Linear Independence and Dependence
- Definition: A set of vectors
and the scalars
has a solution for the vector equation
, namely when
. If this is the only solution to the vector equation, then the set
is said to be Linearly Independent. If there exists other solutions where not all
, then
is said to be Linearly Dependent.
We will now look at some examples of vector sets which are either linearly independent or linearly dependent.
- Note: Another way to write that a set of vectors
is linearly independent is by saying that
, which says that any vector that is a linear combination of the set of vectors
is a unique linear combination.
Example 1
Determine whether or not the vector set
is linearly independent or linearly dependent.
We first must see if there exists only one set of scalars
(the trivial solution) as a solution to the vector equation
or if there exists more solutions.
From this vector equation we get the following system of linear equations:

When we reduce this system to RREF, we get that:

Therefore our only set of scalars
are both equal to zero. Therefore, the vector set
is linearly independent.
Example 2
Determine whether the vector set
is a linearly independent or a linearly dependent set.
Once again, for
to be a linearly independent set then the vector equation
must containing only one set of scalars
. From our vector equation be obtain the following system of linear equations:

When we solve this system of equations we obtain that:

Suppose that
so that
. We see that for any
, the scalars
and
satisfy the vector equation and thus there are infinitely many scalars sets. Therefore
is a linearly dependent set.
Example 3
Show that if
is an
invertible matrix, then the column vectors of
denoted
,
, …,
are linearly independent.
Suppose that
is an invertible matrix and suppose that we have the linear system
. We note that the only solution to this linear system is the trivial solution, namely
. This system corresponds to the following vector equations though:

And since
, we see that this vector equation only have the scalars
, so the column vectors of
are linearly independent.
Example 4
Consider the set of vectors
from the vector space
. Determine if this set of vectors is linearly independent or linearly dependent.
When we expand the vector equation
as follows, notice:

We can clearly see that this equation holds only if
. We can also verify this with the following matrix:

Since this is an upper triangular matrix, its determinant is the product of the entries down the main diagonal and so
which implies
is invertible, and so the homogenous system
has only the trivial solution
.
Example 5
Show that the set of vectors
from
is a linearly independent set for all nonzero
.
We should note that this example is a more general version of example 4. First, let's expand the vector equation
as follows:

Once again, this vector equation is only true if
. The same matrix from example 4 can be used to provide a more extensive argument.
Linear Dependence Lemma
We will now look at a very important lemma known as the linear dependence lemma.
- Lemma (Linear Dependence Lemma): Let
be a set of linearly dependent vectors in the vector space
and
. Then there exists
such that:
- a)
.
- b)
.
The linear dependence lemma tells us that given a linear dependent set of vectors where the first vector is nonzero, then there exists a vector in the set
such that
can be written as a linear combination of
(that is
), and that the set of linear combinations of the set of vectors
is the same set as the combination of the set of vectors
.
- Proof: Let
be a set of vectors that are linearly dependent where
, and let
. Since this set of vectors is not linearly independent, then:

- Where not all
are zero (otherwise the set of vectors would be linearly independent). Now since
, it follows that not all
are equal to zero, and so there exists an
such that
where
is the largest index such that
, in other words,
,
, etc…, and so so:

- Therefore
can be written as a linear combination of the set of vectors
so
which proves A.
- Now let
, and so
can be written as a linear combination of the vectors in this set, and so there exists
such that:

- Now we substitute that
and so:
![{\displaystyle {\begin{aligned}\quad u=b_{1}v_{1}+b_{2}v_{2}+...+b_{j-1}v_{j-1}+b_{j}v_{j}+b_{j+1}v_{j+1}+...+b_{m}v_{m}\\\quad u=b_{1}v_{1}+b_{2}v_{2}+...+b_{j-1}v_{j-1}+b_{j}[-{\frac {a_{1}}{a_{j}}}v_{1}-{\frac {a_{2}}{a_{j}}}v_{2}-...-{\frac {a_{j-1}}{a_{j}}}v_{j-1}]+b_{j+1}v_{j+1}+...+b_{m}v_{m}\\\quad u=(b_{1}-a_{1}b_{j})v_{1}+(b_{2}-a_{2}b_{j})v_{2}+...+(b_{j-1}-a_{j-1}b_{j})v_{j-1}+(b_{j+1}-a_{j+1}b_{j})v_{j+1}+...+(b_{m}-a_{m}b_{j})v_{m}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e0adae14b80cfcf0c98ce62efb1723efb085aaa5)
- Therefore
can be written as a linear combination of the vectors
and so
. Therefore
. 
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