'(x₀)f'(x₀) = lim(x→x₀) [f(x)−f(x₀)]/(x−x₀) ∈ ℝ

Differentiable ⇒ continuous (not conversely). Carathéodory form: ∃ m continuous at x₀ with f(x)=f(x₀)+m(x)(x−x₀), and f'(x₀)=m(x₀) — makes product/chain-rule proofs clean.

Rules(f+g)'=f'+g' · (fg)'=f'g+fg'
(f/g)'=(f'g−fg')/g² · (f∘g)'=f'(g)·g'

Critical point: f'(x₀)=0 or undefined. Extrema are critical, but not every critical point is an extremum (x³ at 0). 2nd-deriv test: f''>0 min, <0 max, =0 inconclusive.

7 · The MVT ChainRolle→…→L'Hôpital

Rolle ⚠️. f cts [a,b], diff (a,b), f(a)=f(b) ⇒ ∃c∈(a,b): f'(c)=0. Proof: EVT gives extrema; an interior extremum forces f'(c)=0 (else f const).

Cauchy MVT ⚠️. ∃c: [f(b)−f(a)]g'(c) = [g(b)−g(a)]f'(c). Proof: apply Rolle to F(x)=[f(b)−f(a)][g(x)−g(a)]−[g(b)−g(a)][f(x)−f(a)] (F(a)=F(b)=0).

MVT (take g(x)=x)∃c∈(a,b): f'(c) = [f(b)−f(a)]/(b−a)

Corollaries: f'≡0 ⇒ f const; f'>0 ⇒ strictly ↑; f'<0 ⇒ strictly ↓ (proof: MVT on subintervals). "How many roots?" → f' one sign ⇒ injective ⇒ ≤1 root, + IVT for existence.

L'Hôpital ⚠️ (from Cauchy MVT). 0/0 form, g'≠0, lim f'/g'=L ⇒ lim f/g = L. Proof: set f(a)=g(a)=0, apply Cauchy MVT on [a,x].

7b · Worked · MVT useinequality

Prove |sin a − sin b| ≤ |a−b|. Apply MVT to f(x)=sin x on [b,a]: sin a − sin b = cos(c)·(a−b) for some c. Since |cos c| ≤ 1, |sin a − sin b| ≤ |a−b|. ∎

Monotone count: "how many roots has 3x − sin x?" f'(x)=3−cos x ≥ 2 > 0 ⇒ strictly increasing ⇒ at most one root; f(0)=0 gives exactly one.

7c · Implicit Diff & L'Hôpitaltechnique

Implicit: differentiate the relation in x treating y=y(x), then solve dy/dx. e.g. x²y² + x sin y = 4 ⇒ dy/dx = −(2xy² + sin y)/(2x²y + x cos y).

L'Hôpital worked: lim(x→0)(1−cos x)/x² = lim sin x/2x = lim cos x/2 = ½ (two applications, each a 0/0 form). Check the indeterminate form before each step.

7d · Diff ⇒ Cts, not converselyclassic trap

f(x)=|x| is continuous at 0 but not differentiable: the left difference quotient → −1, the right → +1, so f'(0) does not exist. Differentiable ⇒ continuous; continuous ⇏ differentiable.

2nd-deriv test inconclusive case: f(x)=x⁴ at 0 has f'=f''=0 yet a min — fall back to the sign change of f'.

Rolle in one line: any polynomial of degree n has at most n real roots — between consecutive roots f'(c)=0 by Rolle, so the derivative (degree n−1) caps the root count by induction.

The product/chain proofs run cleanest through the Carathéodory form: write each factor as f(x₀)+m(x)(x−x₀) and read off the derivative from m(x₀).

Singular critical points (f′ undefined) count too: |x| at 0 is a minimum with no derivative — always check both stationary and singular points.

8 · Taylor + Lagrangehigh-value proof

nth Taylor poly at x₀Tₙ(x) = Σ_{k=0}^{n} f⁽ᵏ⁾(x₀)/k! · (x−x₀)ᵏ

Tₙ is the unique degree-≤n poly P with lim(x→x₀)[f−P]/(x−x₀)ⁿ = 0 — so build it by substitution / multiply / integrate, no repeated differentiation.

