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Online Notes / Linear Algebra / Vector Spaces / Linear Independence
Linear Algebra

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In the previous section we saw several examples of writing a particular vector as a linear combination of other vectors.  However, as we saw in Example 1(b) of that section there is sometimes more than one linear combination of the same set of vectors can be used for a given vector.  We also saw in the previous section that some sets of vectors, , can span a vector space.  Recall that by span we mean that every vector in the space can be written as a linear combination of the vectors in S.  In this section we’d like to start looking at when it will be possible to express a given vector from a vector space as exactly one linear combinations of the set S.

 

We’ll start this section off with the following definition.

 

Definition 1  Suppose  is a non-empty set of vectors and form the vector equation,

                                                     

This equation has at least one solution, namely, , , … , .  This solution is called the trivial solution.

 

If the trivial solution is the only solution to this equation then the vectors in the set S are called linearly independent and the set is called a linearly independent set.  If there is another solution then the vectors in the set S are called linearly dependent and the set is called a linearly dependent set.

 

Let’s take a look at some examples.

 

Example 1  Determine if each of the following sets of vectors are linearly independent or linearly dependent.

(a)  and .   [Solution]

(b)  and .   [Solution]

(c) , , and .   [Solution]

(d) , , and .   [Solution]

 

Solution

To answer the question here we’ll need to set up the equation

                                                     

for each part, combine the left side into a single vector and the set all the components of the vector equal to zero (since it must be the zero vector, 0).  At this point we’ve got a system of equations that we can solve.  If we only get the trivial solution the vectors will be linearly independent and if we get more than one solution the vectors will be linearly dependent.

 

(a)  and .

 

We’ll do this one in detail and then do the remaining parts quicker.  We’ll first set up the equation and get the left side combined into a single vector.

                                                   

 

Now, set each of the components equal to zero to arrive at the following system of equations.

                                                               

 

Solving this system gives to following solution (we’ll leave it to you to verify this),

                                                        

 

The trivial solution is the only solution and so these two vectors are linearly independent.

[Return to Problems]

 

(b)  and .

Here is the vector equation we need to solve.

                                                     

The system of equations that we’ll need to solve is,

                                                              

and the solution to this system is,

                                 

 

We’ve got more than the trivial solution (note however that the trivial solution IS still a solution, there’s just more than that this time) and so these vectors are linearly dependent.

[Return to Problems]

 

(c) , , and .

 

The only difference between this one and the previous two are the fact that we now have three vectors out of .  Here is the vector equation for this part.

                                             

The system of equations to solve for this part is,

                                                                     

 

So, not much solving to do this time.  It is clear that the only solution will be the trivial solution and so these vectors are linearly independent.

[Return to Problems]

 

 

(d) , , and .

 

Here is the vector equation for this final part.

                                          

The system of equations that we’ll need to solve here is,

                                                           

 

The solution to this system is,

                         

 

We’ve got more than just the trivial solution and so these vectors are linearly dependent.

[Return to Problems]

 

Note that we didn’t really need to solve any of the systems above if we didn’t want to.  All we were interested in it was whether or not the system had only the trivial solution or if there were more solutions in addition to the trivial solution.  Theorem 9 from the Properties of the Determinant section can help us answer this question without solving the system.  This theorem tells us that if the determinant of the coefficient matrix is non-zero then the system will have exactly one solution, namely the trivial solution.  Likewise, it can be shown that if the determinant is zero then the system will have infinitely many solutions.

 

Therefore, once the system is set up if the coefficient matrix is square all we really need to do is take the determinant of the coefficient matrix and if it is non-zero the set of vectors will be linearly independent and if the determinant is zero then the set of vectors will be linearly dependent.  If the coefficient matrix is not square then we can’t take the determinant and so we’ll not have a choice but to solve the system.

 

This does not mean however, that the actual solution to the system isn’t ever important as we’ll see towards the end of the section.

 

Before proceeding on we should point out that the vectors from part (c) of this were actually the standard basis vectors for .  In fact the standard basis vectors for ,