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Chapter 5: Problem 30

In Exercises 30 and \(31,\) find the \(\mathcal{B}\) -matrix for the transformation \(\mathbf{x} \mapsto A \mathbf{x}\) when \(\mathcal{B}=\left\\{\mathbf{b}_{1}, \mathbf{b}_{2}, \mathbf{b}_{3}\right\\}\) $$ \begin{array}{l}{A=\left[\begin{array}{rrr}{-14} & {4} & {-14} \\ {-33} & {9} & {-31} \\ {11} & {-4} & {11}\end{array}\right]} \\\ {\mathbf{b}_{1}=\left[\begin{array}{r}{-1} \\ {-2} \\ {1}\end{array}\right], \mathbf{b}_{2}=\left[\begin{array}{r}{-1} \\ {-1} \\ {1}\end{array}\right], \mathbf{b}_{3}=\left[\begin{array}{r}{-1} \\ {-2} \\\ {0}\end{array}\right]}\end{array} $$

Short Answer

Expert verified
The \( \mathcal{B} \)-matrix is formed using the coefficients obtained by expressing each \( A\mathbf{b}_i \) as a linear combination of \( \mathbf{b}_1, \mathbf{b}_2, \mathbf{b}_3 \).

Step by step solution

01

Understanding the Transformation

The transformation is given as \( \mathbf{x} \mapsto A \mathbf{x} \). This means for any vector \( \mathbf{x} \), when we apply matrix \( A \), we transform \( \mathbf{x} \) into \( A\mathbf{x} \). We need to express this transformation with respect to basis \( \mathcal{B} \).
02

Define the Standard Basis Vectors

For a 3-dimensional vector space, the standard basis vectors are \( \mathbf{e}_1 = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix} \), \( \mathbf{e}_2 = \begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix} \), \( \mathbf{e}_3 = \begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix} \). These vectors form the standard basis, \( A\mathbf{e}_i \) are the columns of \( A \).
03

Compute \( A \mathbf{b}_1, A \mathbf{b}_2, A \mathbf{b}_3 \)

We must compute \( A \mathbf{b}_1 \), \( A \mathbf{b}_2 \), and \( A \mathbf{b}_3 \) by multiplying each basis vector with matrix \( A \). This gives us the transformed vectors in the standard basis.- Compute \( A \mathbf{b}_1 \): \[ A \mathbf{b}_1 = A \begin{bmatrix} -1 \ -2 \ 1 \end{bmatrix} = \begin{bmatrix} (-14)(-1) + 4(-2) - 14(1) \ (-33)(-1) + 9(-2) - 31(1) \ (11)(-1) - 4(-2) + 11(1) \end{bmatrix} = \begin{bmatrix} 12 \ 50 \ -4 \end{bmatrix} \]- Compute \( A \mathbf{b}_2 \): \[ A \mathbf{b}_2 = A \begin{bmatrix} -1 \ -1 \ 1 \end{bmatrix} = \begin{bmatrix} (-14)(-1) + 4(-1) - 14(1) \ (-33)(-1) + 9(-1) - 31(1) \ (11)(-1) - 4(-1) + 11(1) \end{bmatrix} = \begin{bmatrix} 4 \ 55 \ -6 \end{bmatrix} \]- Compute \( A \mathbf{b}_3 \): \[ A \mathbf{b}_3 = A \begin{bmatrix} -1 \ -2 \ 0 \end{bmatrix} = \begin{bmatrix} (-14)(-1) + 4(-2) \ \ (-33)(-1) + 9(-2) \ (11)(-1) - 4(-2) \end{bmatrix} = \begin{bmatrix} 6 \ 51 \ -3 \end{bmatrix} \]
04

Express Transformations in \( \mathcal{B} \)-Coordinates

Each transformed vector must be expressed as a linear combination of the basis vectors \( \mathbf{b}_1, \mathbf{b}_2, \mathbf{b}_3 \). We solve for coefficients \( c_1, c_2, c_3 \) in \( A\mathbf{b}_i = c_1\mathbf{b}_1 + c_2\mathbf{b}_2 + c_3\mathbf{b}_3 \) for each \( i \).
05

Construct \( \mathcal{B} \)-Matrix

The matrix of \( A \) in basis \( \mathcal{B} \) is constructed from the computed coefficients. For each vector \( A\mathbf{b}_i \, (i=1,2,3) \), the coefficients form a column in the matrix. Solving for the coefficients requires setting up and solving a system of equations for each \( A\mathbf{b}_i \).
06

Verify Result

Check the correctness by ensuring that when combined, the vectors expressed in \( \mathcal{B} \) using these coefficients map back to their original transformations in the standard basis. This confirms the accuracy of the solution.