Lagrange remainder ⚠️Rₙ(x) = f⁽ⁿ⁺¹⁾(c)/(n+1)! · (x−x₀)ⁿ⁺¹
for some c between x₀ and x

Proof: Cauchy MVT on F(t)=Σ_{k=0}^{n} f⁽ᵏ⁾(t)/k!·(x−t)ᵏ and G(t)=−(x−t)ⁿ⁺¹; the sum telescopes to F'(t)=f⁽ⁿ⁺¹⁾(t)/n!·(x−t)ⁿ. ∎

Error bound (the exam use)|Rₙ(x)| ≤ max|f⁽ⁿ⁺¹⁾| / (n+1)! · |x−x₀|ⁿ⁺¹

e.g. ln(1.1) to <10⁻⁴: smallest n with 1/[(n+1)10ⁿ⁺¹] < 10⁻⁴ ⇒ n=3.

8b · Standard Seriesmemorise · all at 0

eˣ = Σ xᵏ/k! · 1/(1−x) = Σ xᵏ (|x|<1)
sin x = Σ (−1)ᵏx²ᵏ⁺¹/(2k+1)!
cos x = Σ (−1)ᵏx²ᵏ/(2k)!
ln(1+x) = Σ (−1)ᵏxᵏ⁺¹/(k+1) (|x|<1)

Parity: sin/arctan (odd) → odd powers; cos (even) → even powers. ⚠️ Taylor poly always exists; the series represents f only if Rₙ→0 — must be shown (eˣ via Lagrange + |x|ⁿ⁺¹/(n+1)!→0).

8c · Build Without Differentiatingexamined methods

• Substitution: order-n poly of f(axᵐ) = Tₙ(axᵐ) (order mn). e.g. e^{x²} = Σ x²ᵏ/k!.
• Integrate/differentiate: integrate 1/(1+t²)=Σ(−1)ᵏt²ᵏ ⇒ arctan x = Σ(−1)ᵏx²ᵏ⁺¹/(2k+1).
• Multiply: multiply two polys, keep terms up to order n.

Uniqueness (the limit characterisation) guarantees any method gives the same Taylor polynomial. Also arctan x = Σ(−1)ᵏx²ᵏ⁺¹/(2k+1) on |x|<1.

8d · Convergenceradius

Geometric prototype: 1/(1−x) = lim(1+x+…+xⁿ), remainder xⁿ⁺¹/(1−x) → 0 for |x|<1.

Radius: factorial coefficients (eˣ, e^{−x²}, sin, cos) converge ∀x; geometric-type (1/(1+x²)) only on (−1,1). Representing f always needs Rₙ→0, proved case-by-case.

8e · Worked · Estimate eˣLagrange bound

Estimate e^{0.5} with T₃ and bound the error. T₃(0.5) = 1 + 0.5 + 0.125 + 0.0208… = 1.6458. Since f⁽⁴⁾=eˣ and e^{0.5}<2, |R₃| ≤ 2/4! · 0.5⁴ = 2/24 · 0.0625 ≈ 0.0052. So e^{0.5} ≈ 1.646 ± 0.006 (true 1.6487). ∎

Taylor poly of degree 2 at 0 for f(x)=√(1+x): f(0)=1, f'(0)=½, f''(0)=−¼, so T₂ = 1 + x/2 − x²/8. Limits of indeterminate forms can be read straight off the leading Taylor terms (a fast alternative to L'Hôpital).

e.g. lim(x→0)(sin x − x)/x³: sin x = x − x³/6 + …, so the quotient → −1/6. Reading the first non-vanishing term beats repeated L'Hôpital.

⚠️ A function can be infinitely differentiable yet its Taylor series not represent it away from x₀ — convergence of the series and equality with f are separate facts.

9 · Riemann Integraldef + Darboux

Partition P: a=x₀<…<xₙ=b, Δxₖ=xₖ−xₖ₋₁, norm ‖P‖=max Δxₖ. Riemann sum S = Σ f(xₖ*)Δxₖ.

Integrablebounded f is integrable if lim(‖P‖→0) S = A exists (same A ∀ partitions & samples); ∫ₐᵇ f := A

Darboux ⚠️: with U(f,P)=Σ(sup f)Δxₖ, L(f,P)=Σ(inf f)Δxₖ, f integrable ⟺ ∀ε>0 ∃P: U−L < ε; then L ≤ ∫ ≤ U.

Continuous ⇒ integrable; monotone ⇒ integrable. ⚠️ Bounded ⇏ integrable (Dirichlet 𝟙_ℚ: U=1, L=−1 always). Unbounded ⇒ not integrable.

