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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.
Let’s take a look at some examples.
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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]
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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 ,
will be linearly independent.
The vectors in the previous example all had the same number
of components as vectors, i.e. two
vectors from or three vectors from . We should work a couple of examples that does
not fit this mold to make sure that you understand that we don’t need to have
the same number of vectors as components.
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Example 2 Determine
if the following sets of vectors are linearly independent or linearly
dependent.
(a) ,
and . [Solution]
(b) ,
and . [Solution]
(c) ,
and . [Solution]
(d) ,
and . [Solution]
Solution
These will work in pretty much the same manner as the
previous set of examples worked.
Again, we’ll do the first part in some detail and then leave it to you
to verify the details in the remaining parts.
Also, we’ll not be showing the details of solving the systems of
equations so you should verify all the solutions for yourself.
(a) ,
and .
Here is the vector equation we need to solve.
The system of
equations that we need to solve is,
and this has
the solution,
We’ve got more
than the trivial solution and so these vectors are linearly dependent.
Note that we
didn’t really need to solve this system to know that they were linearly
dependent. From Theorem 2 in the solving
systems of equations section we know that if there are more unknowns than
equations in a homogeneous system then we will have infinitely many
solutions.
[Return to Problems]
(b) , and .
Here is the
vector equation for this part.
The system of
equations we’ll need to solve is,
Now,
technically we don’t need to solve this system for the same reason we really
didn’t need to solve the system in the previous part. There are more unknowns than equations so
the system will have infinitely many solutions (so more than the trivial
solution) and therefore the vectors will be linearly dependent.
However, let’s
solve anyway since there is an important idea we need to see in this
part. Here is the solution.
In this case
one of the scalars was zero. There is
nothing wrong with this. We still have
solutions other than the trivial solution and so these vectors are linearly
dependent. Note that was it does say
however, is that and are linearly dependent themselves regardless
of .
[Return to Problems]
(c) ,
and .
Here is the
vector equation for this part.
The system of
equations that we’ll need to solve this time is,
The solution to
this system is,
We’ve got more
solutions than the trivial solution and so these three vectors are linearly
dependent.
[Return to Problems]
(d) ,
and .
The vector
equation for this part is,
The system of
equations is,
This system has only the trivial solution and so these
three vectors are linearly independent.
[Return to Problems]
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We should make one quick remark about part (b) of this problem. In this case we had a set of three vectors
and one of the scalars was zero. This
will happen on occasion and as noted this only means that the vectors with the
zero scalars are not really required in order to make the set linearly
dependent. This part has shown that if
you have a set of vectors and a subset is linearly dependent then the whole set
will be linearly dependent.
Often the only way to determine if a set of vectors is
linearly independent or linearly dependent is to set up a system as above and
solve it. However, there are a couple of
cases were we can get the answer just be looking at the set of vectors.
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Theorem 1 A finite set of
vectors that contains the zero vector will be linearly dependent.
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Proof : This is a
fairly simply proof. Let be any set of vectors that contains the zero
vector as shown. We can then set up the
following equation.
We can see from this that we have a non-trivial solution to
this equation and so the set of vectors is linearly dependent.
We’re not going to prove this one but we will outline the
basic proof. In fact, we saw how to
prove this theorem in parts (a) and (b) from Example 2. If we set up the system of equations
corresponding to the equation,
we will get a system of equation that has more unknowns than
equations (you should verify this) and this means that the system will
infinitely many solutions. The vectors
will therefore be linearly dependent.
To this point we’ve only seen examples of linear
independence/dependence with sets of vectors in . We should now take a look at some examples of
vectors from some other vector spaces.
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Example 3 Determine
if the following sets of vectors are linearly independent or linearly
dependent.
(a) ,
,
and . [Solution]
(b) and . [Solution]
(c) and . [Solution]
Solution
Okay, the basic process here is pretty much the same as
the previous set of examples. It just
may not appear that way at first however.
We’ll need to remember that this time the zero vector, 0, is in fact the zero matrix of the
same size as the vectors in the given set.
(a) ,
,
and .
We’ll first need to set of the “vector” equation,
Next, combine
the “vectors” (okay, they’re matrices so let’s call them that….) on the left
into a single matrix using basic matrix scalar multiplication and addition.
Now, we need
both sides to be equal. This means
that the three entries in the matrix on the left that are not already zero
need to be set equal to zero. This
gives the following system of equations.
Of course this
isn’t really much of a system as it tells us that we must have the trivial
solution and so these matrices (or vectors if you want to be exact) are
linearly independent.
[Return to Problems]
(b) and .
So we can see
that for the most part these problems work the same way as the previous
problems did. We just need to set up a
system of equations and solve. For the
remainder of these problems we’ll not put in the detail that did in the first
part.
