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No. is in but is not in .
Solved Problems for Chapter 5
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. \begin{equation*} \left [ \begin{array}{rrrr} 1 & 0 & 3 & 0 \\ 3 & 1 & 10 & 0 \\ -1 & 1 & -2 & 1 \\ 1 & -1 & 2 & -2 \end{array} \right ] \end{equation*}
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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.
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 .
Bibliography
Some of the problems 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.
2024-09-06 23:26:34