Linear Dependence - Vector Algebra, CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

Linear Dependence and Independence

Linear Dependence 

• Let V be a vector space over F. A subset S of V is said to be dependent if there exists distinct vectors a₁, a₂,......,a_n in S and scalar c₁, c₂,.....,c_n in F, not all of which are 0, such that c₁a₁ + c₂a₂ + ------- + c_na_n = 0
• If the set S contains finitely many vectors x₁, x₂,.........,x_n then sometimes we can say that the vectors x₁, x₂,.........,x_n are dependent.
• In other words if x₁, x₂,.........,x_n are linearly dependent. Then ∃c₁, c₂,.........,c_n not all zero such that c₁x₁ + c₂x₂ + ------- + c_na_n = 0 
• Without loss of generality we can assume c₁ ≠ 0 then x₁ = (-c₁^-1 c₂)x₂ + (-c₁^-1 c₃)x₃ + .... + (-c₁^-1 c_n)x_n, 
  hence we can say x₁ is linear combinations of the others.
  

Clear all your Mathematics doubts with EduRev Join for Free

Join for Free

Linear Independent 

• Any set containing the vectors x₁, x₂, ....... , x_n defined over a field F is said to be linearly independent if it is not linear dependent i.e. if c₁x₁ + c₂x₂ + ......... + c_nx_n = 0 Implies c_i = 0 for all i.
• Example: In ℝ³ the vectors x₁ = (3,0,-3), x₂ = (-1,1,2), x₃ = (4,2,-2), x₄ = (2,1,1) are linearly dependent since 2 x₁ + x₂ - x₃ + 0 x₄ = 0

Did You Know
1. Vector Space Spanned by Subset: V(F), be a Vector Space and S ⊆ V then V is called spanned by S if L(S) = V.
2. Finitely Generated Vector Space (FGVS) and Non-Finitely Generated Vector Space (NFGVS): A Vector Space V(F), is called finitely generated vector space if there exists finite S ⊆ V such that L(S) = V otherwise non-finitely Generated Vector Space (NFGVS).

Example: V = ℝ³ is vector space over ℝ is finitely generated vector space.
As S = {(1,0,0), (0,1,0), (0,0,1)} ⊆ ℝ³ spanned V = ℝ³.

1. ℝ(Q) is non-finitely generated vector space (NFGVS)
2. V = ℝ[x] is polynomial space over field ℝ is non-finitely generated vector space (NFGVS)
3. V = {f : ℝ → ℝ | f is function } then V is vector space over field ℝ.
   Under addition of functions and scalar multiplication to the functions.
   V(ℝ) is non-finitely generated vector space (NFGVS)

Basis and Dimension 

• Let V be a Vector Space, a basis for V is a linearly independent set of vectors in V which spans the space V . 
• The space V is finite-dimensional if it has a finite basis i.e. the basis has finite number of elements. i.e., let V . ⊂ V (F), then S is a basis if

1. S is linearly independent
2. L(S) = V (F)

Dimension of Vector Space: Dimension of a finite-dimensional vector space is the number of vectors in any basis set of the vector space and it is denoted by dim(V)

Example:

1. ℂ(ℂ) is a vector space of dimension one as S = {1} is Linearly independent set which spans C(C)
2. ℂ(ℝ) is a vector space of dimension two as S = {1, i} is linearly independent set which spans C(ℝ)
3. V = ℝ[x] is polynomial space over field ℝ is infinite dimension vector space.
4. V = {f : ℝ → ℝ | f is function } then V is vector space over field ℝ . Under addition of functions and scalar multiplication to the functions. V(ℝ) is infinite dimension vector space.

Ordered Basis: 

• If V is a finite-dimensional vector space, then an ordered basis for V is a finite sequence of vectors that is linearly independent and spans V.
• i.e. let S = {x₁, x₂, ........, x_n } be a basis of V (F) then it is called an ordered basis if the position of vectors in S is fixed i.e. the set S is ordered, hence if S = {x₁, x₂, x₃} is an ordered basis then the basis {x₂, x₁, x₃} is considered as different ordered basis.

Download the notes Linear Dependence - Vector Algebra, CSIR-NET Mathematical Sciences

