08-05 — Inner Product Spaces
Phase: 8 — Linear Algebra (Rigorous) Subject: 08-05 Prerequisites: 08-04 — Matrices as Linear Transformations Next subject: 08-06 — Orthogonal Projections
Learning Objectives
By the end of this subject, you will be able to:
- State the axioms of an inner product and verify whether a given operation satisfies them
- Compute norms, distances, and angles from an inner product, and prove the Cauchy-Schwarz and Triangle inequalities
- Determine whether vectors are orthogonal, and decompose a vector into orthogonal components
- Apply the Gram-Schmidt process to convert any basis into an orthonormal basis
- Recognize orthogonal complements and use the decomposition V = W ⊕ W^⟂
Core Content
1. Definition of Inner Product
An inner product on a real vector space V is a function ⟨·, ·⟩ : V × V → R satisfying:
- Symmetry: ⟨u, v⟩ = ⟨v, u⟩
- Linearity in first argument: ⟨au + v, w⟩ = a⟨u, w⟩ + ⟨v, w⟩
- Positive definiteness: ⟨v, v⟩ ≥ 0, and ⟨v, v⟩ = 0 iff v = 0
CRITICAL — Foundational: An inner product gives a vector space GEOMETRY — length, angle, distance, orthogonality. Cauchy-Schwarz $|⟨u,v⟩| ≤ ‖u‖‖v‖$ is the most important inequality — it justifies everything.
(For complex vector spaces, replace symmetry with conjugate symmetry: ⟨u, v⟩ = ⟨v, u⟩̅.)
Standard inner products: - R^n (dot product): ⟨x, y⟩ = x·y = Σ x_i y_i = x^T y - C[a, b] (functions): ⟨f, g⟩ = ∫_a^b f(x)g(x) dx - M_{m×n} (matrices): ⟨A, B⟩ = tr(A^T B) = Σ_i Σ_j a_ij b_ij - Weighted inner product on R^n: ⟨x, y⟩_W = x^T W y where W is symmetric positive definite
The dot product in R^n is an inner product: - Symmetry: x·y = y·x ✓ - Linearity: (ax+z)·y = a(x·y) + z·y ✓ - Positive definiteness: x·x = Σ x_i² ≥ 0, and = 0 only if all x_i = 0 ✓
2. Norm, Distance, Angle
From an inner product we derive geometric concepts:
Norm (length): ‖v‖ = √⟨v, v⟩
Properties of norm (derived from inner product): - ‖v‖ ≥ 0, and ‖v‖ = 0 iff v = 0 - ‖cv‖ = |c| ‖v‖
Distance: d(u, v) = ‖u − v‖
Angle between u and v (both nonzero):
$cos θ = ⟨u, v⟩ / (‖u‖ ‖v‖) $
This is well-defined because |⟨u, v⟩| ≤ ‖u‖ ‖v‖ (Cauchy-Schwarz).
Cauchy-Schwarz Inequality: |⟨u, v⟩| ≤ ‖u‖ ‖v‖
Proof sketch for general inner product: For any t ∈ R, ‖u + tv‖² = ⟨u+tv, u+tv⟩ = ‖u‖² + 2t⟨u,v⟩ + t²‖v‖² ≥ 0. This quadratic in t has discriminant ≤ 0: 4⟨u,v⟩² − 4‖u‖²‖v‖² ≤ 0 → ⟨u,v⟩² ≤ ‖u‖²‖v‖². ∎
Triangle Inequality: ‖u + v‖ ≤ ‖u‖ + ‖v‖
Proof: ‖u+v‖² = ⟨u+v, u+v⟩ = ‖u‖² + 2⟨u,v⟩ + ‖v‖² ≤ ‖u‖² + 2‖u‖‖v‖ + ‖v‖² = (‖u‖+‖v‖)².
3. Orthogonality
Definition: u and v are orthogonal if ⟨u, v⟩ = 0. Notation: u ⟂ v.
Orthogonal set: A set S = {v₁, ..., v_k} where ⟨v_i, v_j⟩ = 0 for all i ≠ j.
