Source: Friedberg, Insel & Spence, Linear Algebra 4th Ed., Ch. 1.5
Tags: linear dependence, linear independence, trivial representation, nontrivial representation, redundant vectors, Theorem 1.6, Theorem 1.7, Friedberg chapter 1
Difficulty: Intermediate Prerequisites: Vector spaces, subspaces, linear combinations, and span (Sections 1.1–1.4 study notes).
You have learned how to build vectors from others (linear combinations) and how to describe everything you can build (span). The natural next question is efficiency: does your collection of vectors have any redundancy? If one vector in your set can already be expressed as a combination of the others, it contributes nothing new. Linear independence is the formal test for "no redundancy." This concept is the other half of what you need to define a basis, which comes in Section 1.6.
A set S is linearly dependent if the zero vector can be written as a nontrivial linear combination of vectors in S (at least one coefficient nonzero). A set is linearly independent if the only way to get the zero vector is the trivial combination (all coefficients zero). Dependence means redundancy; independence means every vector in the set "earns its place."
Linearly dependent
A subset S of a vector space V is linearly dependent if there exist distinct vectors u₁,...,uₙ in S and scalars a₁,...,aₙ, not all zero, such that a₁u₁ + a₂u₂ + ... + aₙuₙ = 0.
In simple terms, the set is dependent if you can combine its vectors (using at least one nonzero scalar) to get the zero vector.
Linearly independent
A subset S that is not linearly dependent. Equivalently, S is linearly independent if the only representation of 0 as a linear combination of vectors in S is the trivial one (all coefficients zero).
In simple terms, no vector in the set is redundant.
Trivial representation of 0
The linear combination a₁u₁ + ... + aₙuₙ = 0 with all aᵢ = 0.
Nontrivial representation of 0
A linear combination a₁u₁ + ... + aₙuₙ = 0 where at least one aᵢ ≠ 0. Existence of such a representation is precisely what it means for the set to be linearly dependent.
To test whether vectors u₁,...,uₙ are linearly dependent or independent, set up the equation:
a₁u₁ + a₂u₂ + ... + aₙuₙ = 0
Then equate coordinates (or coefficients, for polynomials/matrices) to get a homogeneous system of linear equations. Solve it.
If the only solution is a₁ = a₂ = ... = aₙ = 0, the set is linearly independent.
If there exists a nonzero solution, the set is linearly dependent.
The empty set is linearly independent (vacuously true: no nontrivial combination exists because there are no vectors).
A set containing a single nonzero vector is linearly independent. (If {u} were dependent, au = 0 for some a ≠ 0, so u = a⁻¹(au) = 0, contradicting u ≠ 0.)
Any set containing the zero vector is linearly dependent, because 1·0 = 0 is a nontrivial representation.
Two vectors {u, v} are linearly dependent if and only if one is a scalar multiple of the other (Exercise 9).
Three or more vectors can be linearly dependent even though no single vector is a multiple of another. Example: {(1,0,1), (0,1,1), (1,1,2)} in R³, since (1,0,1) + (0,1,1) − (1,1,2) = (0,0,0).
If S = {u₁,...,uₙ} is linearly dependent with a₁u₁ + ... + aₙuₙ = 0 and aⱼ ≠ 0, then you can solve for uⱼ as a linear combination of the others:
uⱼ = −(a₁/aⱼ)u₁ − ... − (aⱼ₋₁/aⱼ)uⱼ₋₁ − (aⱼ₊₁/aⱼ)uⱼ₊₁ − ... − (aₙ/aⱼ)uₙ
Removing uⱼ from S does not change the span. This is the core practical consequence: in a dependent set, you can always trim without losing coverage.
If S₁ ⊆ S₂ ⊆ V and S₁ is linearly dependent, then S₂ is linearly dependent.
Contrapositive (Corollary): If S₂ is linearly independent, then every subset S₁ of S₂ is linearly independent.
In plain language: adding vectors to a dependent set keeps it dependent. Removing vectors from an independent set keeps it independent.
Let S be a linearly independent subset of V, and let v be a vector not in S. Then S ∪ {v} is linearly dependent if and only if v ∈ span(S).
This gives a clean criterion: the only way adding a new vector to an independent set makes it dependent is if that vector was already "reachable" from S.
Dependent set in R⁴ (Example 1):
S = {(1,3,−4,2), (2,2,−4,0), (1,−3,2,−4), (−1,0,1,0)}
Solving a₁u₁ + a₂u₂ + a₃u₃ + a₄u₄ = 0 yields the nonzero solution a₁ = 4, a₂ = −3, a₃ = 2, a₄ = 0. So S is linearly dependent.
Independent set in R⁴ (Example 3):
S = {(1,0,0,−1), (0,1,0,−1), (0,0,1,−1), (0,0,0,1)}
The system forces a₁ = 0, a₂ = 0, a₃ = 0, and then −a₁ − a₂ − a₃ + a₄ = 0 gives a₄ = 0. Only the trivial solution exists. S is independent.
