Linear Combinations of vectors
Today's Learning Goals:
- we will recognize the role of the vectors i and j as the simplest basis vectors for R2, but also
understand that they are not the only set of basis vectors
- we will understand that any pair of non-zero, non-collinear vectors form a set of basis
vectors in R2
- we will be able to express any vector in R2 as a linear combination of any set of basis
vectors in R2
- we will be able to determine if three vectors in 3-space lie in the same plane
Review of previous class
Spanning Set:
A set of vectors that can be used to generate any vector in R2 is called a
spanning set for R2. The simplest and most useful spanning set for R2 is the
set {i , j }
There are an infinite number of spanning sets of the form { u , v } provided
that u ≠ m v , that is u and v are non-collinear.
Similarly, in R3 the simplest spanning set is { i , j , k } but there are an infinite
number of spanning sets of the form { u , v , w } provided that u ≠ m v + n w ;
that is, u , v and w are not coplanar.
Ex 1
Write [5, 19] as a linear combination of [4, -1] and [1, -7] and also as a combination of [-2, 3]
and [20, 23]
𝑎 4, −1 + 𝑏 1, −7 = [5,19]
4𝑎 + 𝑏 = 5
−𝑎 − 7𝑏 = 19
4𝑎 + 𝑏 = 5
4(−𝑎 − 7𝑏 = 19)
+
−4𝑎 − 28𝑏 = 76
−27𝑏 = 81
𝑏 = −3
𝑎=2
2 4, −1 − 3 1, −7 = [5,19]
Ex 1
Write [5, 19] as a linear combination of [4, -1] and [1, -7] and also as a combination of [-2, 3]
and [20, 23]
𝑎 −2,3 + 𝑏 20,23 = [5,19]
−2𝑎 + 20𝑏 = 5
3(−2𝑎 + 20𝑏 = 5)
−6𝑎 + 60𝑏 = 15
3𝑎 + 23𝑏 = 19
−2(3𝑎 + 23𝑏 = 19)
−6𝑎 − 46𝑏 = −38
−
106𝑏 = 53
𝑏=
1
2
5
𝑎=
2
5
1
−2,3 + 20,23 = [5,19]
2
2
Ex 2
𝑖𝑓
𝑥1 𝑦1
=
𝑥2 𝑦2
2 3
=
4 6
the vectors are not a spanning set
2,3 𝑎𝑛𝑑 4,6 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑠𝑝𝑎𝑛 𝑅2
Consider what happens if we try to write a vector in R3 as a combination of two non-collinear
vectors in R3.
let u = [a,b,c] and v = [d,e,f]
if w = r u + t v then w = [ra, rb, rc] + [td, te, tf]
= [ra + td, rb + te, rc + tf]
At first glance, it would appear that w could be virtually any vector in R3, but that is not the case.
Since the basis vectors u and v must lie in the same plane within R3, this means that the sum of
scalar multiples of u and v would be the third side of a triangle formed by drawing u and v head-totail and then applying the triangle law for vector addition. Since a triangle is a plane figure, the
vector w must lie in the same plane as the vectors u and v .
Hence, the set {u, v} spans a plane in R3.
Ex 3
Determine whether a = [3, -1, 2] , b = [9, -5, -11] and c = [-1, 1, 5] are coplanar vectors.
𝑎 3, −1,2 + 𝑏 9, −5, −11 = [−1,1,5]
3𝑎 + 9𝑏 = −1
−𝑎 − 5𝑏 = 1
3𝑎 + 9𝑏 = −1
3(−𝑎 − 5𝑏 = 1)
2𝑎 − 11𝑏 = 5
2
2
1
− 11 −
3
3
−3𝑎 − 15𝑏 = 3
+
−6𝑏 = 2
1
𝑏=−
3
2
𝑎=
3
4 11 15
+
=
=5
3 3
3
Since
2
1
3, −1,2 − 9, −5, −11 = [−1,1,5]
3
3
The three vectors are co-planar and not part of a spanning set in R3
Success criteria:
- I can write a third vector in R2 as a linear combination of two basis vectors
- I can determine if three vectors in R3 are coplanar