Orthonormal set: An orthogonal set where additionally ‖v_i‖ = 1 for all i.
Theorem: Every orthogonal set of nonzero vectors is linearly independent. Proof: If Σ c_i v_i = 0, take inner product with v_j: ⟨Σ c_i v_i, v_j⟩ = c_j ⟨v_j, v_j⟩ = c_j ‖v_j‖² = 0. Since ‖v_j‖² ≠ 0, c_j = 0. ∎
Orthogonal decomposition in R^n: For any nonzero vector u, any v can be decomposed as:
$v = v_∥ + v_⟂ v_∥ = proj_u(v) = (⟨v, u⟩ / ‖u‖²) u (parallel component) v_⟂ = v − v_∥ (orthogonal component) $
These satisfy v_∥ ⟂ v_⟂.
Pythagorean Theorem (inner product version): If u ⟂ v, then ‖u + v‖² = ‖u‖² + ‖v‖².
4. Gram-Schmidt Orthogonalization
Given a basis {b₁, ..., b_n} of V, Gram-Schmidt produces an ORTHOGONAL basis {v₁, ..., v_n}:
Common Pitfall: Classical Gram-Schmidt is numerically unstable. Use Modified Gram-Schmidt or Householder reflections for computation. Also, the output depends on input ORDER — different orderings give different orthogonal bases.
$v₁ = b₁
v₂ = b₂ − proj_{v₁}(b₂) = b₂ − (⟨b₂, v₁⟩ / ‖v₁‖²) v₁
v₃ = b₃ − proj_{v₁}(b₃) − proj_{v₂}(b₃)
= b₃ − (⟨b₃, v₁⟩ / ‖v₁‖²) v₁ − (⟨b₃, v₂⟩ / ‖v₂‖²) v₂
...
v_k = b_k − Σ_{i=1}^{k-1} proj_{v_i}(b_k)
$
To get an orthonormal basis: normalize each v_k: u_k = v_k / ‖v_k‖.
Why it works: At each step, we subtract the projection of b_k onto all previously computed v_i, which are already mutually orthogonal. The result is orthogonal to all previous v_i.
Example (R^3): b₁ = (1,1,0), b₂ = (1,0,1), b₃ = (0,1,1).
v₁ = (1,1,0)
proj_{v₁}(b₂) = (⟨(1,0,1),(1,1,0)⟩ / 2) (1,1,0) = (1/2)(1,1,0) = (1/2, 1/2, 0) v₂ = (1,0,1) − (1/2, 1/2, 0) = (1/2, −1/2, 1)
proj_{v₁}(b₃) = (⟨(0,1,1),(1,1,0)⟩ / 2)(1,1,0) = (1/2)(1,1,0) = (1/2, 1/2, 0) proj_{v₂}(b₃) = ⟨(0,1,1),(1/2,−1/2,1)⟩ / ‖v₂‖² · v₂. ‖v₂‖² = 1/4+1/4+1 = 3/2. Inner product = 0−1/2+1 = 1/2. proj = (1/2)/(3/2) v₂ = (1/3)(1/2, −1/2, 1) = (1/6, −1/6, 1/3)
v₃ = (0,1,1) − (1/2,1/2,0) − (1/6,−1/6,1/3) = (−2/3, 2/3, 2/3)
Normalize: u₁ = (1,1,0)/√2, u₂ = (1/2,−1/2,1)/√(3/2), u₃ = (−2/3,2/3,2/3)/√(4/3).
5. Orthogonal Complement
Definition: For a subspace W ⊆ V, its orthogonal complement is:
W^⟂ = {v ∈ V : ⟨v, w⟩ = 0 for all w ∈ W}
Properties: - W^⟂ is always a subspace - V = W ⊕ W^⟂ (direct sum; every v ∈ V decomposes uniquely as w + w^⟂) - (W^⟂)^⟂ = W - dim(W^⟂) = dim(V) − dim(W)
Example: In R^3, W = span{(1, 1, 0)}. Then W^⟂ = {(x, y, z) : x + y = 0} — the plane perpendicular to the line.
