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Chapter 6: Problem 54

Let \(F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\) be defined by \(F(x, y)=(x-3 y, \quad 2 x-4 y) .\) Find the matrix \(A\) that represents \(F\) relative to each of the following bases: (a) \(S=\\{(2,5),(3,7)\\}\) (b) \(S=\\{(2,3),(4,5)\\}\)

Short Answer

Expert verified
The matrix \(A\) that represents the linear transformation \(F\) relative to each of the given bases is: (a) For the basis \(S=\{(2,5),(3,7)\}\), the matrix \(A\) is: \[A_{1} = \begin{pmatrix} -11 & 1 \\ -3 & -3 \end{pmatrix}\] (b) For the basis \(S=\{(2,3),(4,5)\}\), the matrix \(A\) is: \[A_{2} = \begin{pmatrix} 1 & 1 \\ -2 & -2 \end{pmatrix}\]

Step by step solution

01

Find the image of basis vectors under \(F\)

To find the matrix \(A\) that represents \(F\), we first need to find the image of the basis vectors under the linear transformation \(F\). (a) For the basis \(S=\{(2,5),(3,7)\}\), we have: \[F(2,5)=(2-3(5),2(2)-4(5)) = (-11,-16)\] \[F(3,7)=(3-3(7),2(3)-4(7)) = (-18,-20)\] (b) For the basis \(S=\{(2,3),(4,5)\}\), we have: \[F(2,3)=(2-3(3),2(2)-4(3))=(-7,-4)\] \[F(4,5)=(4-3(5),2(4)-4(5))=(-11,-12)\] ## Express images as linear combinations ##
02

Write each image as a linear combination of the given basis

Next, we express the images (\(F(2,5)\) and \(F(3,7)\)) as a linear combination of the given basis. (a) For the basis \(S=\{(2,5),(3,7)\}\), we have: \[-11a+3b = -11\] \[-16a+7b = -16\] By solving this system of linear equations, we get: \[a=1\] \[b=-3\] So, \(F(2,5)=-11(2,5)-3(3,7)\) and \(F(3,7)=1(2,5)-3(3,7)\). (b) For the basis \(S=\{(2,3),(4,5)\}\), we have: \[2a+4b = -7\] \[3a+5b = -4\] By solving this system of linear equations, we get: \[a=1\] \[b=-2\] So, \(F(2,3)=1(2,3)-2(4,5)\) and \(F(4,5)=1(2,3)-2(4,5)\). ## Write the matrix A ##
03

Write the coefficients of the linear combination in the form of a matrix \(A\)

Finally, we will write the coefficients of the linear combination in the form of a matrix \(A\): (a) For the basis \(S=\{(2,5),(3,7)\}\), the matrix \(A\) is given by: \[A_{1} = \begin{pmatrix} -11 & 1 \\ -3 & -3 \end{pmatrix}\] (b) For the basis \(S=\{(2,3),(4,5)\}\), the matrix \(A\) is given by: \[A_{2} = \begin{pmatrix} 1 & 1 \\ -2 & -2 \end{pmatrix}\] Thus, matrices \(A_{1}\) and \(A_{2}\) represent the linear transformation \(F\) with respect to the given bases.

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Key Concepts

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

Matrix Representation
In linear algebra, when dealing with transformations, one question is how to represent those transformations as matrices. A matrix representation helps us capture the essence of a transformation succinctly. The given linear transformation, \(F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\), is defined by \(F(x, y) = (x - 3y, 2x - 4y)\). To find its matrix representation, especially regarding different bases, we need to consider how the transformation affects the basis vectors.

Once we determine how the transformation \(F\) acts on each basis vector, we express each resulting vector as a linear combination of those same basis vectors. The coefficients from these combinations directly form the matrix that represents \(F\) relative to the basis we are considering. This matrix encapsulates the entire transformation process and provides a simplified means to apply the transformation to any vector in \(\mathbf{R}^{2}\).
Basis Vectors
Basis vectors are fundamental in understanding vector spaces. They provide a framework to represent almost any vector in that space using linear combinations. For the exercise, we look at the bases \(S = \{(2,5),(3,7)\}\) and \(S = \{(2,3),(4,5)\}\). These sets of vectors span \(\mathbf{R}^{2}\) and allow us to decompose other vectors in this space relative to them.

In the provided example, after applying the transformation \(F\), we can find where each basis vector gets mapped. Each mapped result is expressed as a combination of the original basis vectors. This step is crucial in determining the coefficients that make up the matrix representation of \(F\) relative to any particular basis. The transformation's effectiveness is observed by comparing where it "sends" the basis vectors.
Linear Combinations
A linear combination involves taking multiples of basis vectors to express other vectors in the vector space. It is an elemental concept when decomposing or reconstructing vectors relative to a basis. In this exercise, many systems of linear equations are solved to find the linear combinations that express the resulting vector from the transformation.

