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Chapter 7: Problem 18

Find the eigenvalues \(\lambda_{i}\) and the corresponding eigenspaces of the linear transformation \(T\). Determine whether the linear transformation is diagonalizable. \(T\) defined on \(\mathbb{R}^{2}\) by \(T([x, y])=\) \([x-y,-x+y]\)

Short Answer

Expert verified
Eigenvalues: 0 and 2; Eigenspaces: Span\{[1, 1]\} and Span\{[1, -1]\}. The transformation is diagonalizable.

Step by step solution

01

Represent Linear Transformation as a Matrix

The linear transformation \( T \) can be expressed as a matrix when applied to a vector \([x, y]\). For \( T([x, y]) = [x-y, -x+y] \), we can write this transformation as the matrix \( A = \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix} \).
02

Set Up the Characteristic Equation

To find the eigenvalues, we need to solve the characteristic equation, which is \( \det(A - \lambda I) = 0 \). For our matrix \( A \), this translates to solving \( \det\left(\begin{bmatrix} 1-\lambda & -1 \ -1 & 1-\lambda \end{bmatrix}\right) = 0 \).
03

Calculate the Determinant

The determinant of matrix \( \begin{bmatrix} 1-\lambda & -1 \ -1 & 1-\lambda \end{bmatrix} \) is \((1-\lambda)(1-\lambda) - (-1)(-1) = (1-\lambda)^2 - 1 \). Expanding this gives \( \lambda^2 - 2\lambda = 0 \).
04

Solve the Characteristic Equation

Factor the quadratic equation \( \lambda^2 - 2\lambda = 0 \) to \( \lambda(\lambda - 2) = 0 \). The solutions to this equation, i.e., the eigenvalues, are \( \lambda_1 = 0 \) and \( \lambda_2 = 2 \).
05

Find Eigenspaces for Each Eigenvalue

For \( \lambda_1 = 0 \), solve \( (A - 0I)x = 0 \), leading to \( \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix}\begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \). Solving the system shows all vectors as \([x, x] \), forming the eigenspace \( \text{Span}\{[1, 1]\} \). For \( \lambda_2 = 2 \), solve \( (A - 2I)x = 0 \) to get \( \begin{bmatrix} -1 & -1 \ -1 & -1 \end{bmatrix}\begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \). Simplifying shows all vectors as \([1, -1] \), forming the eigenspace \( \text{Span}\{[1, -1]\} \).
06

Determine if the Transformation is Diagonalizable

A linear transformation is diagonalizable if there are \( n \) linearly independent eigenvectors, where \( n \) is the dimension of the space. Since we have two distinct eigenvalues \( \lambda_1 = 0 \) and \( \lambda_2 = 2 \), each with a corresponding linearly independent eigenvector, the transformation is diagonalizable.

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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 a fundamental concept in linear algebra that involves shifting, scaling, rotating, or flipping a vector in a way that preserves straight lines and parallelism. Consider the transformation
  • It can be defined as a function that maps vectors from one vector space to another.
  • Importantly, the transformation should satisfy two main properties: additivity and homogeneity. In simpler terms, this means the transformation of a sum is the same as the sum of transformations, and that scaling a vector before or after transformation gives the same result.
For the exercise given, we look at the transformation \( T([x, y]) = [x-y, -x+y] \), operating in the space \( \mathbb{R}^2 \). The task involves determining how this transformation affects vectors when expressed in terms of a matrix, which is a crucial step to find its eigenvalues and eigenvectors. This matrix provides a useful mathematical framework for understanding and manipulating linear transformations.
Matrix Representation
Matrix representation is about writing a linear transformation as a matrix, so we can easily manipulate and analyze it mathematically. This involves representing a function in terms of a grid of numbers that translates each vector from its original position to the transformed position.
  • For a transformation like \( T([x, y]) = [x-y, -x+y] \), the matrix form simplifies calculations and provides clarity.
  • The matrix \( A = \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix} \) captures the effect of the transformation.
With this matrix, we can perform operations like finding eigenvalues, which help in understanding the transformation's behavior. The matrix provides a clear structure and allows us to apply standard linear algebra techniques, including forming and solving the characteristic equation.
Characteristic Equation
The characteristic equation is a polynomial equation derived from a matrix that encapsulates information about a linear transformation's eigenvalues. These values are crucial for understanding the dynamics of the transformation.
  • We derive the characteristic equation from \( \det(A - \lambda I) = 0 \). Here, \( \lambda \) is the eigenvalue we are looking for, \( A \) is the matrix, and \( I \) is the identity matrix.
  • For our matrix \( A = \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix} \), the determinant calculation results in \( \lambda^2 - 2\lambda = 0 \).
Solving this equation helps us discover the eigenvalues \( \lambda_1 = 0 \) and \( \lambda_2 = 2 \), key numbers that indicate which directions the transformation simply scales or flips the vector without rotating it.
Diagonalizability
Diagonalizability is a mathematical property that tells us whether we can express a linear transformation as a diagonal matrix in some basis. This property helps simplify matrix operations significantly, including raising matrices to powers, which can be computationally intensive with non-diagonal matrices.
  • For a linear transformation to be diagonalizable, it must have enough linearly independent eigenvectors to fill its space.
  • In our \(2\)-dimensional space, this means we need 2 linearly independent eigenvectors.
In the exercise, we find two such eigenvectors corresponding to eigenvalues \(\lambda_1 = 0\) and \(\lambda_2 = 2\). This confirms that the transformation is indeed diagonalizable. Diagonalizability implies efficiency, as operations become straightforward due to the simple diagonal structure.

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Most popular questions from this chapter

Find the matrix representations \(R_{B}\) and \(R_{B}\) and an invertible matrix \(C\) such that \(R_{B}=C^{-1} R_{B} C\) for the linear trazsjormation \(T\) of the given vector space with the indicated ordered bases \(B\) and \(B^{\prime}\). \(T: W \rightarrow W\), where \(W=\operatorname{sp}\left(e^{x}, x e^{x}\right)\) and \(T\) is the derivative transformation; \(B=\left(e^{x}, x e^{x}\right)\), \(B^{\prime}=\left(2 x e^{x}, 3 e^{x}\right)\)

Find the eigenvalues \(\lambda_{i}\) and the corresponding eigenspaces of the linear transformation \(T\). Determine whether the linear transformation is diagonalizable. \(T\) defined on \(\mathbf{R}^{3}\) by \(T\left(\left[x_{1}, x_{2}, x_{3}\right]\right)=\) \(\left[5 x_{1},-5 x_{1}+3 x_{2}-5 x_{3},-3 x_{1}-2 x_{2}\right]\)

Find the change-of-coordinates matrix (a) from \(B\) to \(B^{\prime}\), and (b) from \(B^{\prime}\) to \(B\). Verify that these matrices are inverses of each other. \(B=([2,3,1],[1,2,0],[2,0,3])\) and \(B^{\prime}=\) \(([1,0,0],[0,1,0],[0,0,1])\) in \(\mathbf{R}^{3}\)

Are a review of Section 3.3. In Exercises 1-6, find the coordinate vector of the given vector relative to the given ordered basis. \(\cos 2 x\) in \(\operatorname{sp}\left(\sin ^{2} x, \cos ^{2} x\right)\) relative to \(\left(\sin ^{2} x, \cos ^{2} x\right)\)

Are a review of Section 3.3. In Exercises 1-6, find the coordinate vector of the given vector relative to the given ordered basis. \(x+x^{4}\) in \(P_{4}\) relative to \(\left(1,2 x-1, x^{3}+x^{4}, 2 x^{3}, x^{2}+2\right)\)
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