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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.

Short Answer

Expert verified
The permutation matrix \( P_{D \leftarrow B} \) is \( \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \).

Step by step solution

01

Understanding the Problem

We want to find the permutation matrix that changes the basis from basis \(B\) to basis \(D\). We are given two different orders of the same set of vectors: \(B = \{ \mathbf{b}_1, \mathbf{b}_2, \mathbf{b}_3, \mathbf{b}_4 \}\) and \(D = \{ \mathbf{b}_2, \mathbf{b}_3, \mathbf{b}_1, \mathbf{b}_4 \}\).
02

Identify the Permutation

To get from basis \(B\) to basis \(D\), observe the new order in \(D\). \(\mathbf{b}_2\) should be the first vector, \(\mathbf{b}_3\) the second, \(\mathbf{b}_1\) the third, and \(\mathbf{b}_4\) remains in the last position.
03

Constructing the Permutation Matrix

A permutation matrix is constructed by rearranging the rows of the identity matrix to fit the new order. Start with a 4x4 identity matrix since there are 4 vectors. To permute according to \(D\), rearrange the rows to reflect the new order of vectors. This results in the following permutation matrix:\[ P_{D \leftarrow B} = \begin{pmatrix} 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 \end{pmatrix} \]
04

Verify the Permutation Matrix

Using the resulting permutation matrix, when you multiply it by a vector in \(B\)'s basis, it should reorder the vector to its position in \(D\). For instance, applying \(P_{D \leftarrow B}\) to \(\mathbf{b}_1\) should put it in the third position (where \(\mathbf{b}_1\) appears in \(D\)), and similarly check for other vectors.

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Key Concepts

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

Linear Algebra
Linear algebra is a branch of mathematics focusing on vector spaces and the linear transformations between them. It involves understanding properties and operations related to vectors, matrices, and systems of linear equations.
In our exercise, linear algebra is utilized to handle the transformation from one basis to another using permutation matrices. These matrices are particularly useful because they provide a systematic way to reorganize vectors according to a new order.
Key components in linear algebra include:
  • Vectors - elements that have both magnitude and direction.
  • Matrices - a rectangular arrangement of numbers or functions which can represent linear transformations.
  • Transformations - a way to systematically convert vectors from one form to another, including changing their basis.
These concepts come together to solve problems requiring rearrangements of data in structured ways, such as your exercise with permutation matrices.
Basis Transformation
Basis transformations involve changing the basis of a vector space, which essentially means changing the set of vectors used to describe the space.
In our exercise, we're given two bases: \( B \) and \( D \). The transformation between these bases is defined by how vectors change position from one basis to the other. This is where permutation matrices come into play, as they reorder vectors to fit a new basis.
Understanding basis transformations requires grasping:
  • The concept of a basis - a set of linearly independent vectors that span a vector space.
  • How transformations can rearrange vector bases while maintaining their mathematical properties.
  • The significance of preserving linear independence and span during the transformation.
These transformations are vital for various applications in physics, computer graphics, and data science.
Matrix Multiplication
Matrix multiplication is a crucial operation in linear algebra, allowing for the combination of two matrices to form another matrix that incorporates the transformations of both.
It's essential in applying permutation matrices to reorder vector bases, as seen in your exercise. The multiplication of the permutation matrix \( P_{D \leftarrow B} \) by any vector from basis \( B \) results in a vector in basis \( D \).
The rules of matrix multiplication are as follows:
  • Multiplying a matrix by a vector involves taking dot products between rows of the matrix and the vector.
  • The resulting matrix (or vector) inherits dimensions from the original matrices, following the rule that the number of columns in the first matrix must equal the number of rows in the second.
  • Matrix multiplication is associative but not generally commutative, meaning \( AB eq BA \) unless both matrices are particularly structured.
This operation is indispensable in linear transformations.
Identity Matrix
The identity matrix is a special type of matrix that acts as a multiplicative identity in matrix multiplication. It is composed of 1s on the diagonal and 0s elsewhere, resembling how the number 1 acts in multiplication of real numbers.
In the exercise, the identity matrix serves as a starting point for constructing the permutation matrix. When rearranging its rows according to a new vector order in a basis transformation, you create a permutation matrix.
Characteristics of the identity matrix include:
  • It retains the original matrix when executed in a multiplication operation, i.e., \( AI = A \) and \( IA = A \).
  • Its dimensions are always square, meaning an \( n \times n \) identity matrix.
  • It is the simplest form of an orthogonal matrix, where the rows and columns are orthonormal vectors.
This matrix is fundamental in maintaining structure within matrix operations.
Vector Spaces
Vector spaces are the fundamental setting for linear algebra. A vector space consists of a collection of vectors that can be scaled and added together, following specific algebraic rules.
The bases \( B \) and \( D \) from your exercise represent different configurations within the same vector space. Each basis provides a distinct framework to express vectors, depending on how it is formed with linearly independent vectors.
Essential properties of vector spaces include:
  • Closure under addition and scalar multiplication, ensuring any linear combination of vectors remains in the space.
  • The existence of a zero vector, serving as the additive identity.
  • The capacity to be spanned by a basis - a minimal set of vectors from which all other vectors in the space can be linearly composed.
Understanding vector spaces is pivotal in tackling questions involving coordinate changes and transformations.

