Source: MATH 416 Exam 1, Abstract Linear Algebra, UIUC
Tags: linear independence, linear dependence, basis, dimension, span, polynomial vector space, P2, P4, vector space, subspace, trivial solution, nontrivial linear combination
Difficulty: Intermediate | Prerequisites: Row reduction (Study Notes 01), basic understanding of what a vector space is.
Linear independence and bases are the conceptual core of MATH 416. Where the first set of topics (row reduction, matrix arithmetic) gives you mechanical tools, this material asks you to reason about structure: can one vector be "built" from the others? How many vectors do you need to describe an entire space? The polynomial spaces P_n are the main non-R^n examples the course uses to test whether you understand these ideas abstractly, not just for columns of numbers.
A set of vectors is linearly independent when no nontrivial combination of them equals zero. A basis is a linearly independent set that spans the whole space, and its size equals the dimension. In P₂ (polynomials of degree at most 2), the dimension is 3, so any three linearly independent polynomials form a basis.
Linearly dependent
A set {u₁, ..., u_k} in a vector space V is linearly dependent if there exist scalars a₁, ..., a_k, not all zero, such that a₁u₁ + ... + a_ku_k = 0_V.
In simple terms, at least one of the vectors can be written as a combination of the others. There is redundancy in the set.
Linearly independent
A set {u₁, ..., u_k} is linearly independent if the only solution to a₁u₁ + ... + a_ku_k = 0_V is a₁ = a₂ = ... = a_k = 0 (the trivial solution).
Think of it as: no vector in the set is "redundant." You cannot build any one of them from the rest.
Span
The span of a set of vectors is the collection of all possible linear combinations of those vectors. Written span({u₁, ..., u_k}).
Basis
A basis for a vector space V is a set of vectors that is both linearly independent and spans V. Every vector in V can be written as a unique linear combination of the basis vectors.
Dimension
The number of vectors in any basis for V. All bases for a given vector space have the same size.
In simple terms, dimension counts the "degrees of freedom" in the space.
P_n (polynomial vector space)
The vector space of all polynomials with real coefficients of degree at most n, together with the zero polynomial. The standard basis is {1, x, x², ..., x^n}, so dim(P_n) = n + 1.
Think of it as: P₂ is a 3-dimensional space (constants, x-terms, x²-terms), just like R³ but with polynomials instead of column vectors.
Isomorphic vector spaces
Two vector spaces are isomorphic if there exists a bijective linear map between them. Finite-dimensional vector spaces over the same field are isomorphic if and only if they have the same dimension.
In simple terms, same dimension means "structurally identical," even if the objects look different (polynomials vs. column vectors, for instance).
The method mirrors what you do in R^n:
Set up the equation a₁f₁ + a₂f₂ + ... + a_kf_k = 0 (the zero polynomial).
Expand and collect terms by powers of x.
Each power of x gives one equation in the unknowns a₁, ..., a_k.
Row-reduce the resulting system. If the only solution is all zeros, the set is linearly independent.
Worked example. Test whether {1 - x, 1 + x, x² + x + 1} is linearly independent in P₂.
Set a₁(1 - x) + a₂(1 + x) + a₃(x² + x + 1) = 0.
Expanding:
Constant terms: a₁ + a₂ + a₃ = 0
Coefficient of x: -a₁ + a₂ + a₃ = 0
Coefficient of x²: a₃ = 0
The augmented matrix of this homogeneous system is:
[ 1 1 1 | 0 ]
[-1 1 1 | 0 ]
[ 0 0 1 | 0 ]
Row-reducing gives REF with 3 leading entries (3 pivots), so the only solution is a₁ = a₂ = a₃ = 0. The set is linearly independent.
Once you know a set is linearly independent, check whether it can be a basis:
Count the vectors: you have k vectors.
Find the dimension of the ambient space: dim(P₂) = 3.
If k = dim(V), a linearly independent set of k vectors automatically spans V, so it is a basis.
In the example above, 3 linearly independent vectors in a 3-dimensional space form a basis for P₂.
This shortcut is one of the most useful facts in the course: in a finite-dimensional space, you only need to verify one of {independence, spanning} if you have exactly dim(V) vectors. The other follows for free.
P₂ is a subspace of P₄, because every polynomial of degree at most 2 is also a polynomial of degree at most 4. The zero polynomial sits in both, and the set is closed under addition and scalar multiplication within P₄.