From definition: for f(x)=x on [a,b], equidistant partition gives Uₙ,Lₙ → (b²−a²)/2 using Σk = n(n+1)/2, so ∫ₐᵇ x dx = (b²−a²)/2.

9b · Integral Propertiesf,g integrable

∫(kf+ℓg) = k∫f + ℓ∫g (linearity)
∫ₐᵇ f = ∫ₐᶜ f + ∫_c^b f (additivity)
f ≤ g ⇒ ∫f ≤ ∫g · |∫f| ≤ ∫|f|

Symmetry tricks: on [−a,a] even ⇒ 2∫₀ᵃ, odd ⇒ 0; ∫₀^π x f(sin x)dx = (π/2)∫₀^π f(sin x)dx.

Why unbounded fails (Darboux): if f is unbounded above on some subinterval, the sup there is +∞, so U(f,P)=+∞ for every P and U−L can't be made < ε ⇒ not integrable (e.g. 1/x on (0,1]). ∎

10 · FTC · both partsproved ⚠️

F(x)=∫ₐˣ f is continuous when f integrable (|f|≤C + triangle inequality + squeeze).

FTC I. f cts ⇒ F(x)=∫ₐˣ f is differentiable, F'(x)=f(x). Proof: [F(x+h)−F(x)]/h = (1/h)∫ₓ^{x+h} f; by EVT/IVT = f(ξ), ξ∈[x,x+h], and ξ→x as h→0 gives f(x). ∎

FTC II. G any primitive (G'=f) ⇒ ∫ₐᵇ f = G(b)−G(a). Proof: F−G has zero derivative ⇒ constant. ∎

Leibniz (variable limits)d/dx ∫_{g₁(x)}^{g₂(x)} f = f(g₂)g₂' − f(g₁)g₁'

From FTC: parts ∫fg' = [fg] − ∫f'g; substitution ∫f(g)g' = ∫_{g(a)}^{g(b)} f(u)du; partial fractions for rationals (∫dt/(1+t²)=arctan t).

Worked Leibniz: H(x)=∫_{x}^{x²} e^{t²} dt ⇒ H'(x) = e^{x⁴}·2x − e^{x²}·1. Improper: ∫₀¹ dx/x diverges (unbounded at 0, handled as lim_{ε→0⁺}∫_ε¹).

Partial fractions: divide first if deg(num) ≥ deg(den); repeated/irreducible-quadratic factors → A/(x−r) + (Bx+C)/(x²+px+q) + …. Recognise ∫dt/(1+t²)=arctan t and ∫dt/t=ln|t|.

Integration by parts and substitution both descend from the product and chain rules via FTC II — so every technique is, at heart, a theorem you can prove.

The integral function F(x)=∫ₐˣ f is the bridge: continuous always, differentiable when f is continuous (FTC I), and its endpoint values give every definite integral (FTC II).

Proof Recapside 1

ε–δ: ∀ε ∃δ, 0<|x−a|<δ ⇒ |f−L|<ε
sup: (1) upper bound (2) least — BOTH
Rolle→Cauchy MVT→MVT→L'Hôpital
Rₙ = f⁽ⁿ⁺¹⁾(c)/(n+1)!·(x−x₀)ⁿ⁺¹
FTC: F'=f · ∫ₐᵇf = G(b)−G(a)

Write proofs in full sentences; state the definition, name the technique, justify each step.

Quick proof-chain map: Completeness (LUB) ⇒ IVT & EVT ⇒ Rolle ⇒ Cauchy MVT ⇒ MVT ⇒ {monotonicity, L'Hôpital, Lagrange remainder}; EVT+IVT ⇒ FTC I ⇒ FTC II. Knowing the arrows lets you rebuild any proof from the one before it.

11 · Complex NumbersCartesian → polar

z = x+iy, i²=−1. Argand plot (x,y). Conjugate z̄ = x−iy; Re z = (z+z̄)/2, Im z = (z−z̄)/2i; |z|² = z z̄.

Polar / exponentialz = r(cos θ + i sin θ) = r e^{iθ}
r = |z| = √(x²+y²), θ = arg z

arg z mod 2π; principal Arg z ∈ (−π,π]. Euler: e^{iθ} = cos θ + i sin θ. Product/quotient: multiply/divide moduli, add/subtract args.

⚠️ Cartesian↔polar is where sign/quadrant errors cost marks — check which quadrant (x,y) sits in before fixing θ.