Here is the
vector equation we need to solve for this part.
The system of
equations we need to solve here is,
We’ll leave it
to you to verify that the only solution to this system is the trivial
solution and so these matrices are linearly independent.
[Return to Problems]
(c) and .
Here is the
vector equation for this part
and the system
of equations is,
The solution to
this system is,
So, we’ve got
solutions other than the trivial solution and so these vectors are linearly
dependent.
[Return to Problems]
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Example 4 Determine
if the following sets of vectors are linearly independent or linearly
dependent.
(a) ,
, and in . [Solution]
(b) ,
, and in . [Solution]
(c) ,
, and in . [Solution]
Solution
Again, these will work in essentially the same manner as
the previous problems. In this problem
set the zero vector, 0, is the
zero function. Since we’re actually
working in for all these parts we can think of this as
the following polynomial.
In other words
a second degree polynomial with zero coefficients.
(a) ,
, and in .
Let’s first set
up the equation that we need to solve.
Now, we could
set up a system of equations here, however we don’t need to. In order for these two second degree
polynomials to be equal the coefficient of each term must be equal. At this point is it should be pretty clear
that the polynomial on the left will only equal if all the coefficients of
the polynomial on the left are zero.
So, the only solution to the vector equation will be the trivial
solution and so these polynomials (or vectors if you want to be precise) are
linearly independent.
[Return to Problems]
(b) ,
, and in .
The vector
equation for this part is,
Now, as with
the previous part the coefficients of each term on the left must be zero in
order for this polynomial to be the zero vector. This leads to the following system of
equations.
The only
solution to this system is the trivial solution and so these polynomials are
linearly independent.
[Return to Problems]
(c) ,
, and in .
In this part
the vector equation is,
The system of
equation we need to solve is,
The solution to
this system is,
So, we have more solutions than the trivial solution and
so these polynomials are linearly dependent.
[Return to Problems]
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Now that we’ve seen quite a few examples of linearly
independent and linearly dependent vectors we’ve got one final topic that we
want to discuss in this section. Let’s
go back and examine the results of the very first example that we worked in
this section and in particular let’s start with the final part.
In this part we looked at the vectors ,
,
and and determined that they were linearly
dependent. We did this by solving the
vector equation,
and found that it had the solution,
We knew that the vectors were linearly dependent because
there were solutions to the equation other than the trivial solution. Let’s take a look at one of them. Say,
In fact, let’s plug these values into the vector equation
above.
Now, if we rearrange this a little we arrive at,
or, in a little more compact form : .
So, we were able to write one of the vectors as a linear
combination of the other two. Notice as
well that we could have just as easily written and a linear combination of and or as a linear combination of and if we’d wanted to.
Let’s see if we can do this with the three vectors from the
third part of this example. In this part
we were looking at the three vectors ,
,
and and in that part we determined that these
vectors were linearly independent. Let’s
see if we can write and a linear combination of and . If we can we’ll be able to find constants and that will make the following equation true.
Now, while we can find values of and that will make the second and third entries
zero as we need them to we’re in some pretty serious trouble with the first
entry. In the vector on the left we’ve
got a 1 in the first entry and in the vector on the right we’ve got a 0 in the
first entry. So, there is no way we can
write the first vector as a linear combination of the other two. You should also verify we also do this in any
of the other combinations either.
So, what have we seen here with these two examples. With a set of linearly dependent vectors we
were able to write at least one of them as a linear combination of the other
vectors in the set and with a set of linearly independent vectors we were not
able to do this for any of the vectors.
This will always be the case.
With a set of linearly independent vectors we will never be
able to write one of the vectors as a linear combination of the other vectors
in the set. On the other hand, if we
have a set of linearly dependent vectors then at least one of them can be
written as a linear combination of the remaining vectors.
In the example of linearly dependent vectors we were looking
at above we could write any of the vectors as a linear combination of the
others. This will not always be the
case, to see this take a look at Example 2(b). In this example we determined that the vectors
,
and were linearly dependent. We also saw that the solution to the
equation,
was given by
and as we saw above we can always use this to determine how
to write at least one of the vectors as a linear combination of the remaining
vectors. Simply pick a value of t and the rearrange as you need to. Doing this in our case we see that we can do
one of the following.
It’s easy in this case to write the first or the third
vector as a combination of the other vectors.
However, because the coefficient of the second vector is zero, there is
no way that we can write the second vector as a linear combination of the first
and third vectors.
What that means here is that the first and third vectors are
linearly dependent by themselves (as we pointed out in that example) but the
first and second are linearly independent vectors as are the second and third
if we just look at them as a pair of vectors (you should verify this).
This can be a useful idea about linearly
independent/dependent vectors on occasion.