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Chapter 6: Problem 13

Find either the nullity or the rank of \(T\) and then use the Rank Theorem to find the other. $$T: \mathscr{P}_{2} \rightarrow \mathbb{R} \text { defined by } T(p(x))=p^{\prime}(0)$$

Short Answer

Expert verified
Rank of \( T \) is 1, and nullity is 2.

Step by step solution

01

Understand the Linear Transformation

The given linear transformation \( T : \mathscr{P}_2 \rightarrow \mathbb{R} \) maps a polynomial \( p(x) \) to the real number \( p'(0) \). \( \mathscr{P}_2 \) represents the space of all polynomials of degree at most 2, i.e., \( a_2x^2 + a_1x + a_0 \). Hence \( T(p(x)) = a_1 \).
02

Determine the Dimension of the Domain

The space \( \mathscr{P}_2 \) consists of all polynomials of degree at most 2. Therefore, the dimension of \( \mathscr{P}_2 \) is 3, because the general polynomial \( a_2x^2 + a_1x + a_0 \) can be described by three coefficients \( a_2, a_1, \) and \( a_0 \).
03

Find the Rank of the Transformation

For a polynomial \( p(x) = a_2x^2 + a_1x + a_0 \), the derivative at zero, \( p'(0) = a_1 \). The image of \( T \) is all possible values of \( a_1 \) which is \( \mathbb{R} \). Since the derivative with respect to \( x \) gives a linear polynomial whose coefficients can be any real number, the image is one-dimensional. Thus, the rank of \( T \) is 1.
04

Use the Rank-Nullity Theorem

The Rank-Nullity Theorem states that \( \text{rank}(T) + \text{nullity}(T) = \text{dim}(\text{Domain}) \). Since \( \text{dim}(\mathscr{P}_2) = 3 \) and we found \( \text{rank}(T) = 1 \), we can find the nullity by: \( \text{nullity}(T) = 3 - 1 = 2 \).

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Key Concepts

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

Linear Transformation
A linear transformation is an essential concept in linear algebra. It defines a function between two vector spaces that preserves the operations of vector addition and scalar multiplication. In the context of this exercise, our linear transformation is denoted by \( T \), mapping from the space of polynomials, \( \mathscr{P}_2 \), to the real numbers, \( \mathbb{R} \).
This specific transformation, \( T(p(x)) = p'(0) \), takes a polynomial \( p(x) \) and maps it to its derivative evaluated at zero. Essentially, for the polynomial \( a_2x^2 + a_1x + a_0 \), the transformation gives the coefficient \( a_1 \).
  • It's called linear because it satisfies two properties: Additivity (\( T(u+v) = T(u) + T(v) \)) and Scalar Multiplicativity (\( T(cu) = cT(u) \)).
Polynomial Space
Polynomial space is the set of all polynomials of a certain degree, equipped with the operations of polynomial addition and scalar multiplication. In our problem, this polynomial space is \( \mathscr{P}_2 \).
\( \mathscr{P}_2 \) includes all polynomials of degree up to 2. Typical elements of this space are of the form \( a_2x^2 + a_1x + a_0 \), where \( a_2, a_1, \) and \( a_0 \) are constants. Each polynomial in this set can be uniquely represented by these coefficient triples.
  • This space is a vector space because it is closed under addition and scalar multiplication.
  • Understanding polynomial spaces is crucial, as they often serve as domain spaces for linear transformations.
Derivative of Polynomial
The derivative of a polynomial is a fundamental concept from calculus. It gives us the rate at which the polynomial's value changes with respect to changes in \( x \).
In the problem, this transformation evaluated the derivative of \( p(x) \) at zero, noted as \( p'(0) \). For \( p(x) = a_2x^2 + a_1x + a_0 \), the derivative \( p'(x) \) is \( 2a_2x + a_1 \). Thus, \( p'(0) = a_1 \).
  • This operation simplifies the polynomial and provides information on its slope at \( x = 0 \).
  • Understanding derivatives is critical as they are used in linear transformations to explore the behavior of polynomial spaces.
Dimension of Polynomial Space
The dimension of a polynomial space is the number of coefficients needed to uniquely describe any polynomial within the space. For \( \mathscr{P}_2 \), which includes polynomials of degree at most 2, the dimension is 3.
This is because a general polynomial in \( \mathscr{P}_2 \) can be written as \( a_2x^2 + a_1x + a_0 \). Each coefficient \( a_2, a_1, \) and \( a_0 \), acts like a vector, forming a coordinate system for the space.
Understanding the dimension:
  • The dimension tells us the minimum number of vectors needed to span the space.
  • In this problem, identifying the dimension helps apply the Rank-Nullity Theorem, which connects the dimensions of domain, kernel (null space), and image of a transformation.

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Most popular questions from this chapter

Find the dimension of the vector space \(V\) and give a basis for \(V\). $$V=\left\\{A \text { in } M_{22}: A B=B A\right\\}, \text { where } B=\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right]$$

Find the dimension of the vector space \(V\) and give a basis for \(V\). \(V=\left\\{A \text { in } M_{22}: A \text { is upper triangular }\right\\}\)

Find the solution of the differential equation that satisfies the given boundary condition(s). $$x^{\prime \prime}+4 x^{\prime}+4 x=0, x(0)=1, x^{\prime}(0)=1$$

Let \(S=\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{n}\right\\}\) be a spanning set for a vector space \(V\). Show that if \(\mathbf{v}_{n}\) is in \(\operatorname{span}\left(\mathbf{v}_{1}, \ldots, \mathbf{v}_{n-1}\right),\) then \(S^{\prime}=\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{n-1}\right\\}\) is still a spanning set for \(V\).

Show that \(T: \mathscr{P}_{n} \rightarrow \mathscr{P}_{n}\) defined by \(T(p(x))=p(x-2)\) is an isomorphism.
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