Arithmetic: (a+ib)(c+id) = (ac−bd) + i(ad+bc); divide by multiplying by the conjugate, z/w = z·w̄/|w|². Quadratics over ℂ: usual formula; for real coefficients complex roots come in conjugate pairs.

11b · de Moivre & Rootsinduction proof ⚠️

de Moivre, n∈ℤ(r(cos θ + i sin θ))ⁿ = rⁿ(cos nθ + i sin nθ)
i.e. (r e^{iθ})ⁿ = rⁿ e^{inθ}

Proof for n∈ℕ by induction: n=1 trivial; assume for k, then (e^{iθ})^{k+1}=(e^{iθ})^k·e^{iθ}=e^{ikθ}e^{iθ}=e^{i(k+1)θ} by the angle-sum identities. ∎ (extend to ℤ via z⁻¹.)

Use: multiple-angle formulas; expand (1+ic)ⁿ by binomial, split Re/Im.

n-th roots of z = r e^{iθ}aₖ = r^{1/n} e^{i(θ+2πk)/n}, k=0,…,n−1

n-th roots of unity: e^{2πik/n} — n equally-spaced points, a regular n-gon on the unit circle. ⚠️ For zⁿ=a give all n roots. Real-coeff polynomials: complex roots in conjugate pairs.

11c · Worked · Roots of Unitysolve zⁿ=a

Solve z³ = 8. Write 8 = 8e^{i·0}. Roots aₖ = 8^{1/3} e^{i(2πk)/3} = 2e^{2πik/3}, k=0,1,2: i.e. 2, 2e^{2πi/3}=−1+i√3, 2e^{4πi/3}=−1−i√3. Three points 120° apart, radius 2. ∎

de Moivre use: (1+i)⁸ — polar 1+i = √2 e^{iπ/4}, so (1+i)⁸ = (√2)⁸ e^{i2π} = 16. Cartesian expansion would be far slower. Assignment 1 used (1+ic)⁵∈ℝ ⇒ sin 5θ = 0 to deduce tan(π/5).

12 · Vectors in ℝⁿops + length

u+v, cu componentwise; ‖cv‖ = |c|‖v‖. Addition is head-to-tail; A=(a₁,…,aₙ) ↔ position vector OA⃗.

Length ‖v‖ = √(v₁²+…+vₙ²); unit vector v/‖v‖; standard basis eᵢ (1 in slot i).

Linear combination: v = c₁v₁+…+cₖvₖ. Dot product u·v = Σuᵢvᵢ; symmetric, bilinear, u·u ≥ 0 (=0 ⟺ u=0), ‖u‖=√(u·u).

The dot-product axioms (symmetry, additivity, (cu)·v=c(u·v), positive-definiteness) are exactly the inner-product axioms — the abstract setting in which Cauchy–Schwarz and the triangle inequality are proved once and reused.

⚠️ ℝⁿ with these operations satisfies the 8 vector-space axioms — the prototype against which every abstract vector space on side 2 is checked.

Geometrically: c>0 keeps direction, c<0 reverses it, and ‖cu‖ scales the length by |c| — the algebra and the picture must agree.

13 · Cauchy–Schwarzthe key proof ⚠️

Cauchy–Schwarz|u·v| ≤ ‖u‖ ‖v‖

Proof: for t∈ℝ, p(t)=(tu+v)·(tu+v)=‖u‖²t² + 2(u·v)t + ‖v‖² ≥ 0. A non-negative quadratic has discriminant ≤ 0: 4(u·v)² − 4‖u‖²‖v‖² ≤ 0 ⇒ |u·v| ≤ ‖u‖‖v‖. ∎

Triangle inequality ⚠️‖u+v‖ ≤ ‖u‖ + ‖v‖

Proof: expand ‖u+v‖² = ‖u‖²+2(u·v)+‖v‖² ≤ ‖u‖²+2‖u‖‖v‖+‖v‖² = (‖u‖+‖v‖)² by Cauchy–Schwarz. ∎

Geometric: u·v = ‖u‖‖v‖cos θ (proved by the cosine rule on ‖u−v‖²). u⊥v ⟺ u·v=0.

Equality in Cauchy–Schwarz holds ⟺ u,v are parallel (the quadratic p(t) then has a real double root). This is the geometric content: cos θ = ±1.