More generally, P_m is a subspace of P_n whenever m ≤ n.
Dimension of P_n
dim(P_n) = n + 1
Standard basis for P₂
{1, x, x²}
Independence criterion (dimension shortcut)
If V has dimension d, then:
Any set of more than d vectors in V is linearly dependent.
Any linearly independent set of exactly d vectors is a basis.
Any spanning set of exactly d vectors is a basis.
Polynomial approximation underpins curve fitting, interpolation, and numerical methods across engineering and data science. Checking that your chosen polynomial basis functions are independent is exactly the step that ensures a unique best-fit curve through a set of data points.
Students often confuse "linearly independent" with "orthogonal." Independence does not require any notion of angle or inner product; it only requires that no nontrivial combination gives zero.
A common error is to test independence in P₂ but forget the x² coefficient equation, effectively working in P₁ by accident. Always collect terms for every power of x up to the degree of the space.
Students sometimes claim that {1, x, x²} is the only basis for P₂. It is the standard basis, but infinitely many other bases exist (any three linearly independent polynomials of degree ≤ 2 will do).
"Dimension 3" does not mean the space is R³. It means the space is isomorphic to R³, which is a structural statement, not a literal identity.
⚠️ "Define linearly dependent/independent" is a standard 2-to-3-mark definition question. Know the exact phrasing: scalars not all zero, linear combination equals the zero vector.
⚠️ The dimension shortcut (independent set of size dim(V) is automatically a basis) is tested repeatedly. State it explicitly when you use it.
⚠️ When testing independence in P_n, you must set up the system by matching coefficients for each power of x. Row-reduce and check for nontrivial solutions.
⚠️ True/false: "Every vector space of dimension 3 is isomorphic to P₂." True, because P₂ has dimension 3, and all 3-dimensional real vector spaces are isomorphic to each other.
True or false: A set of 4 vectors in P₂ can be linearly independent.
Fill in the blank: dim(P₄) = ___.
True or false: If {v₁, v₂, v₃} is linearly independent in a vector space of dimension 3, it is a basis.
True or false: P₂ is a subspace of P₄.
Fill in the blank: To test linear independence of polynomials, set a linear combination equal to the ___ polynomial and solve for the coefficients.
Answers: 1. False (dim(P₂) = 3, so any 4 vectors must be dependent). 2. 5. 3. True. 4. True. 5. Zero.
Q: Define what it means for a set {u₁, ..., u_k} to be linearly dependent in a vector space V.
A: The set is linearly dependent if there exist scalars a₁, ..., a_k, not all zero, such that a₁u₁ + ... + a_ku_k = 0_V.
Q: Is the set {1, 1 + x, 1 + x + x²} linearly independent in P₂? Is it a basis?
A: Set a₁(1) + a₂(1 + x) + a₃(1 + x + x²) = 0. Matching coefficients: a₁ + a₂ + a₃ = 0 (constant), a₂ + a₃ = 0 (x), a₃ = 0 (x²). Back-substituting: a₃ = 0, a₂ = 0, a₁ = 0. Only the trivial solution, so the set is linearly independent. Since it has 3 independent vectors in a 3-dimensional space, it is a basis for P₂.
Q: True or false, with justification: Every vector space of dimension 3 is isomorphic to P₂.
A: True. P₂ has dimension 3. Any two finite-dimensional vector spaces over the same field with the same dimension are isomorphic. An explicit isomorphism can be built by mapping a basis of V to the standard basis {1, x, x²} of P₂.
Q: Explain why P₂ is a subspace of P₄ but P₄ is not a subspace of P₂.
A: Every polynomial of degree ≤ 2 is also of degree ≤ 4, so P₂ ⊆ P₄, and it is closed under addition and scalar multiplication within P₄. However, the polynomial x⁴ belongs to P₄ but not to P₂, so P₄ is not contained in P₂.
Linear independence connects back to row reduction: the test reduces to solving a homogeneous system. It connects forward to coordinate vectors (the unique representation of a vector in a given basis) and to the rank-nullity theorem (the number of linearly independent columns of a matrix equals its rank). Bases are also essential for defining matrix representations of linear maps, covered in Study Notes 03.
linearly dependent, linearly independent, basis, standard basis, dimension, span, spanning set, polynomial vector space, P_n, P2, P4, isomorphism, isomorphic vector spaces, vector space axioms, subspace test, trivial solution, nontrivial linear combination, degree of polynomial