/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Suppose \(P\) is the change-of-b... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 18

Suppose \(P\) is the change-of-basis matrix from a basis \(\left\\{u_{i}\right\\}\) to a basis \(\left\\{w_{i}\right\\},\) and suppose \(Q\) is the change-of-basis matrix from the basis \(\left\\{w_{i}\right\\}\) back to \(\left\\{u_{i}\right\\} .\) Prove that \(P\) is invertible and that \(Q=P^{-1}\).

Short Answer

Expert verified
Since the matrix product \(PQ = I\), where \(I\) is the identity matrix, we can conclude that the change-of-basis matrix \(P\) is invertible and its inverse is the change-of-basis matrix \(Q\), i.e., \(Q = P^{-1}\).

Step by step solution

01

Define the change-of-basis matrices

Given the change-of-basis matrix \(P\) from the basis \(\left\\{u_{i}\right\\}\) to the basis \(\left\\{w_{i}\right\\}\), we can represent the vectors in the basis \(\left\\{w_{i}\right\\}\) in terms of \(\left\\{u_{i}\right\\}\) as: \[w_{i} = P u_{i}\] Similarly, given the change-of-basis matrix \(Q\) from the basis \(\left\\{w_{i}\right\\}\) to the basis \(\left\\{u_{i}\right\\}\), we can represent the vectors in the basis \(\left\\{u_{i}\right\\}\) in terms of \(\left\\{w_{i}\right\\}\) as: \[u_{i} = Q w_{i}\]
02

Consider the product of matrices P and Q

Consider the product of the two matrices \(P\) and \(Q\). To find the result of this product, we will multiply each \(w_{i}\) by \(Q\): \[(PQ)u_{i} = Q(Pu_{i})\] Since \(Pu_{i} = w_{i}\), we have: \[(PQ)u_{i} = Qw_{i}\]
03

Relate the product PQ to the identity matrix

Since \(Qw_{i} = u_{i}\), the result of \((PQ)u_{i}\) is the original vector \(u_{i}\). Thus, the transformation \(PQ\) applied to any vector in the basis \(\left\\{u_{i}\right\\}\) will give back the original vector. Therefore, the matrix product \(PQ\) is equivalent to the identity matrix for the basis \(\left\\{u_{i}\right\\}\): \[PQ = I\]
04

Prove that P is invertible and Q is its inverse

Since the matrix product \(PQ\) resulted in the identity matrix, we can conclude that the matrix \(P\) is invertible and its inverse is the matrix \(Q\), i.e., \(Q = P^{-1}\). In conclusion, the change-of-basis matrix \(P\) is invertible and its inverse is the change-of-basis matrix \(Q\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(A=\left[\begin{array}{rr}1 & 1 \\ 2 & -3\end{array}\right]\) and \(P=\left[\begin{array}{rr}1 & -2 \\ 3 & -5\end{array}\right]\).

Let \(A=\left[\begin{array}{rr}4 & -2 \\ 3 & 6\end{array}\right]\) and \(P=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\).

Find the change-of-basis matrix \(P\) from the usual basis \(E\) of \(\mathbf{R}^{3}\) to a basis \(S\), the change-of-basis matrix \(Q\) from \(S\) back to \(E,\) and the coordinates of \(v=(a, b, c)\) relative to \(S,\) where \(S\) consists of the vectors: (a) \(u_{1}=(1,1,0), u_{2}=(0,1,2), u_{3}=(0,1,1)\). (b) \(u_{1}=(1,0,1), u_{2}=(1,1,2), u_{3}=(1,2,4)\). (c) \(u_{1}=(1,2,1), u_{2}=(1,3,4), u_{3}=(2,5,6)\).