13b · Projection & Crossℝ³

Projection of v onto uproj_u v = (u·v)/(u·u) · u = (u·v)/‖u‖² · u

Decompose v = proj_u v + (v − proj_u v) (parallel + orthogonal).

Cross product (ℝ³)u×v = [u₂v₃−u₃v₂, u₃v₁−u₁v₃, u₁v₂−u₂v₁]

Anticommutative u×v=−(v×u); u×u=0; ⊥ both u and v; right-hand rule; distributive; c(u×v)=(cu)×v. ‖u×v‖ = ‖u‖‖v‖ sin θ (via Lagrange identity ‖u‖²‖v‖² = ‖u×v‖² + (u·v)²).

Area of parallelogram = ‖u×v‖; triangle = ½‖u×v‖; parallelepiped volume = |u·(v×w)| (scalar triple product). Three vectors are coplanar ⟺ the triple product = 0.

14 · Lines & Planesforms

Line: vector x = R + t d; normal n·(x−R)=0 ⇒ ax+by=c. Plane (ℝ³): x = R + s u + t v; normal n·(x−R)=0 ⇒ ax+by+cz=d; from 3 points n = PQ⃗ × PR⃗.

Distance point→line = ‖PQ⃗ × d‖/‖d‖; point→plane = |ax₀+by₀+cz₀−d|/√(a²+b²+c²). ⚠️ Equations are not unique.

Worked plane: through P(1,0,0), Q(0,1,0), R(0,0,1): n = PQ⃗×PR⃗ = [−1,1,0]×[−1,0,1] = [1,1,1], so plane is x+y+z = 1. Distance of origin to it = |−1|/√3 = 1/√3.

Worked line distance: point Q=(1,1,1) to the line through R=(0,0,0) with direction d=[1,0,0]: RQ⃗=[1,1,1], RQ⃗×d=[0,1,−1], ‖·‖=√2, /‖d‖=1 ⇒ distance √2.

Equivalently, distance = ‖perpendicular component of the point-to-line vector‖ — the cross-product formula just packages that. Both routes earn full marks.

For a plane, the same idea gives distance = ‖projection of the point-to-plane vector onto the normal n‖.

The scalar triple product also tests coplanarity and gives signed volume — its sign encodes orientation (right- vs left-handed frame).

14b · Worked · Cauchy–Schwarznumbers

u=[1,2,2], v=[2,−1,2]: u·v = 2−2+4 = 4; ‖u‖=3, ‖v‖=3, so |u·v|=4 ≤ 9 = ‖u‖‖v‖ ✓. Angle cos θ = 4/9. proj_u v = (4/9)[1,2,2].

Cross product: u×v = [2·2−2·(−1), 2·2−1·2, 1·(−1)−2·2] = [6,2,−5]. Check u·(u×v) = 6+4−10 = 0 ✓ (orthogonal). ‖u×v‖ = √65 = area of the parallelogram on u,v.

Orthogonal to a given vector in ℝⁿ: swap two components, negate one, zero the rest — e.g. ⊥ to [a,b,…] is [−b,a,0,…]. Parallel: u = cv for some scalar c.

15 · Gaussian Eliminationsolve · rank

Augmented [A | b]. EROs (preserve the solution set): swap Rᵢ↔Rⱼ; scale Rᵢ→cRᵢ (c≠0); add Rᵢ→Rᵢ+cRⱼ.

REF: zero rows at bottom, each leading entry right of the one above. RREF (Gauss–Jordan): leading 1s with zeros above & below. Back-substitute; free parameters for non-pivot columns. A homogeneous system (b=0) is always consistent (x=0 works) and has nontrivial solutions ⟺ a free variable exists.

Trichotomy ⚠️every system: no solution / unique / ∞ many

Proof idea: two distinct solutions ⇒ their difference solves Ax=0 ⇒ adding scalar multiples gives ∞ many. Inconsistent ⟺ a row [0 … 0 | c], c≠0. Rank = # nonzero rows in REF; # free params = n − rank[A|b].

Worked: x+y+z=6, 2y+z=7, z=3 (already echelon) ⇒ back-substitute z=3, then y=2, then x=1. Unique solution (rank 3 = 3 vars). A row like [0 0 0 | 5] would mark it inconsistent; a non-pivot column would give a free parameter and ∞ solutions.

Matrix product (AB)ᵢⱼ = (row i of A)·(col j of B); A acts as a function L_A(x)=Ax, and AB = composition L_A∘L_B. Not commutative (AB≠BA in general). Transpose reverses: (AB)ᵀ=BᵀAᵀ.

