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Chapter 7: Problem 14

(a) Let \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) be vectors in 3 -dimensional space. Describe geometrically the conditions under which \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) (i) span a plane, (ii) span just a line. (b) Explain geometrically why one needs at least three vectors to span \(\mathbb{R}^{3}\).

Short Answer

Expert verified
(a) Two vectors span a plane if linearly independent; span a line if dependent. (b) Three vectors are needed to span \( \mathbb{R}^3 \).

Step by step solution

01

Vectors Spanning a Plane

Two vectors \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) in \( \mathbb{R}^3 \) span a plane if and only if they are linearly independent. Geometrically, this means the vectors are not parallel. When two vectors are linearly independent, the plane they span includes all linear combinations of these vectors. This scenario occurs when neither vector is a scalar multiple of the other.
02

Vectors Spanning a Line

Two vectors \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) span just a line if they are linearly dependent. This means one vector is a scalar multiple of the other, indicating that all linear combinations of these vectors lie on the line defined by either vector. Thus, in 3-dimensional space, such vectors lack the dimensional complexity to span a plane.
03

Requirement for Spanning \( \mathbb{R}^3 \)

To span \( \mathbb{R}^3 \), a set of vectors must be able to describe any vector within that space. Geometrically, this requires the vectors to cover three independent directions. Since each vector can provide at most one dimension of freedom, at least three linearly independent vectors are needed to span the full 3-dimensional space. Otherwise, the vectors can only describe a line or plane.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vectors in 3-dimensional space
In three-dimensional space, vectors can be thought of as arrows pointing from one point to another. These vectors have three components, generally viewed as
  • \(x\): The component in the direction of the x-axis,
  • \(y\): The component in the direction of the y-axis,
  • \(z\): The component in the direction of the z-axis.
Each vector can be written as \( \mathbf{v} = (x, y, z) \).
This representation helps in visualizing and performing calculations in a three-dimensional framework.
Moving through 3-dimensional space can be visually complex, but vectors simplify navigating and describing positions and movements within this space. When dealing with vectors in \( \mathbb{R}^3 \), understanding their independence and overlap helps describe the geometry of the space.
Spanning a Plane
In three-dimensional space, two vectors \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) can span a plane when they are linearly independent.
This means they do not lie along the same line, or in simpler terms, neither is a scaled version of the other. Geometrically, imagine two non-parallel vectors originating from the same point; the plane they span is shaped by all linear combinations of these vectors.

A plane consists of infinite lines, extending forever in two dimensions, within the 3D space. If \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) span a plane, any combination \( a\mathbf{v}_1 + b\mathbf{v}_2 \) will reside in this plane.
  • This is akin to having a piece of paper, which extends indefinitely but remains flat.
  • The independence of the vectors ensures the plane is fully described and utilized.
Spanning a Line
When vectors \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) are linearly dependent, they span merely a line.
Linearly dependent vectors mean one vector is a scalar multiple of the other, effectively pointing in the same direction.

Visualize this as two points placed along the same line but stretched or compressed versions of each other. In terms of space, they lack the diversity needed to extend across a second dimension.
  • With dependent vectors, any combination \( a\mathbf{v}_1 + b\mathbf{v}_2 \) collapses into a single line.
  • This line constitutes all possibilities of these vectors working together, unable to "escape" into spanning a plane.
Spanning \( \mathbb{R}^3 \)
Spanning the entire three-dimensional space \( \mathbb{R}^3 \) requires at least three linearly independent vectors because each vector contributes uniquely in its own dimension.
Think of it like creating a scaffold; you need three different pieces to stretch across length, width, and height without collapsing onto fewer dimensions.
Each vector must provide a distinct direction:
  • No pair of vectors is a combination or parallel, ensuring each direction in the scaffold is secure.
  • When combined, these vectors enable navigation everywhere within the 3D system, covering any point reachable within the space.
This is crucial when modeling real-world phenomena, where complete coverage of all positional factors ensures full spatial understanding.

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Most popular questions from this chapter

Do the vectors \((1,-1,0,0),(0,1,-1,0),(0,0,1,-1),(-1,0,0,1)\) span \(\mathbb{R}^{4}\) ?

Find a system of homogeneous linear equations in four variables for which the solution space is spanned by the set \(\\{(1,2,1,3),(1,3,2,4)\\}\).

(a) Which of the following subsets of the vector space \(M_{3}(\mathbb{R})\) of \(3 \times 3\) matrices with real entries are subspaces of \(M_{3}(\mathbb{R}) ?\) (i) \(\\{A: \operatorname{det} A=0\\}\); (ii) \(\\{A: \operatorname{det} A\) is an integer \(\\}\); (iii) \(\\{A: A\) is invertible\\}; (iv) \(\\{A: A\) is upper triangular \(\\}\); (v) \(\\{A: A\) is symmetric \\}; (vi) \(\\{A: A\) is skew symmetric \(\\}\). (b) Let \(B\) and \(C\) be fixed matrices in \(M_{3}(\mathbb{R})\). Let \(S\) be the set of all those \(3 \times 3\) matrices \(A\) for which \(B A C=0_{3 \times 3}\). Is \(S\) a subspace of \(M_{3}(\mathbb{R}) ?\)

(a) Let \(S\) be the subset \(\\{(x, 0): x \in \mathbb{R}\) and \(x>0\\}\) of \(\mathbb{R}^{2}\). Define ? and \(\square\) on \(S\) by: \((u, 0)\) ? \((v, 0)=(u v, 0)\) and \(\alpha \square(u, 0)=\left(u^{\alpha}, 0\right)\) for \(\alpha \in \mathbb{R}\). Show that \(S\) is a vector space with respect to ? and \(\square\). [Note, however, that we cannot accept \(S\) as a subspace of \(\mathbb{R}^{2}\) since, for example, the 'sum' \((u v, 0)\) of \((u, 0)\) and \((v, 0)\) regarded as elements of \(S\) is not the same as their sum, \((u+v, 0)\), when they are regarded as elements of \(\mathbb{R}^{2}\). [Similar objections apply to the two distinct scalar multiplications.] (b) \(\mathbb{R}[x]\), the (real) vector space of polynomials in \(x\) with real coefficients, is quite naturally regarded as a subset of \(C[x]\), the vector space of polynomials in \(x\) with complex coefficients. But we may not think of \(\mathbb{R}[x]\) as a subspace of \(\mathbb{C}[x]\) - unless \(\mathbb{C}[x]\) is being thought of (unnaturally?) as a vector space over \(\mathbb{R}\) rather than over \(\mathbb{C}\). Why not?

Find a set of vectors in \(\mathbb{R}^{4}\) which span the solution space of $$ \begin{array}{r} -x_{1}-x_{2}+x_{3}+2 x_{4}=0 \\ 2 x_{1}+x_{2}+x_{3}-x_{4}=0 \end{array} $$ Does the pair \(\\{(-16,33,5,6),(-17,33,6,5)\\}\) also span it?
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