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Linear Dependence and Span
P. Danziger
1 Linear Combination
Definition 1 Given a set of vectors {v 1 , v 2 ,... , vk} in a vector space V , any vector of the form
v = a 1 v 1 + a 2 v 2 +... + akvk
for some scalars a 1 , a 2 ,... , ak, is called a linear combination of v 1 , v 2 ,... , vk.
Example 2
- Let v 1 = (1, 2 , 3), v 2 = (1, 0 , 2).
(a) Express u = (− 1 , 2 , −1) as a linear combination of v 1 and v 2 , We must find scalars a 1 and a 2 such that u = a 1 v 1 + a 2 v 2. Thus a 1 + a 2 = − 1 2 a 1 + 0 a 2 = 2 3 a 1 + 2 a 2 = − 1 This is 3 equations in the 2 unknowns a 1 , a 2. Solving for a 1 , a 2 :
R^2 →^ R^2 −^2 R^1
R 3 → R 3 − 3 R 1
So a 2 = −2 and a 1 = 1. Note that the components of v 1 are the coefficients of a 1 and the components of v 2 are
the coefficients of a 2 , so the initial coefficient matrix looks like
v 1 v 2 u
(b) Express u = (− 1 , 2 , 0) as a linear combination of v 1 and v 2. We proceed as above, augmenting with the new vector.
R^2 →^ R^2 −^2 R^1
R 3 → R 3 − 3 R 1
This system has no solution, so u cannot be expressed as a linear combination of v 1 and v 2. i.e. u does not lie in the plane generated by v 1 and v 2.
- Let v 1 = (1, 2), v 2 = (0, 1), v 3 = (1, 1). Express (1, 0) as a linear combination of v 1 , v 2 and v 3. ( 1 0 1 1 2 1 1 0
R 2 → R 2 − 2 R 1
Let a 3 = t, a 2 = −1 + t, a 3 = 1 − t.
This system has multiple solutions. In this case there are multiple possibilities for the ai. Note that v 3 = v 1 − v 2 , which means that a 3 v 3 can be replaced by a 3 (v 1 − v 2 ), so v 3 is redundant.
2 Span
Definition 3 Given a set of vectors {v 1 , v 2 ,... , vk} in a vector space V , the set of all vectors which are a linear combination of v 1 , v 2 ,... , vk is called the span of {v 1 , v 2 ,... , vk}. i.e.
span{v 1 , v 2 ,... , vk} = {v ∈ V | v = a 1 v 1 + a 2 v 2 +... + akvk}
Definition 4 Given a set of vectors S = {v 1 , v 2 ,... , vk} in a vector space V , S is said to span V if span(S) = V
In the first case the word span is being used as a noun, span{v 1 , v 2 ,... , vk} is an object. In the second case the word span is being used as a verb, we ask whether {v 1 , v 2 ,... , vk} san the space V.
Example 5
- Find span{v 1 , v 2 }, where v 1 = (1, 2 , 3) and v 2 = (1, 0 , 2). span{v 1 , v 2 } is the set of all vectors (x, y, z) ∈ R^3 such that (x, y, z) = a 1 (1, 2 , 3) + a 2 (1, 0 , 2). We wish to know for what values of (x, y, z) does this system of equations have solutions for a 1 and a 2. (^)
1 1 x 2 0 y 3 2 z
R^2 →^ R^2 −^2 R^1
R 3 → R 3 − 3 R 1
1 1 x 0 − 2 y − 2 x 0 − 1 z − 3 x
R 2 → −^1
2 R^2
1 1 x 0 1 x − 12 y 0 − 1 z − 3 x
R 3 → R 3 + R 2
1 1 x 0 1 x − 12 y 0 0 z − 2 x − 12 y
So solutions when 4x + y − 2 z = 0. Thus span{v 1 , v 2 } is the plane 4x + y − 2 z = 0.
Example 7
- Determine whether v 1 = (1, 2 , 3) and v 2 = (1, 0 , 2) are linearly dependent or independent. Consider the Homogeneous system
c 1 (1, 2 , 3) + c 2 (1, 0 , 2) = (0, 0 , 0)
Only solution is the trivial solution a 1 = a 2 = 0, so linearly independent.
- Determine whether v 1 = (1, 1 , 0) and v 2 = (1, 0 , 1) and v 3 = (3, 1 , 2) are linearly dependent. Want to find solutions to the system of equations
c 1 (1, 1 , 0) + c 2 (1, 0 , 1) + c 3 (3, 1 , 2) = (0, 0 , 0)
Which is equivalent to solving
c 1 c 2 c 3
Example 8
Determine whether v 1 = (1, 1 , 1), v 2 = (2, 2 , 2) and v 3 = (1, 0 , 1) are linearly dependent or inde- pendent.
So linearly dependent.
Theorem 9 Given two vectors in a vector space V , they are linearly dependent if and only if they are multiples of one another, i.e. v 1 = cv 2 for some scalar c.
Proof:
av 1 + bv 2 = 0 ⇔ v 2 =
−a b
v 1
Example 10
Determine whether v 1 = (1, 1 , 3) and v 2 = (1, 3 , 1), v 3 = (3, 1 , 1) and v 4 = (3, 3 , 3) are linearly dependent.
Must solve Ax = 0 , where A =
v 1 v 2 v 3 v 4
Since the number of columns is greater then the number of rows, we can see immediately that this system will have infinite solutions.
Theorem 11 Given m vectors in Rn, if m > n they are linearly dependent.
Theorem 12 A linearly independent set in Rn^ has at most n vectors.
