\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*}
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We start with a linear combination equation . This gives us the following: This gives us .
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The last row indicates that there is no solution to the linear combination equation . Therefore, is not a linear combination of and .
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The linear combination equation has only the trivial solution. We can see this as follows. Therefore the given set of vectors is linearly independent.
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The linear combination equation has a non-trivial solution. We can see this as follows. Therefore the given set of vectors is linearly dependent.
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Let all of the coefficients in the linear combination be zero:
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This is technically false because the span of the zero vector is a point. For all non-zero vectors this statement is true.
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.
This is true. A normal to the plane is a vector perpendicular to the plane. Only vectors that lie in the plane spanned by and can be expressed as a linear combination of and . Since is not in the plane, it cannot be expressed as a linear combination of and . Therefore the set is linearly independent.
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