Linear Combination in Vector

Definition.
A vector w is called a linear combination and vectors v1, v2, ..., vr if the vector can be expressed in form
w = k1v1 + k2v2 + ... + krvr
Where k1, k2, ..., kr 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, ..., vr are vectors in vector space V and if each vector in V can be expressed as a linear combination v1, v2, ..., vr 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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