The set {(1,0), (0,1), (1,1)} spans ℝ². Why is it NOT a basis for ℝ²?
AIt contains more than 2 vectors, and a basis for ℝ² must have exactly 2
BThe vector (1,1) is a linear combination of the other two, so the set is linearly dependent — any vector in ℝ² has multiple representations
CThe vectors do not all have length 1, so they cannot form a basis
DThree vectors cannot span a 2-dimensional space
The set spans ℝ² (option A captures something real — 3 vectors in a 2D space is 'too many' — but the precise reason is linear dependence). Because (1,1) = (1,0) + (0,1), it is redundant. Any vector in ℝ² can now be written in infinitely many ways as a combination of these three, destroying the uniqueness property a basis requires. Removing the redundant vector gives the standard basis {(1,0),(0,1)}, which is both independent and spanning.
Question 2 Multiple Choice
A set S is linearly independent in a vector space V but does not span V. What is the most accurate statement about S?
AS is a basis for V, since independence is the harder condition to satisfy
BS is too large — a basis cannot contain vectors that don't span
CS fails to be a basis because it is missing the spanning condition; it is a basis only for the subspace it does span
DS can never be extended to a basis for V by adding more vectors
A basis requires both independence AND spanning. S has independence but lacks spanning — it is 'too small.' It serves as a basis for the subspace it spans (which is a proper subspace of V), but not for V itself. Importantly, S can always be extended to a basis for V by adding vectors; the extension theorem guarantees this. The common error here is thinking independence alone is sufficient — both conditions are necessary.
Question 3 True / False
If a set of vectors spans a vector space, it is automatically a basis for that space.
TTrue
FFalse
Answer: False
Spanning is necessary but not sufficient. A spanning set may contain redundant vectors — vectors that are linear combinations of others — making it linearly dependent. Such a set allows each vector in the space to be represented in multiple ways, which defeats the coordinate system a basis provides. To be a basis, the set must be both spanning and linearly independent. Removing redundant vectors from a spanning set produces a basis.
Question 4 True / False
If a vector space V has one basis consisting of 4 vectors, then every basis of V consists of exactly 4 vectors.
TTrue
FFalse
Answer: True
This is the invariance of dimension theorem: all bases of a finite-dimensional vector space have the same cardinality, which defines the dimension of the space. The proof uses the fact that any independent set has size ≤ any spanning set. This invariance is what makes dimension a well-defined property of the space itself rather than of a particular choice of basis — and it flows directly from the uniqueness-of-representation property that a basis guarantees.
Question 5 Short Answer
Explain why a linearly dependent spanning set fails to be a basis, and what specifically goes wrong with representations.
Think about your answer, then reveal below.
Model answer: When a spanning set contains a redundant vector — one expressible as a combination of the others — every vector in the space can be written as a linear combination in infinitely many ways. A basis requires that every vector have exactly one representation as a combination of basis vectors. That uniqueness is what makes coordinates meaningful: in a basis, the coefficients of the linear combination are the coordinates. Dependency destroys this by introducing free choices in how to distribute weight among the redundant vectors.
The redundancy creates a 'free parameter': you can always shift some weight from the redundant vector to the others and get a different-looking representation of the same vector. This is not a minor issue — it means there is no canonical coordinate system, no way to assign unique numerical labels to vectors. The independence condition eliminates this degree of freedom and restores uniqueness.