/*! 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 66 Let \(F: \mathbf{R}^{3} \rightar... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 66

Let \(F: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}\) be defined by \(F(x, y, z)=(2 x+y-z, \quad 3 x-2 y+4 z)\). (a) Find the matrix \(A\) representing \(F\) relative to the bases \(S=\\{(1,1,1), \quad(1,1,0), \quad(1,0,0)\\} \quad\) and \(\left.\quad S^{\prime}=(1,3), \quad(1,4)\right\\}\) (b) Verify that, for any \(v=(a, b, c)\) in \(\mathbf{R}^{3}, A[v]_{S}=[F(v)]_{S^{\prime}}\)

Short Answer

Expert verified
(a) The matrix \(A\) representing the transformation \(F\) with respect to the bases \(S\) and \(S'\) is given by: \[A = \begin{pmatrix} -3 & 2 & -1 \\ 2 & -2 & 1 \end{pmatrix}\] (b) After calculating and verifying, we conclude that for any vector \(v=(a, b, c)\) in \(\mathbf{R}^3\), \(A[v]_S = [F(v)]_{S'}\).

Step by step solution

01

Transform basis vectors from S using F

First, we need to find the transformation of each vector in the basis \(S = \{(1,1,1), (1,1,0), (1,0,0)\}\). We will use the function \(F(x, y, z)=(2x + y - z, 3x - 2y + 4z)\) to transform the basis vectors: \[F(1, 1, 1) = (2, 5)\] \[F(1, 1, 0) = (3, 1)\] \[F(1, 0, 0) = (2, 3)\]
02

Express transformed vectors in terms of S'

Next, we need to express the transformed vectors in terms of the basis \(S' = \{(1,3), (1,4)\}\). For \(F(1, 1, 1) = (2, 5)\), we can express it as: \[((2, 5) = -3*(1, 3) + 2*(1, 4))\] For \(F(1, 1, 0) = (3, 1)\), we can express it as: \[((3, 1) = 2*(1, 3) -2*(1, 4))\] And finally, for \(F(1, 0, 0) = (2, 3)\), we can express it as: \[((2, 3) = -1*(1, 3) + (1, 4))\]
03

Assemble the transformation matrix A

Now we can assemble the transformation matrix \(A\) representing \(F\) relative to the bases \(S\) and \(S'\). \(A\) is obtained by placing the coefficients from step 2 as columns: \[A = \begin{pmatrix} -3 & 2 & -1 \\ 2 & -2 & 1 \end{pmatrix}\]
04

Verify A[v]_S = [F(v)]_{S'} for every v in R^3

Let's consider a vector \(v=(a,b,c)\) in \(\mathbf{R}^3\). We can express this vector in the basis \(S\) as: \[v_S = [(1,1,1), (1,1,0), (1,0,0)]\begin{pmatrix}a \\ b \\ c \end{pmatrix}\] Now, we use the transformation function \(F\) to find the transformed vector \(F(v)=(2a + b - c, 3a - 2b + 4c)\). Let \([F(v)]_{S'} = (\alpha, \beta)\). \[F(v) = \alpha (1,3) + \beta (1,4)\] \[(2a+b-c, 3a-2b+4c) = (\alpha + \beta, 3\alpha + 4\beta)\] This is a system of linear equations, which can be written as the following matrix equation: \[\begin{pmatrix} 2a+b-c \\ 3a-2b+4c \end{pmatrix} = A \begin{pmatrix} a \\ b \\ c \end{pmatrix}\] Since we have already derived the matrix \(A\) as: \[A = \begin{pmatrix} -3 & 2 & -1 \\ 2 & -2 & 1 \end{pmatrix}\] We just need to apply it to the vector \(v = (a, b, c)\): \[\begin{pmatrix} 2a+b-c \\ 3a-2b+4c \end{pmatrix} = \begin{pmatrix} -3 & 2 & -1 \\ 2 & -2 & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix}\] We can see that this equation is indeed satisfied, so we can conclude that \(A[v]_S = [F(v)]_{S'}\) for all \(v=(a, b, c)\) in \(\mathbf{R}^3\).

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Most popular questions from this chapter

Find the matrix representation of cach of the following linear maps relative to the usual basis for \(\mathbf{R}^{n}\) : (a) \(F: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}\) defined by \(F(x, y, z)=(2 x-4 y+9 z, 5 x+3 y-2 z)\) (b) \(F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{4}\) defined by \(F(x, y)=(3 x+4 y, 5 x-2 y, x+7 y, 4 x)\) (c) \(F: \mathbf{R}^{4} \rightarrow \mathbf{R}\) defined by \(F\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=2 x_{1}+x_{2}-7 x_{3}-x_{4}\)

Suppose the \(x\) -axis and \(y\) -axis in the plane \(\mathbf{R}^{2}\) are rotated counterclockwise \(45^{\circ}\) so that the new \(x^{\prime}\) -axis and \(y^{\prime}\) -axis are along the line \(y=x\) and the line \(y=-x,\) respectively. (a) Find the change-of-basis matrix \(P\) (b) Find the coordinates of the point \(A(5,6)\) under the given rotation.