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Most popular questions from this chapter

Let \(E\) be an \(n \times n\) matrix with \(E^{2}=E .\) Show that \(\mathbf{M}_{n n}=U \oplus W,\) where \(U=\\{A \mid A E=A\\}\) and \(W=\\{B \mid B E=0\\} .[\) Hint \(: X E\) lies in \(U\) for every matrix \(X .]\)

In \(\mathbf{P}_{3}\) find \(P_{D \leftarrow B}\) if \(B=\left\\{1, x, x^{2}, x^{3}\right\\}\) and \(D=\left\\{1,(1-x),(1-x)^{2},(1-x)^{3}\right\\}\). Then express \(p=a+b x+c x^{2}+d x^{3}\) as a polynomial in powers of \((1-x)\)

Let \(T: V \rightarrow V\) be a linear operator where \(V\) is finite dimensional. If \(U \subseteq V\) is a subspace, let \(\bar{U}=\left\\{\mathbf{u}_{0}+T\left(\mathbf{u}_{1}\right)+T^{2}\left(\mathbf{u}_{2}\right)+\cdots+T^{k}\left(\mathbf{u}_{k}\right) \mid \mathbf{u}_{i}\right.\) in \(U, k \geq\) 0\\}. Show that \(\bar{U}\) is the smallest \(T\) -invariant subspace containing \(U\) (that is, it is \(T\) -invariant, contains \(U\), and is contained in every such subspace).

In each case find \(P_{D \leftarrow B},\) where \(B\) and \(D\) are ordered bases of \(V\). Then verify that \(C_{D}(\mathbf{v})=P_{D \leftarrow B} C_{B}(\mathbf{v})\) a. \(V=\mathbb{R}^{2}, B=\\{(0,-1),(2,1)\\},\) \(D=\\{(0,1),(1,1)\\}, \mathbf{v}=(3,-5)\) b. \(V=\mathbf{P}_{2}, B=\left\\{x, 1+x, x^{2}\right\\}, D=\left\\{2, x+3, x^{2}-1\right\\},\) \(\mathbf{v}=1+x+x^{2}\) c. \(V=\mathbf{M}_{22}\) \(B=\left\\{\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right]\right\\}\) \(D=\left\\{\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}1 & 0 \\ 1 & 0\end{array}\right],\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right],\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\right\\}\) \(\mathbf{v}=\left[\begin{array}{rr}3 & -1 \\ 1 & 4\end{array}\right]\)

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 \(\left.\mathbb{R}\right\\}\) \(\begin{aligned} \text { d. } V &=\mathbf{M}_{22}, U=\left\\{\left[\begin{array}{ll}a & a \\ b & b\end{array}\right] \mid a, b \text { in } \mathbb{R}\right\\} \\ W &=\left\\{\left[\begin{array}{rl}a & b \\\ -a & b\end{array}\right] \mid a, b \text { in } \mathbb{R}\right\\} \end{aligned}\)

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