Question

Given a line \(\bar{x}=\bar{a}+t \vec{u}(\vec{u}=\bar{b}-\bar{a} \neq \overrightarrow{0})\) in \(E^{n},\) define \(f: E^{1} \rightarrow E^{n}\) by $$ f(t)=\bar{a}+t \vec{u} \text { for } t \in E^{1} $$ Show that \(L[\bar{a}, \bar{b}]\) is exactly the \(f\) -image of the interval [0,1] in \(E^{1},\) with \(f(0)=a\) and \(f(1)=b\), while \(f\left[E^{1}\right]\) is the entire line. Also show that \(f\) is one to one.

Short answer

The image of \([0,1]\) under \(f\) is the line segment \(L[\bar{a}, \bar{b}]\), and \(f\) is one-to-one.

Step-by-step solution

Understanding the Line Equation

The line is given by the expression \( \bar{x} = \bar{a} + t \vec{u} \), where \( \vec{u} = \bar{b} - \bar{a} \) and \( \vec{u} eq \vec{0} \). This is a parametric form of a line in \( E^n \), starting at point \( \bar{a} \) and extending in the direction of the vector \( \vec{u} \).

Defining f for the Interval [0,1]

The function \( f(t) = \bar{a} + t \vec{u} \) maps \( t \) from \( E^1 \) (real numbers) into \( E^n \). Specifically, for \( t \) in the interval \([0, 1]\), \( f(0) = \bar{a} \) and \( f(1) = \bar{b} \). Thus, \( f \) traces the segment \( L[\bar{a}, \bar{b}] \) from \( \bar{a} \) to \( \bar{b} \).

f-image of [0,1] is the Line Segment

For \( t \) in \([0, 1]\), \( f(t) = \bar{a} + t(\bar{b} - \bar{a}) \) represents all points on the line segment between \( \bar{a} \) and \( \bar{b} \), showing \( L[\bar{a}, \bar{b}] \) is precisely the \( f \)-image of \([0,1] \).

f-Image of Entire E^1

The function \( f(t) = \bar{a} + t \vec{u} \) maps \( t \) from all real numbers \( E^1 \) to points on the entire line in \( E^n \), since \( t \) can take any real value, extending \( L \) beyond \( \bar{a} \) and \( \bar{b} \).

Showing f is One-to-One

\( f \) is one-to-one if different values of \( t \) map to different points on the line. Assume \( f(t_1) = f(t_2) \). Then \( \bar{a} + t_1 \vec{u} = \bar{a} + t_2 \vec{u} \) implies \( t_1 \vec{u} = t_2 \vec{u} \). Because \( \vec{u} eq \vec{0} \), it follows that \( t_1 = t_2 \). Thus, \( f \) is one-to-one.

Key concepts

Image of Function

The *image* of a function is the set of all output values it can produce. When considering the function \( f(t) = \bar{a} + t \vec{u} \), it's important to understand how this maps input values \( t \) from one space to points in a vector space. Specifically, when \( t \) is restricted to the interval \([0, 1]\), the image includes only the segment of the line from \( \bar{a} \) to \( \bar{b} \). - For \( t = 0 \), the output is \( \bar{a} \).- For \( t = 1 \), the output is \( \bar{b} \). Thus, the function's image from inputs in \([0,1]\) exactly traces the path from \( \bar{a} \) to \( \bar{b} \). The complete set of outputs from all real numbers \( t \) represents the full line, as \( t \) can be any real value, extending infinitely in both directions.

Vector Spaces

Vector spaces are essential in understanding functions like \( f(t) = \bar{a} + t \vec{u} \) in their true form. These spaces provide a framework where we can add vectors together and multiply them by scalars. - A vector space in \( E^n \) means the space has \( n \) dimensions, which can be visualized as containing points, lines, planes, and other geometrical entities based on the number of dimensions. - The vector \( \vec{u} = \bar{b} - \bar{a} \) signifies direction and distance from \( \bar{a} \) to \( \bar{b} \), showing the line segment's path within the vector space.This function uses the properties of vector spaces to map a one-dimensional input into an \( n \)-dimensional space. The principles of linearity (such as adding vectors or scaling them) apply here and help explain how the function progresses along the line defined.

One-to-One Mapping

A *one-to-one mapping*, or bijective function, means that each input in the domain has a unique output in the range, and vice versa. For the function \( f(t) = \bar{a} + t \vec{u} \), showing it is one-to-one involves demonstrating that distinct \( t \) values produce distinct points on the line.- Assume that \( f(t_1) = f(t_2) \). If this implies \( t_1 = t_2 \), then the function is indeed one-to-one. - In the equation \( \bar{a} + t_1 \vec{u} = \bar{a} + t_2 \vec{u} \), simplifying this gives \( t_1 \vec{u} = t_2 \vec{u} \).Since \( \vec{u} eq \vec{0} \), we conclude that \( t_1 = t_2 \). Therefore, every different \( t \) produces a unique point on the line, confirming the function's one-to-one nature. This is a crucial property for ensuring the mapped line is functionally distinct at each point.