For instance, after determining \(F(2,5) = (-11, -16)\), we express this result as a combination of \((2,5)\) and \((3,7)\). Solving the linear system \[-11 = -11a + 3b\] and \[-16 = -16a + 7b\] gives us \(a = 1\) and \(b = -3\). Such decomposition is vital as each linear combination provides entries for our transformation matrix.
Systems of Linear Equations
Solving systems of linear equations is integral when working with linear combinations and matrices. In the exercise, each transformation image needed expression in terms of basis vectors, leading us to establish equation systems.

These systems arise from comparing coefficients and constants of vectors once expressed as combinations of basis vectors. For example, to express \(F(2,3) = (-7, -4)\) as a linear combination of the basis \((2,3)\) and \((4,5)\), we solve:
  • \(-7 = 2a + 4b\)
  • \(-4 = 3a + 5b\)
Finding \(a = 1\) and \(b = -2\), these solutions translate directly into matrix coefficients. This structured approach reiterates the utility of linear equations in understanding vector transformations.

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Most popular questions from this chapter

Prove Theorem 6.6: Let \(P\) be the change-of-basis matrix from a basis \(S\) to a basis \(S^{\prime}\) in a vector space \(V .\) Then, for any vector \(v \in V,\) we have \(P[v]_{S^{\prime}}=[v]_{S},\) and hence, \(P^{-1}[v]_{S}=[v]_{S^{-}}\).

Prove Theorem 6.7: Let \(P\) be the change-of-basis matrix from a basis \(S\) to a basis \(S^{\prime}\) in a vector space \(V .\) Then, for any linear operator \(T\) on \(V,[T]_{S^{\prime}}=P^{-1}[T]_{S} P\).

Consider the linear operator \(F\) on \(\mathbf{R}^{2}\) defined by \(F(x, y)=(5 x+y, 3 x-2 y)\) and the following bases of \(\mathbf{R}^{2}\) : \\[ S=\\{(1,2),(2,3)\\} \quad \text { and } \quad S^{\prime}=\\{(1,3),(1,4)\\} \\] (a) Find the matrix \(A\) representing \(F\) relative to the basis \(S\). (b) Find the matrix \(B\) representing \(F\) relative to the basis \(S^{\prime}\) (c) Find the change-of-basis matrix \(P\) from \(S\) to \(S^{\prime}\) (d) How are \(A\) and \(B\) related?

Consider the following linear operator \(G\) on \(\mathbf{R}^{2}\) and basis \(S:\) \\[ G(x, y)=(2 x-7 y, 4 x+3 y) \quad \text { and } \quad S=\left\\{u_{1}, u_{2}\right\\}=\\{(1,3),(2,5)\\} \\] (a) Find the matrix representation \([G]_{S}\) of \(G\) relative to \(S\). (b) Verify \([G]_{S}[v]_{S}=[G(v)]_{S}\) for the vector \(v=(4,-3)\) in \(\mathbf{R}^{2}\) First find the coordinates of an arbitrary vector \(v=(a, b)\) in \(\mathbf{R}^{2}\) relative to the basis \(S .\) We have \\[ \left[\begin{array}{l} a \\ b \end{array}\right]=x\left[\begin{array}{l} 1 \\ 3 \end{array}\right]+y\left[\begin{array}{l} 2 \\ 5 \end{array}\right], \quad \text { and so } \quad \begin{array}{r} x+2 y=a \\ 3 x+5 y=b \end{array} \\] Solve for \(x\) and \(y\) in terms of \(a\) and \(b\) to get \(x=-5 a+2 b, y=3 a-b .\) Thus, \\[ (a, b)=(-5 a+2 b) u_{1}+(3 a-b) u_{2} \\] \\[ \text { and so } \quad[v]=[-5 a+2 b, \quad 3 a-b]^{T} \\] (a) Using the formula for \((a, b)\) and \(G(x, y)=(2 x-7 y, 4 x+3 y)\), we have \\[ \begin{array}{l} G\left(u_{1}\right)=G(1,3)=(-19,13)=121 u_{1}-70 u_{2} \\ G\left(u_{2}\right)=G(2,5)=(-31,23)=201 u_{1}-116 u_{2} \end{array} \quad \text { and so } \quad[G]_{S}=\left[\begin{array}{rr} 121 & 201 \\ -70 & -116 \end{array}\right] \\] (We emphasize that the coefficients of \(u_{1}\) and \(u_{2}\) are written as columns, not rows, in the matrix representation.) (b) Use the formula \((a, b)=(-5 a+2 b) u_{1}+(3 a-b) u_{2}\) to get Then \\[ \begin{array}{c} v=(4,-3)=-26 u_{1}+15 u_{2} \\ G(v)=G(4,-3)=(20,7)=-131 u_{1}+80 u_{2} \\ {[v]_{S}=[-26,15]^{T} \quad \text { and } \quad[G(v)]_{S}=[-131,80]^{T}} \end{array} \\] Accordingly, \\[ [G]_{S}[v]_{S}=\left[\begin{array}{rr} 121 & 201 \\ -70 & -116 \end{array}\right]\left[\begin{array}{r} -26 \\ 15 \end{array}\right]=\left[\begin{array}{r} -131 \\ 80 \end{array}\right]=[G(v)]_{S} \\] (This is expected from Theorem \(6.1 .)\)

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.
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