Connection to fundamental subspaces of a matrix A: - N(A) = C(A^T)^⟂ - C(A) = N(A^T)^⟂
Key Terms
- Legendre polynomials
Worked Examples
Example 1: Verifying an Inner Product
Problem: Show ⟨p, q⟩ = p(0)q(0) + p(1)q(1) + p(2)q(2) defines an inner product on P₂.
Solution:
Symmetry: ⟨p, q⟩ = p(0)q(0) + p(1)q(1) + p(2)q(2) = q(0)p(0) + q(1)p(1) + q(2)p(2) = ⟨q, p⟩. ✓
Linearity: ⟨ap+q, r⟩ = (ap+q)(0)r(0) + (ap+q)(1)r(1) + (ap+q)(2)r(2) = a[p(0)r(0)+p(1)r(1)+p(2)r(2)] + [q(0)r(0)+q(1)r(1)+q(2)r(2)] = a⟨p,r⟩ + ⟨q,r⟩. ✓
Positive definiteness: ⟨p, p⟩ = p(0)² + p(1)² + p(2)² ≥ 0. If ⟨p,p⟩ = 0, then p(0) = p(1) = p(2) = 0. A degree-≤2 polynomial with three distinct roots is the zero polynomial. ✓
Yes, this is a valid inner product.
Example 2: Orthogonal Decomposition
Problem: Decompose v = (5, 2) into components parallel and perpendicular to u = (3, 1).
Solution:
‖u‖² = 3² + 1² = 10 ⟨v, u⟩ = 5·3 + 2·1 = 17
v_∥ = (⟨v,u⟩/‖u‖²) u = (17/10)(3, 1) = (51/10, 17/10) = (5.1, 1.7) v_⟂ = v − v_∥ = (5, 2) − (5.1, 1.7) = (−0.1, 0.3)
Check orthogonality: v_∥ · v_⟂ = 5.1·(−0.1) + 1.7·0.3 = −0.51 + 0.51 = 0. ✓
Example 3: Gram-Schmidt with Non-Standard Inner Product
Problem: On P₂, use inner product ⟨p, q⟩ = ∫₀¹ p(x)q(x) dx. Apply Gram-Schmidt to {1, x, x²}.
Solution:
v₁ = 1, ‖v₁‖² = ∫₀¹ 1 dx = 1.
v₂ = x − proj_{v₁}(x). ⟨x, 1⟩ = ∫₀¹ x dx = 1/2. proj = (1/2)/1 · 1 = 1/2. v₂ = x − 1/2.
v₃ = x² − proj_{v₁}(x²) − proj_{v₂}(x²). ⟨x², 1⟩ = ∫₀¹ x² dx = 1/3 → proj = 1/3. ⟨x², x−1/2⟩ = ∫₀¹ x²(x−1/2) dx = ∫₀¹ (x³ − x²/2) dx = [x⁴/4 − x³/6]₀¹ = 1/4 − 1/6 = 1/12. ‖v₂‖² = ∫₀¹ (x−1/2)² dx = [x³/3 − x²/2 + x/4]₀¹ = 1/3 − 1/2 + 1/4 = 1/12. proj = (1/12)/(1/12) (x−1/2) = x − 1/2.
v₃ = x² − 1/3 − (x − 1/2) = x² − x + 1/6.
These are the first three Legendre polynomials (up to scaling), orthogonal on [0,1].
Example 4: Orthogonal Complement Computation
Problem: In R^4, W = span{(1, 1, 0, 0), (0, 0, 1, 2)}. Find W^⟂.
Solution: v ∈ W^⟂ iff v·w = 0 for all w ∈ W, equivalently v is orthogonal to both basis vectors.
v = (x₁, x₂, x₃, x₄):
$x₁ + x₂ = 0 → x₂ = −x₁ x₃ + 2x₄ = 0 → x₃ = −2x₄ $
Free variables: x₁ = s, x₄ = t. So v = s(1, −1, 0, 0) + t(0, 0, −2, 1).
W^⟂ = span{(1, −1, 0, 0), (0, 0, −2, 1)}. dim(W^⟂) = 2.