16 · Inverses & Determinantsn×n

A invertible if ∃B: AB=I=BA. Inverse unique ⚠️: B = BI = B(AC) = (BA)C = C. Compute via [A | I] → [I | A⁻¹].

2×2A⁻¹ = 1/(ad−bc) · [d −b; −c a], det=ad−bc

(AB)⁻¹ = B⁻¹A⁻¹ ⚠️; (Aᵀ)⁻¹=(A⁻¹)ᵀ. Cofactor (Laplace) expansion: det A = Σₖ aᵢₖCᵢₖ, Cᵢₖ=(−1)^{i+k}·(minor). Triangular ⇒ det = product of diagonal.

A invertible ⟺ det A ≠ 0; det(AB)=det A·det B; |det A| = volume of the parallelepiped of the rows.

16b · Invertibility TFAEcentral ⚠️

For A (n×n) the following are equivalent: (1) A invertible; (2) Ax=b has a unique solution ∀b; (3) Ax=0 ⇒ x=0; (4) RREF(A)=Iₙ; (5) A is a product of elementary matrices; (6) rank A = n; (7) nullity A = 0; (8) columns are a basis of ℝⁿ; (9) det A ≠ 0. Proved by the cycle (1)⇒(2)⇒(3)⇒(4)⇒(5)⇒(1).

16c · Worked · Det & Eigen3×3 / 2×2

3×3 det (expand row 1): det[1 2 3; 0 1 4; 5 6 0] = 1(1·0−4·6) − 2(0·0−4·5) + 3(0·6−1·5) = −24 + 40 − 15 = 1 ⇒ invertible.

2×2 eigenvalues: A=[2 1; 1 2]. χ(λ)=(2−λ)²−1 = λ²−4λ+3 = (λ−1)(λ−3). λ=1 ⇒ eigenvector [1,−1]; λ=3 ⇒ [1,1]. Two distinct λ ⇒ diagonalisable.

Elementary matrices: E = one ERO applied to I; doing an ERO on A = left-multiplying by E, each E invertible (reverse ERO). One-sided inverse suffices: BA=I ⇒ A invertible with A⁻¹=B. (AB)⁻¹=B⁻¹A⁻¹ reverses order.

Worked 2×2 inverse: A=[1 2; 3 4], det = 4−6 = −2 ≠ 0 ⇒ invertible; A⁻¹ = (−1/2)[4 −2; −3 1] = [−2 1; 1.5 −0.5]. Check AA⁻¹ = I.

The cofactor expansion can be taken along any row or column — pick the one with the most zeros to cut work. Row-reduce to triangular and multiply the diagonal for larger n.

det(AB)=det A·det B gives det(A⁻¹)=1/det A and det(Aᵀ)=det A — handy checks when a computed inverse or eigenvalue looks wrong.

17 · Vector Spaces8 axioms ⚠️

A vector space over ℝ is a set V with + and scalar mult (closed), a zero 0, inverses −u, satisfying 8 axioms: commutativity & associativity of +; 0; −u; c(u+v)=cu+cv; (c+d)u=cu+du; c(du)=(cd)u; 1u=u.

⚠️ To PROVE V is a vector space: verify closure + every axiom; to disprove, exhibit one failing axiom. Examples beyond ℝⁿ: functions f:ℝ→ℝ (pointwise); (0,∞) with x⊕y=xy, c⊙x=xᶜ (zero is 1).

An isomorphism is a bijective linear map; isomorphic spaces share dimension and all structural facts. Assignment 1 asked to verify a recurrence-defined function space is isomorphic to ℝ² (a function is determined by f(0),f(1)).

To prove a subset is a subspace it suffices to check the three subspace conditions; you get the other axioms free because they are inherited from the ambient space.

17b · Subspace Test3 conditions ⚠️

S⊆ℝⁿ is a subspace if(1) 0 ∈ S
(2) closed under + (u,v∈S ⇒ u+v∈S)
(3) closed under scalar mult (u∈S ⇒ cu∈S)

Then S is itself a vector space. Lines/planes through the origin are subspaces; {x+y≥0} or {xy=0} are not. span(S) is always a subspace (proof: closure). Subspaces of ℝ³: {0}, lines, planes through 0, ℝ³.

