Solved Problems

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Solved Problems for Chapter 3

If possible, express the vector $\begin{equation*} \vec{v}= \begin{bmatrix} 4 \\ 4 \\ -3 \end{bmatrix} \end{equation*}$ as a linear combination of the vectors

$\begin{equation*} \vec{u}_1 = \begin{bmatrix} 3 \\ 1 \\ -1 \end{bmatrix} \mbox{ and } \vec{u}_2 = \begin{bmatrix} 2 \\ -2\\ 1 \end{bmatrix}. \end{equation*}$

Click the arrow to see answer.

We start with a linear combination equation $a\begin {bmatrix}3\\1\\-1\end {bmatrix}+b\begin {bmatrix}2\\-2\\1\end {bmatrix}=\begin {bmatrix}4\\4\\-3\end {bmatrix}$ . This gives us the following: $\left [ \begin {array}{rr|r} 3 & 2 & 4 \\ 1 & -2 & 4\\ -1 & 1 & -3 \end {array} \right ] \rightsquigarrow \left [ \begin {array}{rr|r}1& 0& 2\\ 0& 1 &-1\\ 0& 0& 0\end {array} \right ]$ This gives us $\vec {v}=2\vec {u}_1 -\vec {u}_2$ .

Decide whether $\vec {v}= \begin {bmatrix} 4 \\ 4 \\ 4 \end {bmatrix}$ is a linear combination of the vectors $\vec {u}_1 = \begin {bmatrix} 3 \\ 1 \\ -1 \end {bmatrix} \mbox { and } \vec {u}_2 = \begin {bmatrix} 2 \\ -2\\ 1 \end {bmatrix}.$

Click the arrow to see answer.

$\left [ \begin {array}{rr|r} 3 & 2 & 4 \\ 1 & -2 & 4\\ -1 & 1 & 4 \end {array} \right ] \rightsquigarrow \left [ \begin {array}{rr|r}1& 0& 0\\ 0& 1 & 0\\ 0& 0& 1\end {array} \right ]$ The last row indicates that there is no solution to the linear combination equation $a\vec {u}_1+b\vec {u}_2=\vec {v}$ . Therefore, $\vec {v}$ is not a linear combination of $\vec {u}_1$ and $\vec {u}_2$ .

Determine whether the following set of vectors is linearly independent. $\left \{\begin {bmatrix} 2\\ -3\\0 \end {bmatrix}, \begin {bmatrix} 1\\0\\1 \end {bmatrix}, \begin {bmatrix} 1\\1\\4 \end {bmatrix}\right \}$

Click the arrow for answer.

The linear combination equation $a\begin {bmatrix} 2\\ -3\\0 \end {bmatrix}+ b\begin {bmatrix} 1\\0\\1 \end {bmatrix}+ c\begin {bmatrix} 1\\1\\4 \end {bmatrix}=\vec {0}$ has only the trivial solution. We can see this as follows. $\text {rref}\left (\begin {bmatrix}2 & 1 & 1\\-3 & 0 & 1\\0 & 1 & 4\end {bmatrix}\right )=\begin {bmatrix} 1 & 0 & 0\\0 & 1 &0\\ 0 & 0 &1 \end {bmatrix}$ Therefore the given set of vectors is linearly independent.

Determine whether the following set of vectors is linearly independent. $\left \{\begin {bmatrix} 2\\ -3\\0 \end {bmatrix}, \begin {bmatrix} 4\\-1\\8 \end {bmatrix}, \begin {bmatrix} 1\\1\\4 \end {bmatrix}\right \}$

Click the arrow for answer.

The linear combination equation $a\begin {bmatrix} 2\\ -3\\0 \end {bmatrix}+ b\begin {bmatrix} 4\\-1\\8 \end {bmatrix}+ c\begin {bmatrix} 1\\1\\4 \end {bmatrix}=\vec {0}$ has a non-trivial solution. We can see this as follows. $\text {rref}\left (\begin {bmatrix}2 & 4 & 1\\-3 & -1 & 1\\0 & 8 & 4\end {bmatrix}\right )=\begin {bmatrix} 1 & 0 & -1/2\\0 & 1 &1/2\\ 0 & 0 &0 \end {bmatrix}$ Therefore the given set of vectors is linearly dependent.

Suppose $\left \{ \vec {x}_{1},\ldots ,\vec {x}_{k}\right \}$ is a set of vectors from $\RR ^{n}.$ Show that $\vec {0}$ is in $\mbox { span}\left \{ \vec {x}_{1},\ldots ,\vec {x}_{k}\right \} .$

Click the arrow to see answer.

Let all of the coefficients in the linear combination be zero: $\sum _{i=1}^{k}0\vec {x}_{k}=\vec {0}$

True or false?

Click the arrow to see each answer.

(a): The span of any one vector in $\RR ^n$ is a line.

This is technically false because the span of the zero vector is a point. For all non-zero vectors this statement is true.
(b): The span of two non-zero vectors in $\RR ^n$ is a plane.

This is false. If two vectors are scalar multiples of each other (i.e. they are linearly dependent), then their span is a line, not a plane.
(c): Given two linearly independent vectors, $\vec {v}$ and $\vec {w}$ , let $\vec {n}$ be a normal vector to the plane spanned by $\vec {v}$ and $\vec {w}$ . The set $\left \{\vec {v}, \vec {w}, \vec {n}\right \}$ is linearly independent.

This is true. A normal to the plane is a vector perpendicular to the plane. Only vectors that lie in the plane spanned by $\vec {v}$ and $\vec {w}$ can be expressed as a linear combination of $\vec {v}$ and $\vec {w}$ . Since $\vec {n}$ is not in the plane, it cannot be expressed as a linear combination of $\vec {v}$ and $\vec {w}$ . Therefore the set $\left \{\vec {v}, \vec {w}, \vec {n}\right \}$ is linearly independent.

Bibliography

Some of 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. 151–152.

2024-09-06 23:25:43

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Solved Problems for Chapter 3

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