You have a set of three vectors {v₁, v₂, v₃} in ℝ⁵ that are mutually orthogonal (each pair has zero dot product) and none is the zero vector. What can you conclude about this set?
ANothing — orthogonality has no implications for linear independence
BThe set is linearly independent, and since ℝ⁵ has dimension 5, the set spans ℝ⁵
CThe set is linearly independent, but three orthogonal vectors in ℝ⁵ do not span ℝ⁵
DThe set may or may not be linearly independent — orthogonality and independence are unrelated
Orthogonality implies linear independence (for nonzero vectors): if Σcᵢvᵢ = 0, taking the inner product with any vⱼ kills all cross-terms and forces cⱼ = 0. So the set is definitely linearly independent. However, three vectors in ℝ⁵ cannot span ℝ⁵, which requires at least 5 linearly independent vectors. They span a 3-dimensional subspace of ℝ⁵. Options A and D confuse the implication direction — orthogonality implies independence, but we cannot go the other way.
Question 2 Multiple Choice
In ℝ⁴, a subspace W has dimension 3. What is the dimension of its orthogonal complement W⊥, and what does every vector in ℝ⁴ have in common with this decomposition?
Adim(W⊥) = 3, and some vectors in ℝ⁴ cannot be written as a sum w + w⊥
Bdim(W⊥) = 1, and every vector in ℝ⁴ can be written uniquely as w + w⊥ with w ∈ W and w⊥ ∈ W⊥
Cdim(W⊥) = 1, but the decomposition x = w + w⊥ is not necessarily unique
Ddim(W⊥) = 4 − 3 = 1, and this decomposition fails when x already belongs to W
The dimension formula always holds: dim(W) + dim(W⊥) = n. With dim(W) = 3 and n = 4, we get dim(W⊥) = 1 — W⊥ is a line through the origin perpendicular to the 3D subspace. The direct sum decomposition V = W ⊕ W⊥ means every vector x decomposes uniquely as x = w + w⊥. The decomposition holds even when x ∈ W (in that case w = x and w⊥ = 0) — there are no exceptions. Uniqueness comes from the fact that W ∩ W⊥ = {0}.
Question 3 True / False
Any set of mutually orthogonal nonzero vectors is automatically linearly independent.
TTrue
FFalse
Answer: True
This is a theorem with a clean proof. Suppose Σcᵢvᵢ = 0 where the vᵢ are mutually orthogonal and nonzero. Take the inner product of both sides with any particular vⱼ: ⟨Σcᵢvᵢ, vⱼ⟩ = Σcᵢ⟨vᵢ, vⱼ⟩ = cⱼ‖vⱼ‖² = 0. Since vⱼ ≠ 0, we have ‖vⱼ‖² > 0, so cⱼ = 0. This holds for every j, proving the set is linearly independent. The zero vector must be excluded because ⟨0, v⟩ = 0 for all v, so a set containing the zero vector is always dependent.
Question 4 True / False
Most linearly independent set of vectors is orthogonal — that is, independence and orthogonality are equivalent properties.
TTrue
FFalse
Answer: False
Orthogonality implies independence, but independence does NOT imply orthogonality. For a simple counterexample in ℝ²: {(1, 0), (1, 1)} is linearly independent (neither vector is a scalar multiple of the other), but their dot product is 1·1 + 0·1 = 1 ≠ 0, so they are not orthogonal. The implication runs only one way: orthogonal ⟹ independent. This is why the Gram-Schmidt process is needed — it converts an independent set into an orthogonal one, a nontrivial transformation.
Question 5 Short Answer
What is the direct sum decomposition V = W ⊕ W⊥, and why is it useful beyond just being a bookkeeping fact about dimensions?
Think about your answer, then reveal below.
Model answer: The decomposition says every vector x in V can be written uniquely as x = w + w⊥ where w ∈ W and w⊥ ∈ W⊥. It is useful because w is the orthogonal projection of x onto W — the closest point in W to x — and the residual w⊥ is the error, orthogonal to W by construction. This decomposition is the geometric foundation of orthogonal projection, least-squares problems, and Fourier series: in all these applications, we find the best approximation of x from within a subspace W, and the residual's orthogonality to W is what makes it the 'best' in the sense of minimizing distance.
The uniqueness is key: there is exactly one way to split x into a component inside W and a component outside W (perpendicular to W). This makes the projection well-defined. Without orthogonality, there would be many ways to decompose x as a sum of two components from two subspaces — orthogonality of the complement is what pins down the unique 'closest point' interpretation.