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Key Concepts

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

Linear Transformations
Linear transformations are like magical functions for vectors. They take a vector from one space and map it into another. The transformation you encountered is represented by the expression - \( \mathbf{x} \mapsto A \mathbf{x} \) This means that any vector \( \mathbf{x} \) is transformed by multiplying it with matrix \( A \). This operation creates a new vector in the same or different space as dictated by \( A \). The aim in many exercises is to express this transformation with respect to a different basis, such as \( \mathcal{B} \), rather than the standard basis.Key properties of linear transformations include:
  • They preserve addition: \( f(\mathbf{u} + \mathbf{v}) = f(\mathbf{u}) + f(\mathbf{v}) \)
  • They maintain scalar multiplication: \( f(c \cdot \mathbf{u}) = c \cdot f(\mathbf{u}) \)
These properties ensure that the linear transformation retains the structure of the vector space while mapping vectors.
Matrix Representation
Matrices are the ultimate tools to capture linear transformations. When you have a linear transformation, representing it with a matrix helps in performing actual calculations like transformations and solving equations. The matrix \( A \) in the exercise represents the linear transformation in the standard basis, which consists of vectors like \( \mathbf{e}_1 = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix} \). To express this transformation in a different basis, like \( \mathcal{B} \), we translate the problem using matrix algebra. This means finding the exact matrix that mimics the same transformation rules, but from the perspective of a new basis. Each column of the matrix corresponds to the transformation of one of the basis vectors.Understanding matrix representation enables:
  • Simplifying complex operations into manageable matrix multiplications
  • Changing perspectives from one basis to another
  • Finding solutions to systems of linear equations more efficiently
Vector Spaces
Vector spaces are foundational to linear algebra. Imagine a vector space as a vast 3D canvas where vectors can exist and interact. In this context, it's a set of vectors equipped with two operations: vector addition and scalar multiplication. The vector space in your exercise is 3-dimensional, spanned by the basis vectors.These operations must satisfy certain properties, like:
  • Commutativity: \( \mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u} \)
  • Existence of an additive identity: There exists a vector \( \mathbf{0} \) such that \( \mathbf{u} + \mathbf{0} = \mathbf{u} \)
  • Existence of additive inverses: For any vector \( \mathbf{u} \), there exists a \( -\mathbf{u} \) such that \( \mathbf{u} + (-\mathbf{u}) = \mathbf{0} \)
Vector spaces give rise to sophisticated mathematical ideas and tools. They help describe solutions to linear systems, enable transformations, and lead to the concept of dimensions, which measures the number of vectors required to span the space.
Basis Vectors
Basis vectors are the building blocks of vector spaces. A basis defines a specific way to describe all vectors in the space through linear combinations. Any vector in the space can be expressed uniquely in terms of these basis vectors. In your exercise, the basis \( \mathcal{B} \) is given as:
  • \( \mathbf{b}_1 = \begin{bmatrix} -1 \ -2 \ 1 \end{bmatrix} \)
  • \( \mathbf{b}_2 = \begin{bmatrix} -1 \ -1 \ 1 \end{bmatrix} \)
  • \( \mathbf{b}_3 = \begin{bmatrix} -1 \ -2 \ 0 \end{bmatrix} \)
To express vectors in terms of a new basis, you use these basis vectors to form linear combinations which have unique coefficients representing each vector in the space. The choice of basis affects how easy it is to perform calculations and understand transformations.Key characteristics of a good basis:
  • Spans the entire vector space.
  • The basis vectors must be linearly independent.
Remember, while the standard basis is often the simplest to use, changing to a different basis like \( \mathcal{B} \) can offer new insights and simplify certain problems.

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Most popular questions from this chapter

It can be shown that the trace of a matrix \(A\) equals the sum of the eigenvalues of \(A .\) Verify this statement for the case when \(A\) is diagonalizable.

Suppose the eigenvalues of a \(3 \times 3\) matrix \(A\) are \(3,4 / 5,\) and \(3 / 5,\) with corresponding eigenvectors \(\left[\begin{array}{r}{1} \\ {0} \\\ {-3}\end{array}\right],\left[\begin{array}{r}{2} \\ {1} \\\ {-5}\end{array}\right],\) and \(\left[\begin{array}{r}{-3} \\ {-3} \\\ {7}\end{array}\right] .\) Let \(\mathbf{x}_{0}=\left[\begin{array}{r}{-2} \\\ {-5} \\ {3}\end{array}\right] .\) Find the solution of the equation \(\mathbf{x}_{k+1}=A \mathbf{x}_{k}\) for the specified \(\mathbf{x}_{0},\) and describe what happens as \(k \rightarrow \infty\)

Find the characteristic polynomial and the eigenvalues of the matrices in Exercises 1鈥�8. $$ \left[\begin{array}{ll}{5} & {3} \\ {3} & {5}\end{array}\right] $$

A particle moving in a planar force field has a position vector \(x\) that satisfies \(\mathbf{x}^{\prime}=A \mathbf{x}\) . The \(2 \times 2\) matrix \(A\) has eigenvalues 4 and \(2,\) with corresponding eigenvectors \(\mathbf{v}_{1}=\left[\begin{array}{r}{-3} \\ {1}\end{array}\right]\) and \(\mathbf{v}_{2}=\left[\begin{array}{r}{-1} \\ {1}\end{array}\right] .\) Find the position of the particle at time \(t\) assuming that \(\mathbf{x}(0)=\left[\begin{array}{r}{-6} \\ {1}\end{array}\right]\)

Exercises \(9-14\) require techniques from Section \(3.1 .\) Find the characteristic polynomial of each matrix, using either a cofactor expansion or the special formula for \(3 \times 3\) determinants described prior to Exercises \(15-18\) in Section \(3.1 .\) INote: Finding the characteristic polynomial of a \(3 \times 3\) matrix is not easy to do with just row operations, because the variable \(\lambda\) is involved. $$ \left[\begin{array}{rrr}{6} & {-2} & {0} \\ {-2} & {9} & {0} \\ {5} & {8} & {3}\end{array}\right] $$
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