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Suppose \(T: \mathbf{P}_{2} \rightarrow \mathbb{R}^{2}\) is a linear transformation. If \(B=\left\\{1, x, x^{2}\right\\}\) and \(D=\\{(1,1),(0,1)\\},\) find the action of \(T\) given: a. \(M_{D B}(T)=\left[\begin{array}{rrr}1 & 2 & -1 \\ -1 & 0 & 1\end{array}\right]\) b. \(M_{D B}(T)=\left[\begin{array}{rrr}2 & 1 & 3 \\ -1 & 0 & -2\end{array}\right]\)

Short Answer

Expert verified
In Case a: \(T(1)=(1,-1)\), \(T(x)=(2,0)\), \(T(x^2)=(-1,1)\); in Case b: \(T(1)=(2,-1)\), \(T(x)=(1,0)\), \(T(x^2)=(3,-2)\).

Step by step solution

01

Understand the Problem

We are given two situations where a linear transformation \(T: \mathbf{P}_{2} \rightarrow \mathbb{R}^{2}\) is described by different matrix representations \(M_{D B}(T)\). Our task is to find the action of \(T\) on the basis elements \(1, x, x^2\) of \(\mathbf{P}_{2}\).
02

Matrix Representation Insights

The matrix \(M_{D B}(T)\) represents how the linear transformation \(T\) acts on the basis \(B\) of \(\mathbf{P}_{2}\) when the output is expressed in terms of the basis \(D\) of \(\mathbb{R}^{2}\). Each column of the matrix corresponds to the image of a basis element \(B\) under \(T\).
03

Calculate the Action in Case a

For \(M_{D B}(T)=\begin{bmatrix}1 & 2 & -1 \ -1 & 0 & 1\end{bmatrix}\):- The first column \(\begin{bmatrix}1\-1\end{bmatrix}\) corresponds to \(T(1) = (1,-1)\).- The second column \(\begin{bmatrix}2\0\end{bmatrix}\) corresponds to \(T(x) = (2,0)\).- The third column \(\begin{bmatrix}-1\1\end{bmatrix}\) corresponds to \(T(x^2) = (-1,1)\).
04

Calculate the Action in Case b

For \(M_{D B}(T)=\begin{bmatrix}2 & 1 & 3 \ -1 & 0 & -2\end{bmatrix}\):- The first column \(\begin{bmatrix}2\-1\end{bmatrix}\) corresponds to \(T(1) = (2,-1)\).- The second column \(\begin{bmatrix}1\0\end{bmatrix}\) corresponds to \(T(x) = (1,0)\).- The third column \(\begin{bmatrix}3\-2\end{bmatrix}\) corresponds to \(T(x^2) = (3,-2)\).

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Key Concepts

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

Matrix Representation in Linear Transformations
A matrix representation of a linear transformation provides a structured way to understand how the transformation acts on vectors or basis elements. In our example, we have a linear transformation \( T: \mathbf{P}_{2} \to \mathbb{R}^2 \). This means we are mapping from a space of polynomials up to degree 2, \( \mathbf{P}_{2} \), to \( \mathbb{R}^2 \), a 2-dimensional vector space. The representation matrix \( M_{D B}(T) \) helps us easily determine the output of our transformation when applied to the basis elements.
  • The matrix \( M_{D B}(T) \) is rectangular and has as many rows as there are dimensions in \( \mathbb{R}^2 \) and as many columns as there are elements in the basis of \( \mathbf{P}_{2} \).
  • Each column in the matrix represents the image of a basis element from \( \mathbf{P}_{2} \) under the transformation \( T \) given in terms of the basis \( D \) of \( \mathbb{R}^2 \).
For example, in scenario (a), the first column \( \begin{bmatrix} 1 \ -1 \end{bmatrix} \) tells us that \( T(1) = (1, -1) \). Breaking down the entries allows a detailed understanding of transformations.
Understanding Basis Elements
Basis elements are the building blocks of vector spaces. A basis is a set of vectors that are linearly independent and span the entire vector space. This means any element in the space can be expressed as a linear combination of these basis elements. For \( \mathbf{P}_{2} \), the basis \( B = \{1, x, x^2 \} \) implies each polynomial of degree at most 2 can be formed as \( a + bx + cx^2 \).
  • Each polynomial of degree 2 can thus be uniquely represented by a combination of these basis elements.
  • The transformation \( T \) acts on each of these basis elements, and the images are recorded in the columns of its matrix representation \( M_{D B}(T) \).
In our problem, knowing how \( 1, x, x^2 \) are transformed helps us understand the entire action of \( T \), since any other polynomial combination can also be computed using these results.
Exploring Polynomial Vector Spaces
Polynomial vector spaces are fascinating structures in linear algebra. They consist of polynomials of a specific degree, forming a vector space through polynomial addition and scalar multiplication. In \( \mathbf{P}_{2} \), we consider polynomials up to degree 2, such as \( a + bx + cx^2 \).
  • \( \mathbf{P}_{2} \) is a 3-dimensional vector space, reflected by the three basis elements \( \{1, x, x^2\} \).
  • This space allows linear transformations to map these polynomials to other vector spaces, like \( \mathbb{R}^2 \) in our question.
Understanding polynomial vector spaces gives insight into how complex transformations like \( T \) are formulated and represented, allowing detailed computations and predictions about their behavior. It's a broad and intriguing field, underscoring many algebraic concepts and practical applications.

