Linear Combinations and Systems of Linear Equations

LINEAR COMBINATIONS AND SYSTEMS OF LINEAR EQUATIONS

Definition : Let V be a vector space and Let S a non-empty sub-set of V.

A vector v ε V is called a linear combination of vectors of S if there exist a finite number of vectors

Note:

(1) If v = a₁ u₁+ a₂ u₂+ ... + a_n u_n then v is a linear combination of u₁, u₂, ..., u_n and a₁, a₂, ..., a_n are called co-efficients of the linear combination.

(2) In any vector space V, 0 v = 0 each v ε V

Thus the zero vector is a linear combination of any non-empty sub-set of V.

(a) Solve Linear equations :

Problem 1.

Solution :

Rearrange the equations

['.' The first non-zero coefficient of (1) must be one]

This can also be expressed as

Problem 2.

For each list of polynomials in P₃(R), determine whether the first polynomial can be expressed as a linear combination of the other two.

[A.U N/D 2019, R-17]

Solution :

Equating the coefficients of

Solving (3) & (4), we get, a₁ = 3, a₂ = -2

a₁ = 3, a₂ = -2 satisfying equations (1) & (2)

Hence, the result.

Equating the co-efficients of

(b) The span of S

Definition: Let S be a non-empty sub-set of a vector space V. The span of S, is the set consisting of all linear combinations of the vectors in S. For convenience, we define span (φ) = {0}

Definition: A sub-set S of a vector space V generates (or spans) V if span (S) = V.

Note: In this case, we also say that the vectors of S generate (or span) V.

Theorem 1. The span of any sub-set S of a vector space V is a sub-space of V. Moreover, any sub-space of V that contains S must also contain the span of S.

(or) The linear span L(S) of any sub-set of a vector space V(F) is a sub-space of V(F). Moreover, 

Solution :

(i) If S = φ because span (φ) = {0} which is a sub-space that is connected in any sub-space of V.

Then there exist vectors

and, for any scalar c,

are clearly linear combinations of the vectors in S.

So x + y and cx are in span (S).

Thus span (S) is a sub-space of V.

(ii) Let W denote any sub-space of V that contains S.

Problem 1.

Show that a sub-set W of a vector space V is a sub-space of V if and only if span (W) = W.

Solution :

Given: W is a sub-space of V.

['.' W is a sub-space of V, hence it is closed for addition and scalar multiplication]

Problem 2.

Solution :

Problem 3.

In each part, determine whether the given vector is in the span of S.

Solution :

The given vector is in the span of S.

Equating the coefficients of

.'. the given vector is not the span of S.

.'. The given vector is in the span of S.

.'. The given vector is not in the span of S.

EXERCISE 4.3

1. Solve the following systems of linear equations by the method introduced in this section.

2. For each of the following lists of vectors in R³, determine whether the first vector can be expressed as a linear combination of the other two.

(a) (1, 2, -3), (-3, 2, 1), (2, -1, -1)

(b) (2, -1, 0), (1, 2, -3), (1, -3, 2)

(c) (5, 1, −5), (1, -2, -3), (-2, 3, -4)

(d) (-2, 2, 2), (1, 2, -1), (-3, -3, 3)

3. For each list of polynomials in P₃(R), determine whether the first polynomial can be expressed as a linear combination of the other two.

4. In each part, determine whether the given vector is in the span S.

(a) (−1, 2, 1), S = {(1, 0, 2), (−1, 1, 1)}

(b)(−1, 1, 1, 2), S = {(1, 0, 1, −1), (0, 1, 1, 1)}

(c) (2, −1, 1, −3), S = {(1, 0, 1, −1), (0, 1, 1, 1)}

(d) 

5. Show that the vectors x₁ = (1, 2, 3), x₂ = (0, 1, 2), x₃ = (0, 0, 1) generate V₃ (R)

6. Write the polynomial V = t² + 4t - 3 over R as a linear combination of polynomials e₁ = t² - 2t + 5, e₂ = 2t² - 3t, e₃ = t + 3

7. Write the vector x =  in vector space of 2 x 2 matrices as a linear combination of 

8. Write the vector  in vector space of 2 x 2 matrices as a linear combination of

9. If S and T are any sub-sets of a vector space V(F), then

10. Show that (1, 1, 1), (0, 1, 1) and (0, 1, -1) generate R³.

11. Write the matrix  as a linear combination of

12. Solve :