Check: dim(W) + dim(W^⟂) = 2 + 2 = 4 = dim(R^4). ✓
Practice Problems
(Answers are below. Try each problem before checking.)
Problem 1: Show that ⟨x, y⟩ = 2x₁y₁ + 3x₂y₂ is an inner product on R^2. What is the norm of (1, 1) under this inner product?
Problem 2: Prove the Cauchy-Schwarz inequality for the dot product in R^n: |x·y| ≤ ‖x‖ ‖y‖.
Problem 3: In R^3, determine whether u = (1, 2, −1) and v = (3, −1, 1) are orthogonal under the standard dot product. Find the angle between them.
Problem 4: Apply Gram-Schmidt to {(1, 1, 1), (1, 2, 0), (0, 1, −1)} to obtain an orthonormal basis of R^3.
Problem 5: In R^4, find the orthogonal complement of W = span{(1, 2, 3, 4), (1, 0, 1, 0)}.
Problem 6: Verify the Pythagorean theorem holds for u = (3, 0, 4) and v = (−4, 0, 3) in R^3 under the standard dot product.
Problem 7: Show that if {v₁, ..., v_n} is an orthonormal basis of V, then for any v ∈ V, v = Σ ⟨v, v_i⟩ v_i and ‖v‖² = Σ ⟨v, v_i⟩².
Answers (click to expand)
**Problem 1:** Symmetry: ⟨x,y⟩ = 2x₁y₁+3x₂y₂ = 2y₁x₁+3y₂x₂ = ⟨y,x⟩. ✓ Linearity: ⟨ax+z, y⟩ = 2(ax₁+z₁)y₁ + 3(ax₂+z₂)y₂ = a(2x₁y₁+3x₂y₂) + (2z₁y₁+3z₂y₂) = a⟨x,y⟩ + ⟨z,y⟩. ✓ Positive definiteness: ⟨x,x⟩ = 2x₁² + 3x₂² ≥ 0, and = 0 only if x₁=x₂=0. ✓ Norm: ‖(1,1)‖ = √(2·1² + 3·1²) = √5. **Problem 2:** The standard proof: For any t ∈ R, ‖x − ty‖² = ‖x‖² − 2t(x·y) + t²‖y‖² ≥ 0. This quadratic in t has its minimum at t = (x·y)/‖y‖², giving ‖x‖² − (x·y)²/‖y‖² ≥ 0 → (x·y)² ≤ ‖x‖²‖y‖² → |x·y| ≤ ‖x‖ ‖y‖. **Problem 3:** u·v = 1·3 + 2·(−1) + (−1)·1 = 3 − 2 − 1 = 0. Yes, orthogonal. Angle: cos θ = 0/(‖u‖‖v‖) = 0 → θ = 90° = π/2. **Problem 4:** v₁ = (1,1,1) v₂ = (1,2,0) − proj_{v₁}(1,2,0) ⟨(1,2,0), v₁⟩ = 1+2+0 = 3, ‖v₁‖² = 3. proj = (3/3)(1,1,1) = (1,1,1). v₂ = (1,2,0) − (1,1,1) = (0,1,−1). v₃ = (0,1,−1) − proj_{v₁}(0,1,−1) − proj_{v₂}(0,1,−1) ⟨(0,1,−1), v₁⟩ = 0+1−1 = 0 → proj₁ = 0. ⟨(0,1,−1), v₂⟩ = 0+1+1 = 2, ‖v₂‖² = 0+1+1 = 2. proj₂ = (2/2)(0,1,−1) = (0,1,−1). v₃ = (0,1,−1) − 0 − (0,1,−1) = (0,0,0). Wait — the third vector is in span of first two! Check: (0,1,−1) + (1,1,1) = (1,2,0). Indeed, (0,1,−1) = (1,2,0) − (1,1,1), so the third vector is dependent. Correct: The set {(1,1,1), (1,2,0), (0,1,−1)} is NOT a basis of R^3 — (0,1,−1) depends on the first two. We need three independent vectors for R^3. Let me use (0,0,1) as the third: v₃ = (0,0,1) − proj_{v₁}(0,0,1) − proj_{v₂}(0,0,1) ⟨(0,0,1), v₁⟩ = 1, proj₁ = (1/3)(1,1,1) = (1/3, 