Solved Problems for Chapter 5

Let \(M=\left \{ \vec {u}=\left [ \begin{array}{c} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \end{array}\right ] \in \mathbb {R}^{4}:|u_{1}| \leq 4\right \} .\) Is \(M\) a subspace? Explain.

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No. \(\left [ \begin{array}{r} 1 \\ 0 \\ 0 \\ 0 \end{array} \right ]\) is in \(M\) but \(10\left [ \begin{array}{r} 1 \\ 0 \\ 0 \\ 0 \end{array} \right ]\) is not in \(M\).
Let \(M=\left \{ \vec {u}=\left [ \begin{array}{c} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \end{array}\right ] \in \mathbb {R}^{4}:u_{i}\geq 0\text { for each }i=1,2,3,4\right \} .\) Is \(M\) a subspace? Explain.

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This is not a subspace. \(\left [ \begin{array}{r} 1 \\ 1 \\ 1 \\ 1 \end{array} \right ] \) is in \(M\). However, \(\left ( -1\right ) \left [ \begin{array}{r} 1 \\ 1 \\ 1 \\ 1 \end{array} \right ] \) is not in \(M\).
Let \(\vec {w}\in \mathbb {R}^{4}\) and let \(M=\left \{ \vec {u} =\left [ \begin{array}{c} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \end{array}\right ] \in \mathbb {R}^{4}:\vec {w}\dotp \vec {u} =0\right \} .\) Is \(M\) a subspace? Explain.

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Yes, this is a subspace. First, \(M\) is not empty because the zero vector is in \(M\). You can verify closure under vector addition and scalar multiplication using the properties of the dot product as follows. If \(\vec {u}\) is in \(M\), then \(\vec {w}\dotp \vec {u}=0\) and \(\vec {w}\dotp k\vec {u}=0\) for all constants \(k\). Finally, if \(\vec {u}_1\) and \(\vec {u}_2\) are in \(M\), then \(\vec {w}\dotp (\vec {u}_1+\vec {u}_2)=\vec {w}\dotp \vec {u}_1+\vec {w}\dotp \vec {u}_2=0\).
Let \(M=\left \{ \vec {u}=\left [ \begin{array}{c} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \end{array}\right ] \in \mathbb {R}^{4}:u_{3}=u_{1}=0\right \} .\) Is \(M\) a subspace? Explain.

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This is a subspace. \(M\) is not empty because the zero vector is in it. \(M\) is closed under vector addition and scalar multiplication.
If you have \(5\) vectors in \(\mathbb {R}^{5}\) and the vectors are linearly independent, can it always be concluded they span \(\mathbb {R}^{5}?\) 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 \(6\) vectors in \(\mathbb {R}^{5},\) is it possible they are linearly independent? Explain.

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A matrix \(A\) 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 \(A\vec {x}=\vec {0}\). This shows that the columns of \(A\) are linearly dependent.
Suppose \(V, W\) are subspaces of \(\mathbb {R}^{n}.\) Let \(V\cap W\) be all vectors which are in both \(V\) and \(W\). Show that \(V \cap W\) is a subspace also.

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If \(\vec {x}, \vec {y}\in V\cap W,\) then for scalars \(\alpha ,\beta ,\) the linear combination \(\alpha \vec {x}+\beta \vec {y}\) must be in both \(V\) and \(W\) 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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\[\text {rref}\left (\begin{bmatrix} 1 & 0 & 3 & 0 \\ 3 & 1 & 10 & 0 \\ -1 & 1 & -2 & 1 \\ 1 & -1 & 2 & -2 \end{bmatrix}\right )=\begin{bmatrix}1 &0 &3 &0\\ 0& 1& 1& 0\\ 0& 0& 0& 1\\ 0& 0& 0& 0\end{bmatrix}\]
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 \(\mbox {null} \left (A \right )\).
\[A = \left [ \begin{array}{rrr} 1 & 0 & -1 \\ -1 & 1 & 3 \\ 3 & 2 & 1 \end{array} \right ]\]

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\[\text {rref}\left (\left [ \begin{array}{rrr} 1 & 0 & -1 \\ -1 & 1 & 3 \\ 3 & 2 & 1 \end{array} \right ]\right )=\begin{bmatrix} 1& 0& -1\\ 0 &1& 2\\ 0& 0& 0 \end{bmatrix}\]

A basis for the null space is \(\left \{\begin{bmatrix} 1\\-2\\1 \end{bmatrix}\right \}\).

Find a basis for \(\mbox {null} \left (A \right )\)
\[ A = \left [ \begin{array}{rrrr} 2 & -1 & 3 & 5 \\ 2 & 0 & 1 & 2 \\ 6 & 4 & -5 & -6 \\ 0 & 2 & -4 & -6 \end{array} \right ]\]

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\[\text {rref}\left (\left [ \begin{array}{rrrr} 2 & -1 & 3 & 5 \\ 2 & 0 & 1 & 2 \\ 6 & 4 & -5 & -6 \\ 0 & 2 & -4 & -6 \end{array} \right ]\right )= \begin{bmatrix} 1 &0 &1/2 &1\\ 0 &1 &-2 &-3\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{bmatrix} \]
A basis for the null space is \(\left \{\begin{bmatrix} -1\\3\\0\\1 \end{bmatrix}, \begin{bmatrix} -1/2\\2\\1\\0 \end{bmatrix}\right \}\).
Suppose matrix \(A\) has linearly independent columns. What is the null space of \(A\)?

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The only solution to \(A\vec {x}=\vec {0}\) is the trivial one. Therefore, the only vector in \(\text {null}(A)\) is the zero vector.
Let \(V=\text {span}\left (\begin{bmatrix}2\\-1\\4\end{bmatrix}, \begin{bmatrix}-1\\1\\3\end{bmatrix}, \begin{bmatrix}4\\-3\\-2\end{bmatrix}\right ) \). Find a basis for \(V\). What is the dimension of \(V\)?

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Form a matrix \(A\) using the given vectors as columns. \(\text {rref}(A)=\begin{bmatrix}1& 0 &1\\ 0& 1& -2\\ 0& 0& 0\end{bmatrix}\). The leading \(1\)’s are in the first two columns. We can use the first two of the given vectors to form a basis of \(V\). Therefore \(\text {dim}(V)=2\).

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.