1
Linear Independence
Given a set of vectors 𝑣1 , … 𝑣𝑛 in a vector space 𝑉, we know that:
Span(𝑣1 , … , 𝑣𝑛 ) = {𝛼1 𝑣1 + ⋯ + 𝛼𝑛 𝑣𝑛 , 𝛼𝑖 ∈ ℝ} = 𝑊 is a subspace of 𝑉.
However, it might not be necessary to have all of the vectors 𝑣1 , 𝑣2 , … , 𝑣𝑛 to
span 𝑊. It's often useful to know the minimum number of vectors needed to span
a vector space.
Ex. Let 𝑣1 = < 1, 0, 0 >, 𝑣2 = < 0, 1, 0 >, 𝑣3 = < 2, 5, 0 >. Show that the
Span {𝑣1 , 𝑣2 , 𝑣3 } = Span {𝑣1 , 𝑣2 }.
Notice that we can write 𝑣3 as a linear combination of 𝑣1 and 𝑣2 :
𝑣3 = 2𝑣1 + 5𝑣2
Thus any vector 𝑤 that can be created as a linear combination of 𝑣1 , 𝑣2 , 𝑣3
𝑤 = 𝛼1 𝑣1 + 𝛼2 𝑣2 + 𝛼3 𝑣3
can also be created as a linear combination of just 𝑣1 and 𝑣2 :
𝑤 = 𝛽1 𝑣1 + 𝛽2 𝑣2 .
For example, let's say:
𝑤 = 3𝑣1 + 2𝑣2 + 4𝑣3 .
, 2
Since we know 𝑣3 = 2 𝑣1 + 5 𝑣2 , we have:
𝑤 = 3𝑣1 + 2𝑣2 + 4𝑣3 = 3𝑣1 + 2𝑣2 + 4(2 𝑣1 + 5 𝑣2 )
𝑤 = 3𝑣1 + 2𝑣2 + 8𝑣1 + 20 𝑣2 = 11 𝑣1 + 22 𝑣2 .
Whenever one vector, 𝑤 , can be written as a linear combination of other vectors,
𝑣1 , 𝑣2 , … , 𝑣𝑛 , we say the vectors 𝑣1 , 𝑣2 , … , 𝑣𝑛 , 𝑤 are linearly dependent.
A set of vectors 𝑣1 , … , 𝑣𝑛 are said to be linearly independent if:
𝑐1 𝑣1 + 𝑐2 𝑣2 + … + 𝑐𝑛 𝑣𝑛 = 0
implies that: 𝑐1 = 𝑐2 = 𝑐3 = ⋯ = 𝑐𝑛 = 0.
Notice that if a set of vectors 𝑣1 , 𝑣2 , … , 𝑣𝑛 are linearly dependent, at least one
vector can be written as a linear combination of the other vectors.
Let’s assume that 𝑣1 can be written as a linear combination of 𝑣2 , … , 𝑣𝑛 , that is:
𝑣1 = 𝑐2 𝑣2 + 𝑐3 𝑣3 + ⋯ + 𝑐𝑛 𝑣𝑛 ; where not all of the 𝑐𝑖 ′𝑠 are 0.
But that means that:
𝑐2 𝑣2 + 𝑐3 𝑣3 + ⋯ + 𝑐𝑛 𝑣𝑛 − 𝑣1 = 0; where not all of the 𝑐𝑖 ′𝑠 are 0.
Hence if the set of vectors 𝑣1 , 𝑣2 , … , 𝑣𝑛 are linearly dependent, then they can’t
be linearly independent.
Conversely, if a set of vectors is linearly independent then they can’t be linearly
dependent since if they were linearly dependent we could find a set of 𝑐𝑖 ′𝑠, not all
0 such that 𝑐1 𝑣1 + 𝑐2 𝑣2 + ⋯ + 𝑐𝑛 𝑣𝑛 = 0.
Linear Independence
Given a set of vectors 𝑣1 , … 𝑣𝑛 in a vector space 𝑉, we know that:
Span(𝑣1 , … , 𝑣𝑛 ) = {𝛼1 𝑣1 + ⋯ + 𝛼𝑛 𝑣𝑛 , 𝛼𝑖 ∈ ℝ} = 𝑊 is a subspace of 𝑉.
However, it might not be necessary to have all of the vectors 𝑣1 , 𝑣2 , … , 𝑣𝑛 to
span 𝑊. It's often useful to know the minimum number of vectors needed to span
a vector space.
Ex. Let 𝑣1 = < 1, 0, 0 >, 𝑣2 = < 0, 1, 0 >, 𝑣3 = < 2, 5, 0 >. Show that the
Span {𝑣1 , 𝑣2 , 𝑣3 } = Span {𝑣1 , 𝑣2 }.
Notice that we can write 𝑣3 as a linear combination of 𝑣1 and 𝑣2 :
𝑣3 = 2𝑣1 + 5𝑣2
Thus any vector 𝑤 that can be created as a linear combination of 𝑣1 , 𝑣2 , 𝑣3
𝑤 = 𝛼1 𝑣1 + 𝛼2 𝑣2 + 𝛼3 𝑣3
can also be created as a linear combination of just 𝑣1 and 𝑣2 :
𝑤 = 𝛽1 𝑣1 + 𝛽2 𝑣2 .
For example, let's say:
𝑤 = 3𝑣1 + 2𝑣2 + 4𝑣3 .
, 2
Since we know 𝑣3 = 2 𝑣1 + 5 𝑣2 , we have:
𝑤 = 3𝑣1 + 2𝑣2 + 4𝑣3 = 3𝑣1 + 2𝑣2 + 4(2 𝑣1 + 5 𝑣2 )
𝑤 = 3𝑣1 + 2𝑣2 + 8𝑣1 + 20 𝑣2 = 11 𝑣1 + 22 𝑣2 .
Whenever one vector, 𝑤 , can be written as a linear combination of other vectors,
𝑣1 , 𝑣2 , … , 𝑣𝑛 , we say the vectors 𝑣1 , 𝑣2 , … , 𝑣𝑛 , 𝑤 are linearly dependent.
A set of vectors 𝑣1 , … , 𝑣𝑛 are said to be linearly independent if:
𝑐1 𝑣1 + 𝑐2 𝑣2 + … + 𝑐𝑛 𝑣𝑛 = 0
implies that: 𝑐1 = 𝑐2 = 𝑐3 = ⋯ = 𝑐𝑛 = 0.
Notice that if a set of vectors 𝑣1 , 𝑣2 , … , 𝑣𝑛 are linearly dependent, at least one
vector can be written as a linear combination of the other vectors.
Let’s assume that 𝑣1 can be written as a linear combination of 𝑣2 , … , 𝑣𝑛 , that is:
𝑣1 = 𝑐2 𝑣2 + 𝑐3 𝑣3 + ⋯ + 𝑐𝑛 𝑣𝑛 ; where not all of the 𝑐𝑖 ′𝑠 are 0.
But that means that:
𝑐2 𝑣2 + 𝑐3 𝑣3 + ⋯ + 𝑐𝑛 𝑣𝑛 − 𝑣1 = 0; where not all of the 𝑐𝑖 ′𝑠 are 0.
Hence if the set of vectors 𝑣1 , 𝑣2 , … , 𝑣𝑛 are linearly dependent, then they can’t
be linearly independent.
Conversely, if a set of vectors is linearly independent then they can’t be linearly
dependent since if they were linearly dependent we could find a set of 𝑐𝑖 ′𝑠, not all
0 such that 𝑐1 𝑣1 + 𝑐2 𝑣2 + ⋯ + 𝑐𝑛 𝑣𝑛 = 0.