Prove Theorem 6.1: Let \(T: V \rightarrow V\) be a linear operator, and let \(S\) be a (finite) basis of \(V\). Then, for any vector \(v\) in \(V,[T]_{S}[v]_{S}=[T(v)]_{S}\) Suppose \(S=\left\\{u_{1}, u_{2}, \ldots, u_{n}\right\\},\) and suppose, for \(i=1, \ldots, n\) \\[ T\left(u_{i}\right)=a_{i 1} u_{1}+a_{i 2} u_{2}+\cdots+a_{i n} u_{n}=\sum_{j=1}^{n} a_{i j} u_{j} \\] Then \([T]_{S}\) is the \(n\) -square matrix whose \(j\) th row is \\[ \left(a_{1 j}, a_{2 j}, \ldots, a_{n j}\right) \\] Now suppose \\[ v=k_{1} u_{1}+k_{2} u_{2}+\cdots+k_{n} u_{n}=\sum_{i=1}^{n} k_{i} u_{i} \\] Writing a column vector as the transpose of a row vector, we have \\[ [v]_{S}=\left[k_{1}, k_{2}, \ldots, k_{n}\right]^{T} \\] Furthermore, using the linearity of \(T\) \\[ \begin{aligned} T(v) &=T\left(\sum_{i=1}^{n} k_{i} u_{i}\right)=\sum_{i=1}^{n} k_{i} T\left(u_{i}\right)=\sum_{i=1}^{n} k_{i}\left(\sum_{j=1}^{n} a_{i j} u_{j}\right) \\ &=\sum_{j=1}^{n}\left(\sum_{i=1}^{n} a_{i j} k_{i}\right) u_{j}=\sum_{j=1}^{n}\left(a_{1 j} k_{1}+a_{2 j} k_{2}+\cdots+a_{n j} k_{n}\right) u_{j} \end{aligned} \\] Thus, \([T(v)]_{S}\) is the column vector whose \(j\) th entry is \\[ a_{1 j} k_{1}+a_{2 j} k_{2}+\dots+a_{n j} k_{n} \\] On the other hand, the \(j\) th entry of \([T]_{S}[v]_{S}\) is obtained by multiplying the \(j\) th row of \([T]_{S}\) by \([v]_{S}-\) that is (1) by \((2) .\) But the product of (1) and (2) is \((3) .\) Hence, \([T]_{S}[v]_{S}\) and \([T(v)]_{S}\) have the same entries. Thus, \([T]_{S}[v]_{S}=[T(v)]_{S}\).

Find the matrix representation of each of the following linear operators \(F\) on \(\mathbf{R}^{3}\) relative to the usual basis \(E=\left\\{e_{1}, e_{2}, e_{3}\right\\}\) of \(\mathbf{R}^{3} ;\) that is, find \([F]=[F]_{E}\) (a) \(F\) defined by \(F(x, y, z)=(x+2 y-3 z, 4 x-5 y-6 z, 7 x+8 y+9 z)\) (b) \(F\) defined by the \(3 \times 3\) matrix \(A=\left[\begin{array}{lll}1 & 1 & 1 \\ 2 & 3 & 4 \\ 5 & 5 & 5\end{array}\right]\) (c) \(F\) defined by \(F\left(e_{1}\right)=(1,3,5), F\left(e_{2}\right)=(2,4,6), F\left(e_{3}\right)=(7,7,7) .\) (Theorem 5.2 states that a linear map is completely defined by its action on the vectors in a basis.) (a) Because \(E\) is the usual basis, simply write the coefficients of the components of \(F(x, y, z)\) as rows: \\[ [F]=\left[\begin{array}{rrr} 1 & 2 & -3 \\ 4 & -5 & -6 \\ 7 & 8 & 9 \end{array}\right] \\] (b) Because \(E\) is the usual basis, \([F]=A\), the matrix \(A\) itself. (c) Here \\[ \begin{array}{llll} F\left(e_{1}\right)=(1,3,5)=e_{1}+3 e_{2}+5 e_{3} & & & {\left[\begin{array}{lll} 1 & 2 & 7 \\ 3 & 4 & 7 \\ 5 & 6 & 7 \end{array}\right]} \end{array} \\] That is, the columns of \([F]\) are the images of the usual basis vectors.
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.