18 · Span · Independencebasis

Span(S) = {c₁v₁+…+cₖvₖ}. b ∈ span(v₁…vₖ) ⟺ [v₁…vₖ | b] consistent.

Linear independencec₁v₁+…+cₖvₖ = 0 ⇒ c₁=…=cₖ=0
(only the trivial solution)

Otherwise dependent (any set containing 0 is dependent). ⚠️ n vectors in ℝⁿ l.i. ⟺ [v₁…vₙ] invertible; k l.i. vectors in ℝⁿ ⇒ k ≤ n.

Basis of S = spans S and linearly independent. Basis Theorem: any two bases have the same size = dim S (dim ℝⁿ = n).

S spans ℝⁿ ⇒ |S| ≥ n; S l.i. in ℝⁿ ⇒ |S| ≤ n. Corollary: exactly n vectors form a basis ⟺ they are l.i. ⟺ they span. b∈span(v₁…vₖ) ⟺ the augmented system [v₁…vₖ | b] is consistent.

18b · Worked · Independencetrivial-only test

Are v₁=[1,2,1], v₂=[2,1,0], v₃=[1,−1,−1] l.i.? Row-reduce [v₁ v₂ v₃] (as columns): det = 1(1·−1−0·−1) − 2(2·−1−0·1) + 1(2·−1−1·1) = −1 + 4 − 3 = 0 ⇒ dependent. Indeed v₃ = v₂ − v₁.

⚠️ A zero determinant / a free column means the only-trivial-solution test fails — there is a nontrivial dependence relation.

18c · Worked · Subspace?check 3 conditions

Is W = {(x,y,z): x+y+z=0} a subspace of ℝ³? 0=(0,0,0) ✓; sum of two solutions solves it ✓; scalar multiple solves it ✓ ⇒ yes (a plane through 0, dim 2). But {x+y+z=1} fails (0∉it), and {xy=0} fails closure under +.

Basis of W above: solve x=−y−z ⇒ vectors [−1,1,0],[−1,0,1] span and are l.i. ⇒ dim W = 2. The matrix-space M_{m×n}(ℝ) and the polynomial space Pₙ are also vector spaces (the abstract examples examined in Assignment 1).

19 · Rank–Nullitythe theorem ⚠️

For A (m×n): row(A)⊆ℝⁿ, col(A)⊆ℝᵐ, null(A)={x: Ax=0}⊆ℝⁿ — all subspaces (proof for null: closure).

Finding bases (R = RREF A): row space → nonzero rows of R; column space → columns of A in R's pivot positions (col(A)≠col(R) but same dim); null space → solve Rx=0, separate parameters.

Rank–Nullity ⚠️rank(A) + nullity(A) = n

rank = # leading 1s = dim row = dim col; nullity = dim null. Proof: the n columns split into pivot (→ rank) vs free (→ nullity). Corollary: in dim-n space, n vectors are l.i. ⟺ they span.

20 · Linear Transformationsstandard matrix

T:ℝⁿ→ℝᵐ linear if T(u+v)=T(u)+T(v) and T(cu)=cT(u). Hence T(0)=0.

Standard-matrix thm ⚠️: every linear T = L_A with A = [T(e₁) … T(eₙ)] (m×n, unique). Proof: x=Σxᵢeᵢ + linearity. Composition: A_{S∘T} = A_S A_T.

⚠️ Prove T NOT linear: exhibit a failing case — absolute values, constants or squares break it (e.g. T with a |z| term fails T(cu)=cT(u) at c=−1). T:ℝⁿ→ℝⁿ bijective ⟺ A_T invertible.

21 · EigenvaluesAx = λx

Nonzero x, scalar λ with Ax=λx: λ eigenvalue, x eigenvector. Eigenspace E_λ = null(A−λI) (a subspace; 0∈E_λ but is never an eigenvector).

Characteristic polynomialλ eigenvalue ⟺ χ_A(λ)=det(A−λI)=0

Method: solve det(A−λI)=0 for λ, then for each λ find null(A−λI). ⚠️ Eigenvectors must be nonzero; report a basis of each eigenspace. Real A can have complex eigenvalues (rotation) — links to the complex stream.

21b · Worked · Rank–Nullitycheck n

A is 3×4 with RREF having pivots in columns 1,2 (2 pivots). Then rank = 2, and nullity = 4 − 2 = 2 (two free variables). Column space dim 2 ⊆ ℝ³; null space dim 2 ⊆ ℝ⁴. Sum 2+2 = 4 = n ✓.

