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Chapter 6: Problem 20

Show that there is no linear transformation \(T: \mathbb{R}^{3} \rightarrow \mathscr{P}_{2}\) such that \\[\begin{aligned}T\left[\begin{array}{l}2 \\\1 \\\0\end{array}\right]=1+x, \quad T\left[\begin{array}{l}3 \\ 0 \\\2\end{array}\right]=2-x+x^{2} \\\T\left[\begin{array}{r}0 \\\6 \\\\-8\end{array}\right]=-2+2 x^{2}\end{aligned}\\]

Short Answer

Expert verified
No such linear transformation exists due to inconsistency in dimensions between model and image basis vectors.

Step by step solution

01

Define the transformation

A linear transformation \( T: \mathbb{R}^3 \to \mathscr{P}_2 \) assigns a unique polynomial of degree at most 2 to each vector in \( \mathbb{R}^3 \). Given transformation rules for some vectors, we can express any vector \( \begin{bmatrix} a \ b \ c \end{bmatrix} \) as a linear combination: \( a \begin{bmatrix} 2 \ 1 \ 0 \end{bmatrix} + b \begin{bmatrix} 3 \ 0 \ 2 \end{bmatrix} + c \begin{bmatrix} 0 \ 6 \ -8 \end{bmatrix} \) and find \( T \).
02

Use Linear Independence

The images given by the transformation are polynomials: \( 1+x \), \( 2-x+x^2 \), and \(-2+2x^2 \). Check if they are linearly independent. Assume \( \alpha(1 + x) + \beta(2 - x + x^2) + \gamma(-2 + 2x^2) = 0 \). This translates to polynomial equations: \( \alpha + 2\beta - 2\gamma = 0 \), \( \alpha - \beta = 0 \), and \( \beta + 2\gamma = 0 \).
03

Solve the System of Equations

The equations \( \alpha + 2\beta - 2\gamma = 0 \), \( \alpha - \beta = 0 \), and \( \beta + 2\gamma = 0 \) must be solved to check for consistency. Substitute \( \alpha = \beta \) (from \( \alpha - \beta = 0 \)) into the others: \( \beta + 2\gamma = 0 \) becomes \( \alpha = -2\gamma \). Substitute these into the first equation: \( \alpha + 2\alpha + 4\alpha = 0 \) implies \( 7\alpha = 0 \), thus \( \alpha = \beta = \gamma = 0 \).
04

Analyze the Dimension Problem

Since the polynomials \( 1+x \), \( 2-x+x^2 \), and \(-2+2x^2 \) are linearly independent, they form a basis for a vector space of dimension 3. However, \( \mathscr{P}_2 \) (polynomials of degree 鈮� 2) has dimension 3, while \( \mathbb{R}^3 \) also has dimension 3. Therefore, a linear transformation matching these mappings doesn't exist since the given set of target images span \( \mathscr{P}_2 \) completely with a different set of basis vectors.

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Key Concepts

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

Vector Spaces
In mathematics, a vector space is a collection of vectors, which may be added together and multiplied by numbers (scalars). The set of all vectors of a vector space satisfies the rules of addition and scalar multiplication, making it a fundamental structure in linear algebra. For example, in our problem,
  • \( \mathbb{R}^3 \) represents the vector space of all 3-dimensional vectors,
  • \( \mathscr{P}_2 \) represents the space of all polynomials with a degree at most 2.
Both these vector spaces are defined by their dimensionality and the operations (addition and scalar multiplication) they support. As such, understanding vector spaces means recognizing these spaces are not only about numbers or polynomials but also about system behaviors where those operations are defined.
Linear Independence
Linear independence is a crucial concept in linear algebra, especially when discussing vector spaces. In essence, a set of vectors is said to be linearly independent if none of the vectors can be expressed as a linear combination of the others. In our exercise,
  • The polynomials \( 1+x \), \( 2-x+x^2 \), and \(-2+2x^2 \) were checked for linear independence.
This was done by assuming a linear relation among them and forming a system of linear equations. The solution was that all coefficients were zero (\( \alpha = \beta = \gamma = 0 \)), confirming their independence. This means each polynomial brings new information or structure to the vector space \( \mathscr{P}_2 \), showing how independence is vital in understanding the basis of a space.
Polynomials
Polynomials are expressions of finite length constructed from variables and constants using only the operations of addition, subtraction, multiplication, and non-negative integer exponents. In our exercise, we refer to polynomials in the space \( \mathscr{P}_2 \), which contains polynomials of degree at most 2.
  • Examples include \( 1+x \), \( 2-x+x^2 \), and \(-2+2x^2 \).
Each polynomial can be viewed as a vector in the vector space of polynomials, making the operations on these polynomials analogous to vector operations like addition and scalar multiplication. Understanding polynomials as vectors helps in visualizing and solving problems related to polynomial algebra in higher dimensions.
Dimension
Dimension in the context of a vector space refers to the number of vectors in a basis for the space. It gives a measure of the "size" of the space in terms of freedom or directionality. For example,
  • \( \mathbb{R}^3 \) has a dimension of 3 because it takes three vectors to describe any vector in this space.
  • \( \mathscr{P}_2 \) also has a dimension of 3, as it can be spanned by three linearly independent polynomials like \( 1+x \), \( 2-x+x^2 \), and \(-2+2x^2 \).
In our exercise, the given polynomials, being linearly independent, confirm they fully span the space of \( \mathscr{P}_2 \), thus matching its dimension. However, they form a basis that does not align with the basis defined over \( \mathbb{R}^3 \), explaining why a linear transformation aligning these dimensions is impossible. Understanding dimension clarifies the underlying structure and limits of transformations between spaces.

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