1/3, 1/3) ⟨(0,0,1), v₂⟩ = −1, proj₂ = (−1/2)(0,1,−1) = (0, −1/2, 1/2) v₃ = (0,0,1) − (1/3,1/3,1/3) − (0,−1/2,1/2) = (−1/3, 1/6, 1/6) Normalize: u₁ = (1,1,1)/√3 u₂ = (0,1,−1)/√2 u₃ = (−2,1,1)/√6 (scaling v₃ by 6) **Problem 5:** Solve x·(1,2,3,4) = 0 and x·(1,0,1,0) = 0: x₁+2x₂+3x₃+4x₄ = 0 x₁+0+x₃+0 = 0 → x₁ = −x₃ Substitute: −x₃+2x₂+3x₃+4x₄ = 0 → 2x₃+2x₂+4x₄ = 0 → x₂ = −x₃−2x₄. Free: x₃=s, x₄=t. x = (−s, −s−2t, s, t) = s(−1, −1, 1, 0) + t(0, −2, 0, 1). W^⟂ = span{(−1, −1, 1, 0), (0, −2, 0, 1)}. **Problem 6:** Check orthogonality: 3·(−4) + 0·0 + 4·3 = −12+12 = 0. ✓ ‖u‖² = 9+0+16 = 25, ‖v‖² = 16+0+9 = 25, ‖u+v‖² = ‖(−1,0,7)‖² = 1+49 = 50. Pythagoras: 25+25 = 50. ✓ **Problem 7:** Since {v_i} is a basis, v = Σ c_i v_i. Take inner product with v_j: ⟨v, v_j⟩ = ⟨Σ c_i v_i, v_j⟩ = Σ c_i ⟨v_i, v_j⟩ = c_j⟨v_j, v_j⟩ = c_j·1 = c_j. So c_j = ⟨v, v_j⟩. ‖v‖² = ⟨v, v⟩ = ⟨Σ c_i v_i, Σ c_j v_j⟩ = Σ_i Σ_j c_i c_j ⟨v_i, v_j⟩ = Σ_i c_i² = Σ_i ⟨v, v_i⟩². This is Parseval's identity.Summary
- An inner product generalizes the dot product to abstract vector spaces via three axioms: symmetry, linearity, and positive definiteness; it endows a vector space with geometry (length, angle, distance)
- Cauchy-Schwarz (|⟨u,v⟩| ≤ ‖u‖ ‖v‖) and the Triangle Inequality (‖u+v‖ ≤ ‖u‖+‖v‖) are the foundational inequalities of inner product spaces and underpin all of analysis
- Orthogonal vectors (⟨u,v⟩ = 0) behave like perpendicular vectors in geometry; every vector decomposes uniquely into parallel + perpendicular components relative to any subspace
- Gram-Schmidt converts any basis to an orthogonal (then orthonormal) one: at each step, subtract the projection onto all previously computed orthogonal vectors
- The orthogonal complement W^⟂ together with W decomposes the space: V = W ⊕ W^⟂; this is fundamental to least-squares, projections, and the four fundamental subspaces of a matrix
Pitfalls
-
Forgetting to verify positive definiteness. The third inner product axiom (⟨v,v⟩ ≥ 0 with equality iff v = 0) is the most frequently overlooked. An operation that satisfies symmetry and linearity but fails positive definiteness — such as ⟨x,y⟩ = x₁y₁ on R² — is NOT an inner product.
-
Treating Cauchy-Schwarz as an equality. |⟨u,v⟩| ≤ ‖u‖‖v‖ is an INEQUALITY. Equality holds only when u and v are linearly dependent (collinear). Students often write |⟨u,v⟩| = ‖u‖‖v‖ by reflex, leading to incorrect angle calculations.