Standard matrices: rotation by θ = [cos θ −sin θ; sin θ cos θ]; RₐRᵦ = R_{α+β} recovers the angle-sum identities. Reflection, projection, scaling each have their own A.

21c · The Multiplicity Trapwhen diag fails

A=[2 1; 0 2]: χ(λ)=(2−λ)², so λ=2 with algebraic mult 2. But A−2I = [0 1; 0 0] has nullity 1 ⇒ geometric mult 1 < 2. So A is NOT diagonalisable — only one independent eigenvector.

Contrast A=[2 0; 0 2] (=2I): same λ=2 twice but every vector is an eigenvector, geometric mult 2 ⇒ already diagonal.

Eigenvectors for distinct λ are l.i. — the lemma behind "n distinct eigenvalues ⇒ diagonalisable". Trace = Σλ and det = Πλ give fast sanity checks on a computed spectrum.

A real matrix can have complex eigenvalues (a rotation [cos θ −sin θ; sin θ cos θ] has λ = e^{±iθ}) — this is exactly where the complex-number stream feeds back into linear algebra.

Always compute χ_A(λ) and factor it before hunting eigenvectors — the marks for "eigenvalue reasoning" live in showing det(A−λI)=0, not just stating the answers.

22 · DiagonalisationA = PDP⁻¹

A is diagonalisable if A=PDP⁻¹ (D diagonal) ⟺ A has n linearly independent eigenvectors (columns of P), eigenvalues on D's diagonal.

Sufficient: n distinct eigenvalues ⇒ diagonalisable (eigenvectors for distinct λ are l.i.). Use: Aᵏ = PDᵏP⁻¹; Markov chains (e.g. PageRank).

⚠️ The multiplicity trap: a repeated eigenvalue may still diagonalise — check geometric mult (dim E_λ) = algebraic mult (repetition). If any eigenspace is too small, A is not diagonalisable.

Powers worked: with A=[2 1;1 2] (λ=1,3; P=[1 1;−1 1]), Aᵏ = P diag(1,3ᵏ) P⁻¹. Markov chains: the steady state is the eigenvector for λ=1, found as 3ᵏ-terms wash out under repeated multiplication.

Diagonalising also solves linear recurrences and decouples systems of ODEs — same machine: change to the eigenbasis, act diagonally, change back.

Final reflex: every "compute" question in 1961 hides a "justify" — name the theorem you are invoking and you bank the method marks even if arithmetic slips.

23 · Proof Pattern Beltpick the recipe

• "Prove ∀…" → arbitrary element OR induction (strong if order ≥2).
• "Prove a limit" → ε–δ: fix ε, bound, pick δ.
• "Prove c = sup" → both clauses; clause 2 via c−ε.
• "∃ a root / value" → IVT (sign change / limits ±∞).
• "How many roots" → MVT monotonicity + IVT.
• "Bound the error" → Lagrange remainder, solve for n.
• "Is it a subspace / VS" → check 0 + closure / 8 axioms.
• "l.i.?" → trivial-solution-only of Σcᵢvᵢ=0.
• "Diagonalisable?" → n l.i. eigenvectors / geom = alg mult.
• "Prove an inequality" → Cauchy–Schwarz, or MVT, or a non-negative square.
• "Find all roots of zⁿ=a" → polar form + the n-root formula.
• "Disprove" → one counterexample suffices.

24 · High-Yield Trapswhere rigour wins

• sup proof needs both clauses, not just an upper bound.
• Limit DNE needs two explicit sequences.
• EVT needs closed + bounded + cts.
• Bounded ⇏ integrable (Dirichlet).
• Taylor series reps f only if Rₙ→0.
• Critical point ⇏ extremum (x³).
• Repeated eigenvalue ⇒ check geom = alg mult.
• col(A) uses A's columns, not RREF's.
• Eigenvector must be nonzero; report a basis of E_λ.
• zⁿ=a has all n roots — don't stop at one.
• Proofs in full sentences (LO2), not symbol-soup.

LA Recapside 2

z = re^{iθ} · (re^{iθ})ⁿ = rⁿe^{inθ}
|u·v| ≤ ‖u‖‖v‖ · proj = (u·v)/‖u‖²·u
rank + nullity = n · det≠0 ⟺ invertible
Ax=λx · χ_A(λ)=det(A−λI)=0
diag ⟺ n l.i. eigenvectors