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Most popular questions from this chapter

Find \(P_{D \leftarrow B}\) if \(B=\left\\{\mathbf{b}_{1}, \mathbf{b}_{2}, \mathbf{b}_{3}, \mathbf{b}_{4}\right\\}\) and \(D=\left\\{\mathbf{b}_{2}, \mathbf{b}_{3}, \mathbf{b}_{1}, \mathbf{b}_{4}\right\\} .\) Change matrices arising when the bases differ only in the order of the vectors are called permutation matrices.

In each case, show that \(V=U \oplus W\). a. \(\begin{aligned} V &=\mathbb{R}^{4}, U=\operatorname{span}\\{(1,1,0,0),(0,1,1,0)\\}, \\ W &=\operatorname{span}\\{(0,1,0,1),(0,0,1,1)\\} \end{aligned}\) b. \(V=\mathbb{R}^{4}, U=\\{(a, a, b, b) \mid a, b\) in \(\mathbb{R}\\}\) \(W=\\{(c, d, c,-d) \mid c, d\) in \(\mathbb{R}\\}\) c. \(V=\mathbf{P}_{3}, U=\\{a+b x \mid a, b\) in \(\mathbb{R}\\},\) \(W=\left\\{a x^{2}+b x^{3} \mid a, b\right.\) in \(\left.\mathbb{R}\right\\}\) \(\begin{aligned} \text { d. } V &=\mathbf{M}_{22}, U=\left\\{\left[\begin{array}{ll}a & a \\ b & b\end{array}\right] \mid a, b \text { in } \mathbb{R}\right\\} \\ W &=\left\\{\left[\begin{array}{rl}a & b \\\ -a & b\end{array}\right] \mid a, b \text { in } \mathbb{R}\right\\} \end{aligned}\)

Let \(T: V \rightarrow V\) be an operator such that \(T^{2}=c^{2}, c \neq 0\) a. Show that \(V=U_{1} \oplus U_{2}\), where $$\begin{array}{l} U_{1}=\\{\mathbf{v} \mid T(\mathbf{v})=c \mathbf{v}\\} \text { and } U_{2}=\\{\mathbf{v} \mid T(\mathbf{v})=-c \mathbf{v}\\} \\\\{\left[\text { Hint: } \mathbf{v}=\frac{1}{2 c}\\{[T(\mathbf{v})+c \mathbf{v}]-[T(\mathbf{v})-c \mathbf{v}]\\} .\right]}\end{array}$$ b. If \(\operatorname{dim} V=n\), show that \(V\) has a basis \(B\) such that \(M_{B}(T)=\left[\begin{array}{cc}c l_{k} & 0 \\ 0 & -c I_{n-k}\end{array}\right]\) for some \(k\) c. If \(A\) is an \(n \times n\) matrix such that \(A^{2}=c^{2} I, c \neq 0\), show that \(A\) is similar to \(\left[\begin{array}{cc}c I_{k} & 0 \\ 0 & -c I_{n-k}\end{array}\right]\) for some \(k\).

Let \(T: V \rightarrow V\) be an operator where \(T^{3}=0 .\) If \(\mathbf{u} \in V\) and \(U=\operatorname{span}\left\\{\mathbf{u}, T(\mathbf{u}), T^{2}(\mathbf{u})\right\\},\) show that \(U\) is \(T\) -invariant and has dimension \(3 .\)

Let \(T: \mathbf{P}_{2} \rightarrow \mathbb{R}^{3}\) be defined by \(T(p)=(p(0), p(1), p(2))\) for all \(p\) in \(\mathbf{P}_{2}\). Let \(B=\left\\{1, x, x^{2}\right\\}\) and \(D=\\{(1,0,0),(0,1,0),(0,0,1)\\}\) . a. Show that \(M_{D B}(T)=\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 2 & 4\end{array}\right]\) and conclude that \(T\) is an isomorphism. b. Generalize to \(T: \mathbf{P}_{n} \rightarrow \mathbb{R}^{n+1}\) where \(T(p)=\left(p\left(a_{0}\right), \quad p\left(a_{1}\right), \ldots, \quad p\left(a_{n}\right)\right) \quad\) and \(a_{0}, a_{1}, \ldots, a_{n}\) are distinct real numbers.

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