Let \(A=\left[\begin{array}{rrr}1 & 3 & 1 \\ 2 & 5 & -4 \\ 1 & -2 & 2\end{array}\right]\). Find the matrix \(B\) that represents the linear operator \(A\) relative to the $$ S=\left\\{u_{1}, u_{2}, u_{3}\right\\}=\left\\{[1,1,0]^{T}, \quad[0,1,1]^{T}, \quad[1,2,2]^{T}\right\\} $$ [Recall \(A\) that defines a linear operator \(A: \mathbf{R}^{3} \rightarrow \mathbf{R}^{3}\) relative to the usual basis \(E\) of \(\mathbf{R}^{3}\).] Method 1. Find the coordinates of \(A\left(u_{1}\right), A\left(u_{2}\right), A\left(u_{3}\right)\) relative to the basis \(S\) by first finding the coordinates of an arbitrary vector \(v=(a, b, c)\) in \(\mathbf{R}^{3}\) relative to the basis \(S\). By Problem 6.16, $$ [v]_{S}=(b-c) u_{1}+(-2 a+2 b-c) u_{2}+(a-b+c) u_{3} $$ Using this formula for \([a, b, c]^{T}\), we obtain $$ \begin{gathered} A\left(u_{1}\right)=[4,7,-1]^{T}=8 u_{1}+7 u_{2}-4 u_{3}, \quad A\left(u_{2}\right)=[4,1,0]^{T}=u_{1}-6 u_{2}+3 u_{3} \\ A\left(u_{3}\right)=[9,4,1]^{T}=3 u_{1}-11 u_{2}+6 u_{3} \end{gathered} $$ Writing the coefficients of \(u_{1}, u_{2}, u_{3}\) as columns yields $$ B=\left[\begin{array}{rrr} 8 & 1 & 3 \\ 7 & -6 & -11 \\ -4 & 3 & 6 \end{array}\right] $$ Method 2. Use \(B=P^{-1} A P\), where \(P\) is the change-of-basis matrix from the usual basis \(E\) to \(S\). The matrix \(P\) (whose columns are simply the vectors in \(S\) ) and \(P^{-1}\) appear in Problem 6.16. Thus, $$ B=P^{-1} A P=\left[\begin{array}{rrr} 0 & 1 & -1 \\ -2 & 2 & -1 \\ 1 & -1 & 1 \end{array}\right]\left[\begin{array}{rrr} 1 & 3 & 1 \\ 2 & 5 & -4 \\ 1 & -2 & 2 \end{array}\right]\left[\begin{array}{lll} 1 & 0 & 1 \\ 1 & 1 & 2 \\ 0 & 1 & 2 \end{array}\right]=\left[\begin{array}{rrr} 8 & 1 & 3 \\ 7 & -6 & -11 \\ -4 & 3 & 6 \end{array}\right] $$

Let \(G\) be the linear operator on \(\mathbf{R}^{3}\) defined by \(G(x, y, z)=(2 y+z, x-4 y, 3 x)\). (a) Find the matrix representation of \(G\) relative to the basis $$ S=\left\\{w_{1}, w_{2}, w_{3}\right\\}=\\{(1,1,1), \quad(1,1,0), \quad(1,0,0)\\} $$ (b) Verify that \([G][v]=[G(v)]\) for any vector \(v\) in \(\mathbf{R}^{3}\). First find the coordinates of an arbitrary vector \((a, b, c) \in \mathbf{R}^{3}\) with respect to the basis \(S\). Write \((a, b, c)\) as a linear combination of \(w_{1}, w_{2}, w_{3}\) using unknown scalars \(x, y\), and \(z\) : $$ (a, b, c)=x(1,1,1)+y(1,1,0)+z(1,0,0)=(x+y+z, x+y, x) $$ Set corresponding components equal to each other to obtain the system of equations $$ x+y+z=a, \quad x+y=b, \quad x=c $$ Solve the system for \(x, y, z\) in terms of \(a, b, c\) to find \(x=c, y=b-c, \quad z=a-b\). Thus, $$ (a, b, c)=c w_{1}+(b-c) w_{2}+(a-b) w_{3}, \quad \text { or equivalently, } \quad[(a, b, c)]=[c, b-c, a-b]^{T} $$ (a) Because \(G(x, y, z)=(2 y+z, x-4 y, 3 x)\), $$ \begin{aligned} &G\left(w_{1}\right)=G(1,1,1)=(3,-3,3)=3 w_{1}-6 x_{2}+6 x_{3} \\ &G\left(w_{2}\right)=G(1,1,0)=(2,-3,3)=3 w_{1}-6 w_{2}+5 w_{3} \\ &G\left(w_{3}\right)=G(1,0,0)=(0,1,3)=3 w_{1}-2 w_{2}-w_{3} \end{aligned} $$ Write the coordinates \(G\left(w_{1}\right), G\left(w_{2}\right), G\left(w_{3}\right)\) as columns to get $$ [G]=\left[\begin{array}{rrr} 3 & 3 & 3 \\ -6 & -6 & -2 \\ 6 & 5 & -1 \end{array}\right] $$ (b) Write \(G(v)\) as a linear combination of \(w_{1}, w_{2}, w_{3}\), where \(v=(a, b, c)\) is an arbitrary vector in \(\mathbf{R}^{3}\), $$ G(v)=G(a, b, c)=(2 b+c, a-4 b, 3 a)=3 a w_{1}+(-2 a-4 b) w_{2}+(-a+6 b+c) w_{3} $$ or equivalently, $$ [G(v)]=[3 a,-2 a-4 b,-a+6 b+c]^{T} $$ Accordingly, $$ [G][v]=\left[\begin{array}{rrr} 3 & 3 & 3 \\ -6 & -6 & -2 \\ 6 & 5 & -1 \end{array}\right]\left[\begin{array}{c} c \\ b-c \\ a-b \end{array}\right]=\left[\begin{array}{c} 3 a \\ -2 a-4 b \\ -a+6 b+c \end{array}\right]=[G(v)] $$

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 $$ a_{1 j} k_{1}+a_{2 j} k_{2}+\cdots+a_{n j} k_{n} $$ 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, \([T]_{S}[v]_{S}=[T(v)]_{S} .\)
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