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Describe all linear transformations \(T: \mathbb{R} \rightarrow V\).

Short Answer

Expert verified
Each linear transformation corresponds to a vector \( v \in V \), with \( T(x) = x \cdot v \).

Step by step solution

01

Understanding Linear Transformations

A linear transformation is a function between two vector spaces that preserves vector addition and scalar multiplication. In this problem, we are asked to find linear transformations from the real numbers \( \mathbb{R} \) to a vector space \( V \).
02

Characterizing the Domain

The domain of the linear transformation \( T \) is \( \mathbb{R} \), the set of real numbers. Any element \( x \in \mathbb{R} \) can be expressed as \( x = 1 \cdot x \), indicating that every element is essentially a scalar multiple of the real number 1.
03

Exploring the Mapping from the Domain

For a transformation \( T: \mathbb{R} \rightarrow V \), the function needs to map each real number to an element in \( V \). Since \( T \) is linear, for \( x \in \mathbb{R} \), it follows that \( T(x) = T(1 imes x) = x \cdot T(1) \).
04

Determining the Image of 1

To fully determine \( T \), it is enough to determine \( T(1) \) because every \( x \in \mathbb{R} \) can be written as \( x = 1 imes x \), and thus \( T(x) = x \cdot T(1) \). The vector \( T(1) \) can be any vector \( v \in V \), as \( T \) is defined to be linear.
05

Formulating the General Linear Transformation

Since \( T \) is determined entirely by \( T(1) \), for any vector \( v \in V \), \( T \) can be defined by \( T(x) = x imes v \) for all \( x \in \mathbb{R} \). Thus, every linear transformation from \( \mathbb{R} \) to \( V \) corresponds to a choice of vector \( v \in V \).

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Key Concepts

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

Vector Spaces
Vector spaces are fundamental concepts in linear algebra, and they provide a framework for doing mathematics in a structured way. A vector space consists of a collection of vectors, which can be added together and multiplied by scalars. Importantly, this set must satisfy a few specific rules or properties, which are designed to ensure consistency and predictability.

First, vector addition must be commutative and associative. This means that the order in which you add vectors doesn't matter, and you can group them in any way you like without changing the result. There must also be a zero vector that acts as an additive identity, such that adding it to any vector doesn't change the original vector.

  • Commutative Property: For any two vectors \( extbf{u} \) and \( extbf{v} \), \( extbf{u} + extbf{v} = extbf{v} + extbf{u} \).
  • Associative Property: For any three vectors \( extbf{u} \), \( extbf{v} \), and \( extbf{w} \), \( ( extbf{u} + extbf{v}) + extbf{w} = extbf{u} + ( extbf{v} + extbf{w}) \).
  • Zero Vector: There exists a vector \( extbf{0} \) such that for any vector \( extbf{u} \), \( extbf{u} + extbf{0} = extbf{u} \).
Vector spaces are critical when dealing with linear transformations because these transformations map one vector space onto another, preserving the vector space structure.
Scalar Multiplication
Scalar multiplication is a key operation in the context of vector spaces and is pivotal in understanding linear transformations. A scalar in this context is typically a real number when working with real vector spaces, but it can be extended to other fields as well. This operation allows us to scale vectors by multiplying them by scalars.

  • When you multiply a vector by a scalar, the direction of the vector remains unchanged, unless the scalar is negative, in which case the direction is reversed.
  • Scalar multiplication is consistent with the distributive property, meaning \( a(\textbf{u} + \textbf{v}) = a\textbf{u} + a\textbf{v} \).
  • It is also associative with respect to the scalar factors: \( (ab)\textbf{v} = a(b\textbf{v}) \).
In linear transformations, scalar multiplication ensures that when you scale an input vector, the output is also scaled in a predictable way. Specifically, if you know how a transformation \( T \) acts on one element of \( \mathbb{R} \) (like 1, since every number is a multiple of 1), you can determine how it acts on any other scalar multiple of that element by simply adjusting the scale accordingly.
Real Numbers
Real numbers, denoted as \( \mathbb{R} \), are a fundamental part of mathematics and serve as the domain for many transformations and functions, especially in linear algebra. They include all the numbers that can be found on the number line—integers, fractions, and irrational numbers.