Download as PDF

Download as PDF

Properties on a Linearly Independent and Dependent Basis

1. Every subset of a linearly independent set is linearly independent. That is, let V(F) be a vector space, and let S₁ ⊆ S₂ ⊆ V. If S₂ is linearly independent, then S₁ is linearly independent
2. The empty set is linearly independent; hence every linearly dependent set must be nonempty.
3. A set consisting of a single nonzero vector is linearly independent. For if {u} is linearly dependent, then au = 0 for some nonzero scalar a
4. A set is linearly independent if and only if the only representations of 0 as linear combinations of its vectors are trivial representations
5. A set linearly dependent if and only if the only representations of 0 as linear combinations of its vectors are non-trivial representations
6. Any set containing 0 is linearly dependent.
7. Let V be a vector space, and let S₁ ⊆ S₂ ⊆ V. If S₁ is linearly dependent, then S₂ is linearly dependent
8. Let S be a linearly independent subset of a vector space and let v be a vector in V that is not in S. Then S∪{v} is linearly dependent if and only if v ∈ span(S)
9. Let u and v be distinct vectors in vector space V, then {u,v} is linearly dependent if and only if u or v is a multiple of the other
10. Let V be a vector space over a field of characteristic not equal to two.
    a. Let u and v be distinct vectors in V then {u,v} is linearly independent if and only if {u+v,u-v} is linearly independent.
    b. Let u,v and w be distinct vectors in V. Then {u,v,w} is linearly independent if and only if {u+v, u+w, v+w} is linearly independent
11. Let S={u₁,u₂,...,u_n} be a finite set of vectors. Then S is linearly dependent if and only if u_i=0 or u_i ∈ span({u₁,u₂,...,u_k}) for some k(1≤k<n)
12. A set S of vectors is linearly independent if and only if each finite subset of S is linearly independent
13. If S be a set of nonzero polynomials in P(F) such that the same degree, then S is linearly independent
14. If {A₁, A₂, ..., Aₖ} is a linearly independent subset of Mₙₓₙ(F), then {A₁ᵀ, A₂ᵀ, ..., Aₖᵀ} is also linearly independent
15. Let f, g ∈ F(R, R) be the functions defined by f(t) = eʳᵗ and g(t) = eˢᵗ where r ≠ s Then f and g are linearly independent in F(R, R)
16. Let V be a vector space and β = {u₁, u₂, ..., uₙ} be a subset of V. Then β is a basis for V. If and only if each v ∈ V can be uniquely expressed as a linear combination of vectors of β, that is, can be expressed in the form v = a₁u₁ + a₂u₂ + ... + aₙuₙ
17. If a vector space V is generated by a finite set S, then some subset of S is a basis for V, i.e. V has a finite basis
18. A vector space V over a field K is said to be finite dimensional if there are a finite number of elements x₁, ..., xₙ in V such that V = ⟨x₁, ..., xₙ⟩. If no such finite number of elements exist in V, V is called infinite dimensional.
19. If vector space of continuously differentiable functions is taken then any subset of this vector space is linearly independent if Wronskian of the functions is non-zero.
20. Let V be a vector space having a finite basis. Then every basis for V contains the same number of vectors
21. Let V be a vector space with dimension n
    a. Any finite generating set for V contains at least n vectors, and generating set V that contains exactly n vectors is a basis for V.
    b. Any linearly independent subset of V that contains exactly n vectors is a basis for V.
    c. Every linearly independent subset of V can be extended to a basis for V.
22. Let W be a subspace of finite-dimensional vector space V. Then W is finite-dimensional and dim(W) ≤ dim(V). Moreover, if dim(W) = dim(V), then V = W
23. If W is a subspace of finite-dimensional vector space V, then any basis for W can be extended to basis for V.
24. If W₁ and W₂ are finite-dimensional subspaces of a vector space V, then
    a. The subspace W₁ + W₂ is finite-dimensional, and
    b. W₁ + W₂ is the smallest subspace of V containing W₁ ∪ W₂ i.e. L(W₁ ∪ W₂) = W₁ + W₂
    c. dim(W₁ + W₂) = dim(W₁) + dim(W₂) - dim(W₁ ∩ W₂)
25. Let W₁ and W₂ be subspaces of vector space V having dimensions m and n, respectively, where m ≥ n Then
    a. dim(W₁ ∩ W₂) ≤ n
    b. dim(W₁ + W₂) ≤ m + n
26. Let W₁ and W₂ be subspaces of a vector space V such that V = W₁ ⊕ W₂
    a. If β₁ and β₂ are bases for W₁ and W₂, respectively, then β₁ ∩ β₂ = φ and β₁ ∪ β₂ is a basis for V
    b. Let β₁ and β₂ be disjoint bases for subspaces W₁ and W₂, respectively, of a vector space V if β₁ ∪ β₂ is a basis for V, then V = W₁ ⊕ W₂
27. Let W be a subspace of a (not necessarily finite-dimensional) vector space V then any basis for W is a subset of basis for V
28. Let V be a finite dimensional vector space over field K, and let X and Y be finite subsets of V If Y is linearly independent and V = ⟨X⟩, then |Y| ≤ |X|
29. Let V be finite dimensional vector space over field K. If X is a linearly independent subset of V is finite.
30. Let V be finite dimensional vector space over K. Then the following statements are equivalent for a subset β of V :
    a. β is a basis
    b. β is a minimal generating set, that is, no subset of β can generate V
    c. β is a maximal linearly independent set
31. Let V be a vector space of dimension n. Then every linearly independent subset of V with n elements is a basis of V
32.  but never contained in V.
33. If xᵢ's are linearly independent in V then it is not necessary that (W + xᵢ)'s are linearly independent
34. If xᵢ's are linearly dependent in V then (W + xᵢ)'s are also linearly dependent in V/W.
35. For every FDVS V there exists subspaces W₁ and W₂ s.t. V = W₁ ⊕ W₂.
36. dim(W₁ ∩ W₂) ≤ min(dim(W₁), dim(W₂))
37. dim(W₁ ∩ W₂) ≤ max(0, dim(W₁) + dim(W₂) - dim(V))
38. max{dim(W₁), dim(W₂)} ≤ dim(W₁ + W₂) ≤ min{dim(W₁) + dim(W₂), dim(V)}
39. If P_n(x, y) denotes vector space of polynomial with degree less than or equal to n in indeterminate x and y then following is true
    
     and can be generalized.

