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Chapter 23: Problem 597

Show how the set of solutions to \(2 \mathrm{x}+\mathrm{y}=4\) can be expressed as a one-dimensional subspace of \(\mathrm{R}^{2}\).

Short Answer

Expert verified
The set of solutions to the equation \(2x + y = 4\) can be expressed as a one-dimensional subspace of \(\mathbb{R}^2\) using the general form of the line and vector operations. In this case, the line can be represented as: \(\begin{bmatrix} 0 \\ 4 \end{bmatrix} + t \begin{bmatrix} 1 \\ -2 \end{bmatrix}\), where \(t\) is a real number. This subspace consists of all scalar multiples of the direction vector added to a fixed point on the line.

Step by step solution

01

Rewrite the equation in the general form.

First, we need to rewrite the equation in the general form: \(y = -2x + 4\).
02

Determine a vector representing a point on the line.

Any point on the line described by the equation \(y = -2x + 4\) can be represented by a vector. One such point is when \(x = 0\), which gives us the vector \(\begin{bmatrix} 0 \\ 4 \end{bmatrix}\).
03

Determine a direction vector for the line.

A direction vector for the line can be determined by subtracting the coordinates of one point on the line from the coordinates of another point on the line. For example, the point \((1, 2)\) is on the line because \(2(1) + 2 = 4\). The direction vector can then be calculated using the two points: \[\begin{bmatrix} 1 - 0 \\ 2 - 4 \end{bmatrix} = \begin{bmatrix} 1 \\ -2 \end{bmatrix}\]
04

Express the line as a one-dimensional subspace.

To represent the line as a one-dimensional subspace, we can use the vector addition and scalar multiplication operations. The general form of the line can be written as: \[\begin{bmatrix} 0 \\ 4 \end{bmatrix} + t \begin{bmatrix} 1 \\ -2 \end{bmatrix},\] where \(t\) is a real number. This is a one-dimensional subspace of \(\mathbb{R}^2\), as it consists of all scalar multiples of the direction vector added to a fixed point on the line.

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Key Concepts

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

One-Dimensional Subspace
A one-dimensional subspace is a line that passes through the origin in a vector space. In this context, we focus on solutions to the equation \(2x + y = 4\). Although the solutions do not initially pass through the origin, we can express them as a one-dimensional subspace by rethinking the problem. We use vector equations to understand this better. In \(\mathbb{R}^2\), any line can be represented as a one-dimensional subspace through the process of combining vector addition and scalar multiplication. This method allows us to create any point on the line using
  • one fixed vector (a point on the line), and
  • scalar multiples of a direction vector.
This approach simplifies the analysis of linear systems and representation of lines within vector spaces.
Vector Addition
Vector addition is a fundamental concept in linear algebra. It allows us to combine vectors to form a new vector. When considering the line defined by \(y = -2x + 4\), vector addition combines our fixed vector \[ \begin{bmatrix} 0 \ 4 \end{bmatrix} \] with scalar multiples of our direction vector. This addition helps build the entire set of solutions for the equation. If you visualize the process:
  • The fixed vector points to a known position on the graph.
  • Adding helps shift along the direction of the line through the origin.
This operation is key to forming the one-dimensional subspace and understanding the behavior of linear equations in vector spaces.
Direction Vector
A direction vector provides insight into the orientation of the line. For \(y = -2x + 4\), it points in the direction where the line extends. From the solution, our direction vector is \[ \begin{bmatrix} 1 \ -2 \end{bmatrix} \]. This vector is obtained by examining two points on the line:
  • Using initial point \[ \begin{bmatrix} 0 \ 4 \end{bmatrix} \]
  • Comparing it to another point \[ \begin{bmatrix} 1 \ 2 \end{bmatrix} \]
The subtraction of these vectors gives the direction vector. It indicates how to move from any point on the line to another. In combining it with vector addition, one can traverse the entire line, demonstrating how direction vectors facilitate understanding of linear paths within vector spaces.
Scalar Multiplication
Scalar multiplication scales a vector by a factor, stretching or shrinking it. Within our context, to express the line as a subspace, we multiply the direction vector \[ \begin{bmatrix} 1 \ -2 \end{bmatrix} \]by a real number \(t\). This operation creates vectors that point along the same direction but vary in magnitude:
  • Positive values of \(t\) move us forward along the line.
  • Negative values of \(t\) move us backward.
  • Zero results in no movement away from the initial point.
This flexibility is what enables us to form any point on the line from the fixed point using vector addition. Scalar multiplication is crucial for creating line representations and understanding their spans within a vector space.

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Most popular questions from this chapter

Give an example to show that two non-parallel lines in \(\mathrm{R}^{3}\) need not intersect.

Find an equation of the hyperplane \(\mathrm{H}\) in the vector space \(\mathrm{R}^{4}\) if: (i) \(\mathrm{H}\) passes through \(\mathrm{P}=[3,-2,1,-4]\) and is normal to \(\mathrm{u}=[2,5,-6,-2] ;\) ii) \(\mathrm{H}\) passes through \(\mathrm{P}=[1,-2,3,5]\) and is parallel to the hyperplane \(\mathrm{H}^{\prime}\) determined by \(4 \mathrm{x}-5 \mathrm{y}+2 \mathrm{z}+\mathrm{w}=11\).

Determine the nearest point in \(\mathrm{U}=\) span \(\left\\{\mathrm{V}_{1}, \mathrm{~V}_{2}\right\\}\) to \(\mathrm{Y}\), where \(\mathrm{V}_{1}=(2,1,0) \mathrm{V}_{2}=(-1,2,0) \mathrm{Y}=(1,2,3)\)Geometrical Problems

Prove that the lines joining the mid-points of the opposite edges of a tetrahedron bisect each other.

Find a parametric representation of the line passing through \(\mathrm{P}\) and in the direction of u where (a) \(\mathrm{P}=(2,5)\) and \(\mathrm{u}=(-3,4)\); (b) \(\mathrm{P}=(4,-2,3,1)\) and \(\mathrm{u}=(2,5,-7,11)\).
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