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Mathematical Expression Editor
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Solved Problems for Chapter 5
Let Is a subspace? Explain.
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No. is in but is not in .
Let Is a subspace? Explain.
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This is not a subspace. is in . However, is not in .
Let and let Is a subspace? Explain.
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Yes, this is a subspace. First, is not empty because the zero vector is in . You can
verify closure under vector addition and scalar multiplication using the properties of
the dot product as follows. If is in , then and for all constants . Finally, if and are
in , then .
Let Is a subspace? Explain.
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This is a subspace. is not empty because the zero vector is in it. is closed under
vector addition and scalar multiplication.
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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A matrix with five rows and six columns will contain a non-pivot column, giving rize
to a non-trivial solution (infinitely many of them) to the equation . This shows that
the columns of are linearly dependent.
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.
Find the rank of the following matrix, and find a basis for the column spaces.
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The rank of the matrix is 3. We can use columns 1, 2, and 4 as a basis for the
column space.
Find a basis for .
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A basis for the null space is .
Find a basis for
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A basis for the null space is .
Suppose matrix has linearly independent columns. What is the null space of
?
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The only solution to is the trivial one. Therefore, the only vector in is the zero
vector.
Let . Find a basis for . What is the dimension of ?
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Form a matrix using the given vectors as columns. . The leading ’s are in the first
two columns. We can use the first two of the given vectors to form a basis of .
Therefore .