Take a Practice Test Test yourself on topics from Mathematics exam

Practice Now

Practice Now

Solved Examples

Example 1: Are the vectors v₁ = (2, 5, 3), v₂ = (1, 1, 1), and v₃ = (4, −2, 0) linearly independent?

Ans: If none of these vectors can be expressed as a linear combination of the other two, then the vectors are independent; otherwise, they are dependent. 
If, for example, v₃ were a linear combination of v₁ and v₂, then there would exist scalars k₁ and k₂ such that k v + k₂ v_2b = v₃. This equation reads

which is equivalent to 

However, this is an inconsistent system.
For instance, subtracting the first equation from the third yields k₁ = −4, and substituting this value into either the first or third equation gives k₂ = 12. 
However, (k₁, k₂) = (−4, 12) does not satisfy the second equation. 
The conclusion is that v₃ is not a linear combination of v₁ and v₂. A similar argument would show that v₁ is not a linear combination of v₂ and v₃ and that v₂ is nota linear combination of v₁ and v₃. 
Thus, these three vectors are indeed linearly independent.

An alternative—but entirely equivalent and often simpler—definition of linear independence reads as follows.
A collection of vectors v₁, v₂, …, v _r from Rⁿ is linearly independent if the only scalars that satisfy
are k₁ = k₂ = ....= k_r = 0. 
This is called the trivial linear combination. If, on the other hand, there exists a nontrivial linear combination that gives the zero vector, then the vectors are dependent.

Example 2: Use this second definition to show that the vectors from Example 1— v₁ = (2, 5, 3), v₂ = (1, 1, 1), and v₃ = (4, −2, 0)—are linearly independent

Ans: These vectors are linearly independent if the only scalars that satisfy

are k₁ = k₂ = k₃ = 0. But (*) is equivalent to the homogeneous system

Row‐reducing the coefficient matrix yields 

This echelon form of the matrix makes it easy to see that k₃ = 0, from which follow k₂ = 0 and k₁ = 0. Thus, equation (**)—and therefore (*)—is satisfied only by k₁ = k₂ = k₃ = 0, which proves that the given vectors are linearly independent.

Example 3: Are the vectors v₁ = (4, 1, −2), v₂ = (−3, 0, 1), and v₃ (1, −2, 1) linearly independent?

Ans: The equation k₁ v₁ + k₂ v₂ + k₃ v₃ = 0 is equivalent to the homogeneous system

Row‐reduction of the coefficient matrix produces a row of zeros:

Since the general solution will contain a free variable, the homogeneous system (*) has nontrivial solutions. This shows that there exists a nontrivial linear combination of the vectors v₁, v₂, and v₃ that give the zero vector: v₁, v₂, and v₃ are dependent.

Example 4: There is exactly one value of c such that the vectors 

are linearly dependent. Find this value of c and determine a nontrivial linear combination of these vectors that equals the zero vector.

Ans: As before, consider the homogeneous system

and perform the following elementary row operations on the coefficient matrix: 

In order to obtain nontrivial solutions, there must be at least one row of zeros in this echelon form of the matrix. If c is 0, this condition is satisfied. Since c = 0, the vector v₄ equals (1, 1, 1, 0). Now, to find a nontrivial linear combination of the vectors v₁, v₂, v₃, and v₄ that gives the zero vector, a particular nontrivial solution to the matrix equation

is needed. From the row operations performed above, this equation is equivalent to

The last row implies that k₄ can be taken as a free variable; let k₄ = t. The third row then says 

The second row implies

and, finally, the first row gives

Thus, the general solution of the homogeneous system (**)—and (*)—is

for any t in R. Choosing t = 1 , for example, gives k₁, k₂, k₃, k₄ so

is a linear combination of the vectors v₁, v₂, v₃, and v₄ that equals the zero vector. To verify that

simply substitute and simplify: 

Infinitely many other nontrivial linear combinations of v₁, v₂, v₃, and v₄ that equal the zero vector can be found by simply choosing any other nonzero value of t in (***) and substituting the resulting values of k₁, k₂, k₃, and k₄ in the expression k₁ v₁ + k₂ v₂ + k₃ v₃ + k₄ v₄.

If a collection of vectors from Rⁿ contains more than n vectors, the question of its linear independence is easily answered. If C = { v₁, v₂, …, v_m } is a collection of vectors from Rⁿ and m > n, then C must be linearly dependent. To see why this is so, note that the equation

is equivalent to the matrix equation

Since each vector v _j contains n components, this matrix equation describes a system with m unknowns and n equations. Any homogeneous system with more unknowns than equations has nontrivial solutions, a result which applies here since m > n. Because equation (*) has nontrivial solutons, the vectors in C cannot be independent.