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Mathematical Expression Editor
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Solved Problems for Chapter 10
Let be subspaces of a vector space and consider defined as the set of all where
and . Show that is a subspace of .
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We need to show that is closed under addition and scalar multiplication. (See
Theorem th:subspacetestabstract.) Suppose and are in . Then and , where is in and is in . To see
that is closed under scalar multiplication, consider . Since is in , and is
in , is in . Next, we show that is closed under vector addition. . Since
is an element of , and is an element of we have closure under addition.
Let be subspaces of a vector space . Then consists of all vectors which are in both
and . Show that is a subspace of .
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If a vector is in both subspaces and , then its scalar multiple must also be in both.
Same is true about the sum of two elements of .
Let be subspaces of a vector space Then consists of all vectors which are in either
or . Show that is not necessarily a subspace of by giving an example where fails to
be a subspace.
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As our example, let to be the first quadrant, and let be the third quadrant.
Define by .
(a)
Find .
(b)
Is a linear transformation? If so, prove it. If not, give a counterexample.
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(a)
(b)
. is a linear transformation.
Let . Is in ? Consider the question in two different ways, then compare your work
for each approach.
(a)
Do this directly using the definition of span (Definition def:lincombabstract).
(b)
Do this using isomorphisms.
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(a)
Apply the definition of span directly. We are looking for coefficients , and
such that
Collecting like terms on the left and setting their coefficients equal to
their counterparts on the right gives us the following system of equations.
This gives rise to the augmented matrix
This shows that our system has a unique solution and gives us the specific
coefficients to express as a linear combination of the vectors in the given
set.
(b)
We can also look at this problem in light of isomorphisms. Let’s start by
mapping
(Why is this an isomorphism?)
Is the vector in the span of the three vectors above? Set up an augmented
matrix to answer this question. Compare this matrix to the matrix in part
(a).
Let . Is in ?
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is not in the span of the given vectors. The system
is infeasible.
Consider the vector space of polynomials of degree at most , . Determine whether
the following is a basis for .
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There is a isomorphism from to , defined as follows:
Thus,
It follows that if
is a basis for then the polynomials will be a basis for because they will be
independent. Recall that an isomorphism takes a linearly independent set to a
linearly independent set. Also, since is an isomorphism, it preserves all linear
relations.
Find a basis in for the subspace
If the above three vectors do not yield a basis, exhibit one of them as a linear
combination of the others.
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This is the situation in which you have a spanning set and you want to cut it down to
form a linearly independent set which is also a spanning set. Use the same
isomorphism as above. Since is an isomorphism, it preserves all linear relations
so if such can be found in , the same linear relations will be present in .
Define as follows.
Show that is one to one.
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We need to show that implies that . Suppose
Performing matrix-vector multiplication on both sides shows that , and . Therefore, .
Define as follows.
Is onto?
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is not onto. One way to show this is by finding a counter-example. We
need a vector in that is not an image of any element of under . Try .
Find the coordinates of with respect to the ordered basis of .
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We need to find coefficients , , , and such that
This leads to the following system of equations:
We get , , , . Therefore .
Define by . (Verify that is a linear transformation.) Find the matrix of if the basis
for the domain and the codomain is .
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We start with a diagram:
Based on where the standard unit vectors of map to, we have the following
matrix.
