Additional Exercises for Ch 5: Subspaces of
Review Exercises
Let Is a subspace? Explain.
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This is not a subspace. is in it. However, is not.
Let be given vectors in and define
Is a subspace? Explain.
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This is a subspace because it is closed with respect to vector addition and scalar multiplication.
Let and let Is a subspace? Explain.
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Yes, this is a subspace because it is closed with respect to vector addition and scalar multiplication.
Let Is a subspace? Explain.
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This is not a subspace. is in it. However is not.
Let Is a subspace? Explain.
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This is a subspace. It is closed with respect to vector addition and scalar multiplication.
Consider the set of vectors given by
Is a subspace of If so, explain why, give a basis for the subspace and find its
dimension.
Consider the set of vectors given by
Is a subspace of If so, explain why, give a basis for the subspace and find its
dimension.
Consider the set of vectors given by
Is this set of vectors a subspace of If so, explain why, give a basis for the subspace
and find its dimension.
Consider the vectors of the form
Is this set of vectors a subspace of If so, explain why, give a basis for the subspace
and find its dimension.
Consider the set of vectors given by
Is a subspace of If so, explain why, give a basis for the subspace and find its
dimension.
If you have vectors in and the vectors are linearly independent, can it always be
concluded they span Explain.
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Yes. If not, there would exist a vector not in the span. But then you could add in this vector and obtain a linearly independent set of vectors with more vectors than a basis.
If you have vectors in is it possible they are linearly independent? Explain.
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They can’t be.
Suppose are subspaces of Let be all vectors which are in both and . Show that is
a subspace also.
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If then for scalars the linear combination must be in both and since they are both subspaces.
Suppose and both have dimension equal to and they are subspaces of What are
the possibilities for the dimension of ?
Let
Demonstrate that the Rank-Nullity Theorem (Theorem th:matrixranknullity of Subspaces of Associated
with Matrices) holds for .
Show that if is an matrix, then is a subspace of .
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If then and so is closed under linear combinations. Hence it is a subspace.
Challenge Exercises
In each case either show that the statement is true or give an example showing that
it is false. Throughout, denote vectors in .
- (a)
- If is a subspace of and is in , then and are both in . TRUE FALSE
- (b)
- If is a subspace of and is in , then is in . TRUE FALSE
- (c)
- If is a nonempty set and is in for any and whenever and are in , then is a subspace. TRUE FALSE
- (d)
- If is a subspace of and is in , then is in . TRUE FALSE
- (e)
- If is linearly independent, then is linearly independent. TRUE FALSE
- (f)
- If is linearly independent, then is linearly independent. TRUE FALSE
- (g)
- If is linearly dependent, then is linearly dependent. TRUE FALSE
- (h)
- If all of are nonzero, then is linearly independent. TRUE FALSE
- (i)
- If one of is zero, then is linearly dependent. TRUE FALSE
- (j)
- If where , , and are in , then is linearly independent. TRUE FALSE
- (k)
- If is linearly independent, then for some , , and in . TRUE FALSE
- (l)
- If is linearly dependent, then
for in not all zero. TRUE FALSE - (m)
- If is linearly independent, then
for some in . TRUE FALSE - (n)
- Every set of four non-zero vectors in is a basis. TRUE FALSE
- (o)
- No basis of can contain a vector with a component . TRUE FALSE
- (p)
- has a basis of the form where and are vectors. TRUE FALSE
- (q)
- Every basis of contains one column of . TRUE FALSE
- (r)
- Every nonempty subset of a basis of is again a basis of . TRUE FALSE
- (s)
- If and are bases of , then is also a basis of . TRUE FALSE
Suppose has dimension and has dimension and they are each contained in a
subspace, which has dimension equal to where What are the possibilities for the
dimension of ?
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Let be a basis for Then there is a basis for and which are respectively It follows that you must have and so you must have
Octave Exercises
Use Octave to check your work on Problems prb:5.1 to prb:5.8. Problem prb:5.1 is in the code
below.
To use Octave, go to the Sage Math Cell Webpage, copy the code below into the cell, select OCTAVE as the language, and press EVALUATE.
% Find a basis for column vectors
v1=transpose([2 1 1 1]);
v2=transpose([-1 0 -1 -1]);
v3=transpose([5 2 3 3]);
v4=transpose([-1 1 -2 -2]);
A=[v1 v2 v3 v4];
% We can answer all of these questions by performing Gauss-Jordan elimination.
rref(A)
Use Octave to check your work on Problems prb:5.32 to prb:5.38. Problem prb:5.36 is started in the code
below. See if you can interpret the result to answer the question.
To use Octave, go to the Sage Math Cell Webpage, copy the code below into the cell, select OCTAVE as the language, and press EVALUATE.
% Find a basis for row(A) and column(A)
A=[0 0 -1 0 1; 1 2 3 -2 -18; 1 2 2 -1 -11; -1 -2 -2 1 11]
% We can answer all of these questions by performing Gauss-Jordan elimination.
rref(A)
Bibliography
The Review Exercises come from the end of Chapter 4 of Ken Kuttler’s A First Course in Linear Algebra. (CC-BY)
Ken Kuttler, A First Course in Linear Algebra, Lyryx 2017, Open Edition, pp. 227–232.
The Challenge Exercises come from the end of Chapter 5 of Keith Nicholson’s Linear Algebra with Applications. (CC-BY-NC-SA)
W. Keith Nicholson, Linear Algebra with Applications, Lyryx 2018, Open Edition, pp. 327–328.