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Example 1 Determine
if each of the sets of vectors will be a basis for  .
(a) ,
and . [Solution]
(b) ,
and . [Solution]
(c) and . [Solution]
(d) ,
and . [Solution]
Solution
(a) ,
and .
Now, let’s see what we’ve got to do here to determine
whether or not this set of vectors will be a basis for . First, we’ll need to show that these
vectors span and from the section on Span
we know that to do this we need to determine if we can find scalars ,
,
and so that a general vector from can be expressed as a linear combination of
these three vectors or,
As we saw in
the section on Span all we need to do is convert this to a system of
equations, in matrix form, and then determine if the coefficient matrix has a
non-zero determinant or not. If the
determinant of the coefficient matrix is non-zero then the set will span the
given vector space and if the determinant of the coefficient matrix is zero
then it will not span the given vector space.
Recall as well that if the determinant of the coefficient matrix is
non-zero then there will be exactly one solution to this system for each u.
The matrix form
of the system is,
Before we get
the determinant of the coefficient matrix let’s also take a look at the other
condition that must be met in order for this set to be a basis for . In order for these vectors to be a basis
for then they must be linearly independent. From the section on Linear Independence we know that to
determine this we need to solve the following equation,
If this system
has only the trivial solution the vectors will be linearly independent and if
it has solutions other than the trivial solution then the vectors will be
linearly dependent.
Note however,
that this is really just a specific case of the system that we need to solve
for the span question. Namely here we
need to solve,
Also, as noted
above, if these vectors will span then there will be exactly one solution to
the system for each u. In this case we know that the trivial
solution will be a solution, our only question is whether or not it is the
only solution.
So, all that we
need to do here is compute the determinant of the coefficient matrix and if
it is non-zero then the vectors will both span and be linearly independent and hence the
vectors will be a basis for . On the other hand, if the determinant is
zero then the vectors will not span and will not be linearly independent and so
they won’t be a basis for .
So, here is the
determinant of the coefficient matrix for this problem.
So, these
vectors will form a basis for .
[Return to Problems]
(b) ,
and .
Now, we could
use a similar path for this one as we did earlier. However, in this case, we’ve done all the
work for this one in previous sections.
In Example 4(a) of the section on
Span we determined that the standard basis vectors (Interesting name isn’t
it? We’ll come back to this in a bit) ,
and will span . Notice that while we’ve changed the
notation a little just for this problem we are working with the standard
basis vectors here and so we know that they will span .
Likewise, in Example 1(c) from the section on
Linear Independence we saw that these vectors are linearly independent.
Hence based on
all this previous work we know that these three vectors will form a basis for
.
[Return to Problems]
(c) and .
We can’t use
the method from part (a) here because the coefficient matrix wouldn’t be
square and so we can’t take the determinant of it. So, let’s just start this out be checking
to see if these two vectors will span . If these two vectors will span then for each in there must be scalars and so that,
However, we can
see right away that there will be problems here. The third component of the each of these
vectors is zero and hence the linear combination will never have any non-zero
third component. Therefore, if we
choose to be any vector in with we will not be able to find scalars and to satisfy the equation above.
Therefore,
these two vectors do not span and hence cannot be a basis for .
Note however,
that these two vectors are linearly independent (you should verify
that). Despite this however, the
vectors are still not a basis for since they do not span .
(d) ,
and
[Return to Problems]
In this case
we’ve got three vectors with three components and so we can use the same
method that we did in the first part.
The general equation that needs solved here is,
and the matrix
form of this is,
We’ll leave it to you to verify that and so these three vectors do not span and are not linearly independent. Either of which will mean that these three
vectors are not a basis for .
[Return to Problems]
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