-
Losing orthogonality in Gram-Schmidt due to numerical instability. Classical Gram-Schmidt subtracts projections against the original a_j vectors, accumulating round-off error. In practice, use Modified Gram-Schmidt (subtract projections against the current v sequentially) or Householder reflections for better numerical stability.
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Applying Gram-Schmidt to dependent sets. Gram-Schmidt requires a linearly independent input set. If the input vectors are dependent, one of the v_k will become zero — this is a signal, not an error to ignore. The process naturally detects linear dependence.
-
Forgetting that orthogonal vectors must be nonzero. Every orthogonal set of nonzero vectors is linearly independent, but the zero vector is orthogonal to everything. A set containing the zero vector cannot be a basis, even if all nonzero members are mutually orthogonal.
Quiz
Answer each question, then read the explanation for your choice.
Q1: Which of the following is NOT required for an inner product on a real vector space?
A) ⟨u, v⟩ = ⟨v, u⟩ B) ⟨u + v, w⟩ = ⟨u, w⟩ + ⟨v, w⟩ C) ⟨u, u⟩ = 1 for all u D) ⟨u, u⟩ ≥ 0 with equality only when u = 0
Answer and Explanations
**Correct: C) ⟨u, u⟩ = 1 for all u** Positive definiteness requires ⟨u, u⟩ ≥ 0 and ⟨u, u⟩ = 0 iff u = 0. Only unit vectors have norm 1. - A) Symmetry — required. - B) Additivity in first argument — required. - C) ✓ Not required. This would mean all vectors have length 1, which is false. - D) Positive definiteness — required.Q2: The Cauchy-Schwarz inequality guarantees that:
A) ⟨u, v⟩ ≤ ‖u‖ + ‖v‖ B) |⟨u, v⟩| ≤ ‖u‖ ‖v‖ C) ‖u + v‖ ≥ ‖u‖ + ‖v‖ D) ⟨u, v⟩ = ‖u‖ ‖v‖ always
Answer and Explanations
**Correct: B) |⟨u, v⟩| ≤ ‖u‖ ‖v‖** Cauchy-Schwarz bounds the absolute value of the inner product by the product of norms. - A) This is not Cauchy-Schwarz; ⟨u,v⟩ can be larger than ‖u‖+‖v‖. - B) ✓ Correct. This is the statement of Cauchy-Schwarz. - C) This is the reverse triangle inequality, which is actually ‖u+v‖ ≥ |‖u‖ − ‖v‖|. - D) Only holds when u and v are parallel (collinear).Q3: Two vectors are orthogonal if:
A) They are linearly independent B) ⟨u, v⟩ = 0 C) ‖u‖ = ‖v‖ D) u + v = 0
Answer and Explanations
**Correct: B) ⟨u, v⟩ = 0** Orthogonality is defined directly by ⟨u, v⟩ = 0. - A) Independent vectors need not be orthogonal: (1,0) and (1,1) are independent but not orthogonal. - B) ✓ Correct. This is the definition. - C) Equal length doesn't imply orthogonality. - D) v = −u → they're antiparallel, not orthogonal (unless u = 0).Q4: In the Gram-Schmidt process, at step k we compute v_k by:
A) Adding all previous v_i to b_k B) Subtracting from b_k its projections onto v₁, ..., v_{k-1} C) Computing the cross product of b_k with all previous vectors D) Computing the eigenvalues of b_k
Answer and Explanations
**Correct: B) Subtracting from b_k its projections onto v₁, ..., v_{k-1}** v_k = b_k − Σ_{i=1}^{k-1} proj_{v_i}(b_k). This ensures v_k is orthogonal to all previously computed v_i. - A) Adding would increase non-orthogonal components. - B) ✓ Correct. Subtract projections to achieve orthogonality. - C) Cross product is only defined in R^3. - D) Eigenvalues aren't involved in Gram-Schmidt.Q5: In R^3, W^⟂ for W = span{(1, 0, 0), (0, 1, 0)} is:
A) The origin B) The z-axis C) The xy-plane D) All of R^3
Answer and Explanations
**Correct: B) The z-axis** W is the xy-plane. W^⟂ consists of vectors orthogonal to (1,0,0) and (0,1,0), i.e., dot products x=0 and y=0. So W^⟂ = {(0,0,z)} = z-axis. - A) Only {0} is orthogonal to everything. - B) ✓ Correct. Perpendicular to the xy-plane is the z-axis. - C) That's W itself. - D) (1,0,0) is not orthogonal to itself.Q6: What is the dimension of W^⟂ if W is a 3-dimensional subspace of R^7?