Real numbers are used extensively in linear transformations as they allow us to map elements from one vector space to another in a continuous and consistent way. Each real number can be thought of as a scalar, which can modify vectors in vector spaces through scalar multiplication.

  • Continuous Set: Real numbers form a continuous line without gaps, which makes them ideal for modeling physical phenomena.
  • Field Properties: Real numbers follow the usual rules of arithmetic, including addition, subtraction, multiplication, and division (except by zero).
When dealing with linear transformations from \( \mathbb{R} \) to a vector space, every real number can be seen as a potential scaling factor—each "scales" how the transformation \( T \) will alter a vector, depending on how \( T[1] \) is defined in the vector space \( V \). Thus, understanding the behavior of real numbers is key to understanding the behavior of linear transformations that involve these numbers.

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Most popular questions from this chapter

Show that every polynomial \(f(x)\) in \(\mathbf{P}_{n-1}\) can be written as \(f(x)=p(x+1)-p(x)\) for some polynomial \(p(x)\) in \(\mathbf{P}_{n} .\) [Hint : Define \(T: \mathbf{P}_{n} \rightarrow \mathbf{P}_{n-1}\) by \(T[p(x)]=p(x+1)-p(x) .]\)

Let \(S: V \rightarrow W\) and \(T: V \rightarrow W\) be linear transformations. Given \(a\) in \(\mathbb{R},\) define functions \((S+T): V \rightarrow W\) and \((a T): V \rightarrow W\) by \((S+T)(\mathbf{v})=\) \(S(\mathbf{v})+T(\mathbf{v})\) and \((a T)(\mathbf{v})=a T(\mathbf{v})\) for all \(\mathbf{v}\) in \(V\). Show that \(S+T\) and \(a T\) are linear transformations.

Let \(\left\\{\mathbf{e}_{1}, \mathbf{e}_{2}\right\\}\) be the standard basis of \(\mathbb{R}^{2}\). Is it possible to have a linear transformation \(T\) such that \(T\left(\mathbf{e}_{1}\right)\) lies in \(\mathbb{R}\) while \(T\left(\mathbf{e}_{2}\right)\) lies in \(\mathbb{R}^{2}\) ? Explain your answer.

For each matrix \(A,\) find a basis for the kernel and image of \(T_{A}\), and find the rank and nullity of \(T_{A}\) a. \(\left[\begin{array}{rrrr}1 & 2 & -1 & 1 \\ 3 & 1 & 0 & 2 \\ 1 & -3 & 2 & 0\end{array}\right]\) b. \(\left[\begin{array}{rrrr}2 & 1 & -1 & 3 \\ 1 & 0 & 3 & 1 \\ 1 & 1 & -4 & 2\end{array}\right]\) c. \(\left[\begin{array}{rrr}1 & 2 & -1 \\ 3 & 1 & 2 \\ 4 & -1 & 5 \\ 0 & 2 & -2\end{array}\right]\) d. \(\left[\begin{array}{rrr}2 & 1 & 0 \\ 1 & -1 & 3 \\ 1 & 2 & -3 \\ 0 & 3 & -6\end{array}\right]\)

If \(T: \mathbf{M}_{n n} \rightarrow \mathbb{R}\) is any linear transformation satisfying \(T(A B)=T(B A)\) for all \(A\) and \(B\) in \(\mathbf{M}_{n n}\) show that there exists a number \(k\) such that \(T(A)=k \operatorname{tr} A\) for all A. (See Lemma 5.5.1.) [Hint: Let \(E_{i j}\) denote the \(n \times n\) matrix with 1 in the \((i, j)\) position and zeros elsewhere. Show that \(E_{i k} E_{l j}=\left\\{\begin{array}{cl}0 & \text { if } k \neq l \\\ E_{i j} & \text { if } k=l\end{array}\right.\). Use this to show that \(T\left(E_{i j}\right)=0\) if \(i \neq j\) and \(T\left(E_{11}\right)=T\left(E_{22}\right)=\cdots=T\left(E_{n n}\right) .\) Put \(k=T\left(E_{11}\right)\) and use the fact that \(\left\\{E_{i j} \mid 1 \leq i, j \leq n\right\\}\) is a basis of \(\mathbf{M}_{n n}\).

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