Linear Combination in Vector

Definition.

A vector w is called a linear combination and vectors v₁, v₂, ..., v_r if the vector can be expressed in form

w = k₁v₁ + k₂v₂ + ... + k_rv_r

Where k₁, k₂, ..., k_r is scalar.

Example 1.

Review the vectors u = (1, 2, -1) and v = (6, 4, 2) on R³. Show that w = (9, 2, 7) is a linear combination u and v and w’= (4, -1, 8) is not a linear combination u and v.

Solution:

So that w is a linear combination of u and v, there must be a scalar of k₁ and k₂ so that

w = k₁u + k₂v

that is

(9, 2, 7) = k₁(1, 2, -1) + k₂(6, 4, 2)

or

(9, 2, 7) = (k₁ + 6k₂, 2k₁ + 4k₂, -k₁ + 2k₂)

Equation of the appropriate components gives

  k₁ + 6k₂ = 9

2k₁ + 4k₂ = 2

- k₁ + 2k₂ = 7

By solving this system it will produce k₁ = -3, k₂ = 2 so that

w = -3u + 2v

likewise, so that w’ is a linear combination of u and v, there must be a scalar of k₁ and k₂ so that w’ = k₁u + k₂v, i.e.

(4, -1, 8) = k₁ (1, 2, -1) + k₂ ( 6, 4, 2)

or

(4, -1, 8) = (k₁ + 6k₂, 2k₁ + 4k₂, -k₁ + 2k₂)

Equation of the appropriate components gives

  k₁ + 6k₂ = 4

2k₁ + 4k₂ = -1

- k₁ + 2k₂ = 8

This system of equations is inconsistent, so there are no such scalars. As a consequence, w' is not a linear combination of u and v.

Definition.

If v₁, v₂, ..., v_r are vectors in vector space V and if each vector in V can be expressed as a linear combination v₁, v₂, ..., v_r we say that these vectors span V .

Example 2.

The vectors i = (1, 0, 0), j = (0, 1, 0) and k = (0, 0, 1) span R³ because each vector (a, b, c) on R³ can be written as

(a, b, c) = ai + bj + ck

Which is a linear combination i, j, k.

Example 3.

Determine whether u = (1, 1, 2), v = (1, 0, 1) and w = (2, 1, 3) span R3?

Solution:

We must determine whether any vector b = (b₁, b₂, b₃) in R³ can be expressed as a linear combination.

b = k₁u + k₂v  + k₃w

From vectors u, v, and w. By expressing this equation in the components it will give

(b₁, b₂, b₃) = k₁(1, 1, 2) + k₂(1, 0, 1)  + k₃(2, 1, 3)

or

(b₁, b₂, b₃) = k₁ + k₂ + 2k₃, k₁ + k₃, 2k₁ + k₂ + 3k₃

By equating the corresponding components giving

    k₁ + k₂ + 2k₃ = b₁

  k₁ +          k₃ = b₂

2k₁ + k₂ + 3k₃ = b₃

This system will be consistent for all values ​​b₁, b₂, and b₃ if and only if the matrix coefficients,

Can be reversed. But det (A) = 0, so that A cannot be reversed, and as a consequence, u, v and w do not stretch R³.

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