A) 0 B) 3 C) 4 D) 7
Answer and Explanations
**Correct: C) 4** dim(W^⟂) = dim(V) − dim(W) = 7 − 3 = 4. - A) Only if W = R^7. - B) That's dim(W). - C) ✓ Correct. 7 − 3 = 4. - D) That's dim(R^7).Q7: Under the standard dot product, ‖u + v‖² = ‖u‖² + ‖v‖² always holds when:
A) u and v are parallel B) u and v are orthogonal C) u and v are linearly dependent D) u = v
Answer and Explanations
**Correct: B) u and v are orthogonal** ‖u+v‖² = ⟨u+v, u+v⟩ = ‖u‖² + 2⟨u,v⟩ + ‖v‖² = ‖u‖² + ‖v‖² iff ⟨u,v⟩ = 0. - A) Then ⟨u,v⟩ = ±‖u‖‖v‖, giving ‖u‖²±2‖u‖‖v‖+‖v‖². - B) ✓ Correct. This is the Pythagorean theorem for inner product spaces. - C) Same as A. - D) Then ‖2u‖² = 4‖u‖² = ‖u‖²+‖u‖² only if ‖u‖=0.Q8: After Gram-Schmidt on a basis of R^4, how many orthonormal vectors do we obtain?
A) 2 B) 3 C) 4 D) Can't say without knowing the basis
Answer and Explanations
**Correct: C) 4** Gram-Schmidt on any basis of an n-dimensional space always produces n orthonormal vectors. Each original basis vector is orthogonalized, and an independent set stays independent. - A, B) Wrong — we get exactly as many as the dimension. - C) ✓ Correct. R^4 is 4-dimensional. - D) The number depends only on the dimension of V, not the specific basis.Q9: For the inner product ⟨f, g⟩ = ∫₀¹ f(x)g(x) dx on C[0,1], what is the distance between f(x) = x and g(x) = x²?
A) 1/12 B) √(1/30) C) 1/6 D) √(1/6)
Answer and Explanations
**Correct: B) √(1/30)** d(f,g)² = ⟨f−g, f−g⟩ = ∫₀¹ (x − x²)² dx = ∫₀¹ (x² − 2x³ + x⁴) dx = [x³/3 − x⁴/2 + x⁵/5]₀¹ = 1/3 − 1/2 + 1/5 = (10−15+6)/30 = 1/30. d(f,g) = √(1/30). - A) 1/12 = ∫₀¹ x(x²)dx, not squared. - B) ✓ Correct. - C) 1/6 is d² without the sqrt (wrong sign: 1/3−1/2+1/5 = 1/30, not 1/6). - D) 1/6 ≠ 1/30.Q10: Parseval's identity for an orthonormal basis {v_i} states:
A) Σ ⟨v, v_i⟩ = ‖v‖ B) Σ ⟨v, v_i⟩² = ‖v‖² C) ⟨v, v⟩ = Σ v_i D) Σ ‖v_i‖² = dim(V)
Answer and Explanations
**Correct: B) Σ ⟨v, v_i⟩² = ‖v‖²** This generalizes the Pythagorean theorem: the squared length of v equals the sum of squared lengths of its projections onto an orthonormal basis. - A) Wrong — sum of inner products is not the norm. - B) ✓ Correct. Generalized Pythagorean theorem. - C) Not meaningful — summing vectors, not scalars. - D) True (each ‖v_i‖² = 1, sum is n), but not Parseval's identity.Next Steps
Move on to 08-06 — Orthogonal Projections to learn about projecting vectors onto lines and subspaces, orthogonal matrices, least-squares solutions, and QR decomposition — powerful computational tools built on inner product geometry.