You have three vectors in R²: v₁=(1,0), v₂=(0,1), v₃=(2,3). Which is true?
A{v₁, v₂, v₃} is a valid basis for R² because all three are needed to reach every point
B{v₁, v₂, v₃} spans R² but is not a basis, because v₃ is a linear combination of v₁ and v₂
C{v₁, v₂} is not a basis because it doesn't include the large vector v₃=(2,3)
DAny two of these three vectors form a linearly independent set and thus a basis for R²
Since v₃ = 2v₁ + 3v₂, v₃ is linearly dependent on v₁ and v₂. The set {v₁,v₂,v₃} spans R² correctly, but fails the independence requirement — v₃ is redundant. A basis must be both spanning AND independent. Option A is the classic misconception: 'we need v₃ to reach (2,3)' confuses span (you can reach (2,3) using v₁ and v₂ alone) with independence. Option D is also false: (1,0) and (2,0) are two vectors in R² that are NOT independent — being two vectors doesn't guarantee independence.
Question 2 Multiple Choice
Five vectors span a subspace V, but some are linearly dependent. After removing the redundant ones, 3 linearly independent vectors remain that still span V. What can you conclude?
AV has dimension 3, and any basis for V must contain exactly 3 vectors
BV has dimension between 3 and 5, since the original set needed 5 vectors
CThe dimension of V depends on which 3 vectors were kept, not just their count
DV is 5-dimensional because 5 vectors were originally required to describe it
Once you have 3 linearly independent vectors that span V, you have a basis. The fundamental theorem of dimension states that every basis for the same subspace has the same cardinality. So V has dimension exactly 3 — any other basis will also have 3 vectors. Option B is wrong because the original 5-vector spanning set was overcomplete (had redundancy); it's the size of a *minimal* spanning set (a basis) that defines dimension, not the size of any spanning set. Option D conflates the size of an over-complete spanning set with dimension.
Question 3 True / False
Two vectors are linearly dependent if and mainly if at least one of them is the zero vector.
TTrue
FFalse
Answer: False
Linear dependence means one vector can be written as a linear combination of the others — equivalently, there exist scalars c₁, c₂, not all zero, such that c₁v₁ + c₂v₂ = 0. Two nonzero vectors like (1,2) and (2,4) are linearly dependent (the second is 2× the first) without either being zero. Geometrically in R², two nonzero vectors are dependent if and only if they are parallel (collinear through the origin). The zero vector does always create dependence, but dependence does not require a zero vector.
Question 4 True / False
If you add a vector w to a linearly independent set {v₁, ..., vₖ}, the result is still linearly independent only if w is not in the span of {v₁, ..., vₖ}.
TTrue
FFalse
Answer: True
If w is already a linear combination of v₁,...,vₖ, then adding w creates linear dependence: you can write w − (c₁v₁ + ... + cₖvₖ) = 0 with non-trivial coefficients. If w is outside the span, no such combination exists, and independence is preserved. This is the key link between span and independence: a vector extends the span if and only if it extends the independent set — the two concepts are dual in exactly this sense.
Question 5 Short Answer
Why must every basis for the same subspace have the same number of vectors? What would go wrong if two bases had different cardinalities?
Think about your answer, then reveal below.
Model answer: If V had a basis B₁ with k vectors and a basis B₂ with m > k vectors, then B₂ is a set of m linearly independent vectors all living in a space spanned by k vectors. But any independent set in a k-dimensional space has at most k vectors — a contradiction. The Exchange Lemma formalizes this: any independent set has at most as many vectors as any spanning set, forcing all minimal spanning sets (bases) to share the same size.
The equal-cardinality result is why 'dimension' is a well-defined property of the subspace itself, not of any particular basis. Without it, dimension would be ambiguous — the same plane through the origin might have 'dimension 2' with one basis and 'dimension 3' with another, making the concept useless. The result says dimension measures the degrees of freedom in a subspace — how many truly independent directions it contains — a fact about the geometry, not the coordinates.