Independent polynomials (Example 4):
For pₖ(x) = xᵏ + xᵏ⁺¹ + ... + xⁿ, the set {p₀, p₁, ..., pₙ} is linearly independent in Pₙ(F). Equating coefficients of xᵏ gives a triangular system whose only solution is all zeros.
A set of nonzero polynomials in P(F) where no two have the same degree is automatically linearly independent (Exercise 18). The reason: in any supposed nontrivial combination equalling zero, the highest-degree term can only come from one polynomial, forcing its coefficient to be zero, and then you cascade downward.
The linear dependence equation:
a₁u₁ + a₂u₂ + ... + aₙuₙ = 0
If the only solution is all aᵢ = 0, the vectors are independent. If a nonzero solution exists, they are dependent.
In signal processing, a set of independent signals means each carries information that the others cannot reproduce. Dependence means one signal is redundant and can be removed without losing information. In statistics, linearly independent predictor variables means no predictor is a perfect linear function of the others (the absence of perfect multicollinearity).
"If S is linearly dependent, then every vector in S is a combination of the others." This is false. In a dependent set, at least one vector with a nonzero coefficient in the dependence relation can be written as a combination of the others, but a vector whose coefficient is zero in every dependence relation cannot.
"Subsets of linearly dependent sets are linearly dependent." False. Theorem 1.6's corollary says subsets of independent sets are independent. A dependent set can contain independent subsets.
Students sometimes confuse "a₁x₁ + ... + aₙxₙ = 0 and the vectors are independent, therefore all aᵢ = 0" with "all aᵢ = 0, therefore the vectors are independent." The latter is circular. You must show the only solution is trivial.
The empty set is linearly independent, not dependent. This trips students up because it feels like it should be neither.
⚠️ Testing a given set for dependence or independence by setting up and solving the homogeneous system is a bread-and-butter exam skill.
⚠️ Theorem 1.7 links independence to span and is used directly in the proof of the basis theorems in Section 1.6.
⚠️ Know the quick facts: singleton nonzero vectors are independent; any set containing 0 is dependent; for two vectors, dependence means one is a scalar multiple of the other.
⚠️ In Fⁿ, a set of more than n vectors is always linearly dependent (this is proved via the replacement theorem in Section 1.6, but the principle is often tested here).
⚠️ Be able to state and apply Theorem 1.6 and its corollary in both directions.
True or false: if S is a linearly dependent set, then each vector in S is a linear combination of other vectors in S.
True or false: any set containing the zero vector is linearly dependent.
True or false: the empty set is linearly dependent.
True or false: subsets of linearly independent sets are linearly independent.
Fill in the blank: {u, v} is linearly dependent if and only if u or v is a ______ of the other.
Answers: 1. False. 2. True. 3. False (it is independent). 4. True (Corollary of Thm 1.6). 5. Scalar multiple.
Q: Determine whether {(1, −1, 2), (1, −2, 1), (1, 1, 4)} is linearly dependent or independent in R³.
A: Solve a(1,−1,2) + b(1,−2,1) + c(1,1,4) = (0,0,0). System: a + b + c = 0, −a − 2b + c = 0, 2a + b + 4c = 0. Adding the first two: −b + 2c = 0, so b = 2c. From the first equation, a = −b − c = −3c. Check the third: 2(−3c) + 2c + 4c = −6c + 6c = 0 ✓. Taking c = 1 gives a = −3, b = 2, c = 1. Nonzero solution exists, so the set is linearly dependent.
Q: Prove that {e₁, e₂, ..., eₙ} is linearly independent in Fⁿ.
A: Suppose a₁e₁ + ... + aₙeₙ = 0. The left side equals (a₁, a₂, ..., aₙ) = (0, 0, ..., 0). So aᵢ = 0 for all i. Only the trivial solution exists, hence the set is linearly independent.
Q: Let S be linearly independent and v ∉ S with v ∉ span(S). Is S ∪ {v} linearly independent?
A: Yes. By Theorem 1.7, S ∪ {v} is dependent if and only if v ∈ span(S). Since v ∉ span(S), the set S ∪ {v} is linearly independent.
Q: Give an example of three linearly dependent vectors in R³ where no one of them is a scalar multiple of another.
A: Take (1, 0, 0), (0, 1, 0), (1, 1, 0). None is a scalar multiple of another, but (1,0,0) + (0,1,0) − (1,1,0) = (0,0,0), so they are dependent.
Linear independence is one half of the definition of a basis (Section 1.6): a basis is a linearly independent set that spans V. The ideas here also underpin the replacement theorem (Section 1.6), which limits how large an independent set can be and leads to the concept of dimension. In Chapter 2, the linear independence of images under a linear transformation determines whether the transformation is injective (one-to-one).
linearly dependent, linearly independent, trivial representation, nontrivial representation, homogeneous system, Theorem 1.6, Theorem 1.7, redundant vector, scalar multiple test, Friedberg chapter 1 section 5, abstract linear algebra independence