/*! 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 34 Consider the linear transformati... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 34

Consider the linear transformation \(T\) on \(\mathbf{R}^{2}\) defined by \(T(x, y)=(2 x-3 y, \quad x+4 y)\) and the following bases of \(\mathbf{R}^{2}\) : $$ E=\left\\{e_{1}, e_{2}\right\\}=\\{(1,0),(0,1)\\} \quad \text { and } \quad S=\left\\{u_{1}, u_{2}\right\\}=\\{(1,3),(2,5)\\} $$ (a) Find the matrix \(A\) representing \(T\) relative to the bases \(E\) and \(S\). (b) Find the matrix \(B\) representing \(T\) relative to the bases \(S\) and \(E\). (We can view \(T\) as a linear mapping from one space into another, each having its own basis.) (a) From Problem 6.2, \((a, b)=(-5 a+2 b) u_{1}+(3 a-b) u_{2} .\) Hence, $$ \begin{aligned} &T\left(e_{1}\right)=T(1,0)=(2,1)=-8 u_{1}+5 u_{2} \\ &T\left(e_{2}\right)=T(0,1)=(-3,4)=23 u_{1}-13 u_{2} \end{aligned} \quad \text { and so } \quad A=\left[\begin{array}{rr} -8 & 23 \\ 5 & -13 \end{array}\right] $$ (b) We have $$ \begin{aligned} &T\left(u_{1}\right)=T(1,3)=(-7,13)=-7 e_{1}+13 e_{2} \\ &T\left(u_{2}\right)=T(2,5)=(-11,22)=-11 e_{1}+22 e_{2} \end{aligned} \quad \text { and so } \quad B=\left[\begin{array}{rr} -7 & -11 \\ 13 & 22 \end{array}\right] $$

Short Answer

Expert verified
(a) The matrix A representing T relative to the bases E and S is \(A = \begin{bmatrix} 5 & 7 \\ -1.5 & -5 \end{bmatrix}\) (b) The matrix B representing T relative to the bases S and E is \(B = \begin{bmatrix} -7 & -9 \\ 13 & 22 \end{bmatrix}\)

Step by step solution

01

Find T(e鈧�) and T(e鈧�)

Apply the linear transformation T to the basis vectors e鈧� and e鈧�: \(T(e_1)=(2(1)-3(0),1(1)+4(0))=(2,1) = 2e_1 + 1e_2\) \(T(e_2)=(2(0)-3(1),1(0)+4(1))=(-3,4) = -3e_1 + 4e_2\)
02

Express T(e鈧�) and T(e鈧�) in terms of the basis S

Now we need to express T(e鈧�) and T(e鈧�) in terms of the basis vectors u鈧� and u鈧� in S. We can set up the following system of equations: \(T(e_1)=c_1u_1 + c_2u_2 = (2,1)\) \(T(e_2)=d_1u_1 + d_2u_2 = (-3,4)\) Find the coefficients c鈧�, c鈧�, d鈧�, and d鈧�: \(c_1 (1,3)+c_2 (2,5)=(2,1)\) \(d_1 (1,3)+d_2 (2,5)=(-3,4)\) Solve these systems: \(c_1 + 2c_2 = 2\) \(3c_1 + 5c_2 = 1\) \(d_1 + 2d_2 = -3\) \(3d_1 + 5d_2 = 4\) From the first system, we get \(c_1=5\) and \(c_2=-1.5.\) From the second system, we get \(d_1=7\) and \(d_2=-5.\)
03

Form the matrix A

Using the coefficients c鈧�, c鈧�, d鈧�, and d鈧�, we can form the matrix A: \(A = \begin{bmatrix} c_1 & d_1 \\ c_2 & d_2 \end{bmatrix} = \begin{bmatrix} 5 & 7 \\ -1.5 & -5 \end{bmatrix}\)
04

Find T(u鈧�) and T(u鈧�)

Apply the linear transformation T to the basis vectors u鈧� and u鈧�: \(T(u_1)=(2(1)-3(3),1(1)+4(3))=(-7,13) = -7e_1 + 13e_2\) \(T(u_2)=(2(2)-3(5),1(2)+4(5))=(-9,22) = -9e_1 + 22e_2\)
05

Express T(u鈧�) and T(u鈧�) in terms of the basis E

Now we need to express T(u鈧�) and T(u鈧�) in terms of the basis vectors e鈧� and e鈧� in E. We can set up the following system of equations: \(T(u_1)=a_1e_1 + a_2e_2 = (-7,13)\) \(T(u_2)=b_1e_1 + b_2e_2 = (-9,22)\) Find the coefficients a鈧�, a鈧�, b鈧�, and b鈧�: \(a_1 (1,0)+a_2 (0,1)=(-7,13)\) \(b_1 (1,0)+b_2 (0,1)=(-9,22)\) These systems are straightforward to solve. We get \(a_1=-7\), \(a_2=13\), \(b_1=-9\), and \(b_2=22\).
06

Form the matrix B

Using the coefficients a鈧�, a鈧�, b鈧�, and b鈧�, we can form the matrix B: \(B = \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix} = \begin{bmatrix} -7 & -9 \\ 13 & 22 \end{bmatrix}\) The final results are: (a) The matrix A representing T relative to the bases E and S is \(A = \begin{bmatrix} 5 & 7 \\ -1.5 & -5 \end{bmatrix}\) (b) The matrix B representing T relative to the bases S and E is \(B = \begin{bmatrix} -7 & -9 \\ 13 & 22 \end{bmatrix}\)

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Key Concepts

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

Matrix Representation
Understanding the matrix representation of a linear transformation is like decoding a secret message that reveals how space is reshaped. A linear transformation, in essence, takes vectors from one space and outputs them into another, potentially altered, in orientation or size. The matrix representation is a systematic way of keeping track of how each of these transformations takes place.

In the given exercise, we compute the matrix representation of the transformation based on different bases. The standard basis, denoted by E, is intuitive鈥攁s if your graph paper came to life and moved points around. But the other basis, S, is like a personalized coordinate system that only you understand. The matrix A, which represents the transformation from the standard basis E to the unique basis S, gives us a meticulous record of how a transformation alters vectors when viewed under the lens of these two bases.

Similarly, the matrix B offers reversed insight鈥攊t encapsulates the transformation considering S as the starting point and E as the destination. Both matrices A and B are paramount as they answer how we can map vectors back and forth between different perspectives of our space, much like translating between languages, with each basis offering a new dialect in the world of linear algebra.
Basis Vectors
Imagine standing in a vast open field with no discernible landmarks. How would you describe your location to a friend? Basis vectors are the mathematical answer to this dilemma; these are the landmarks in the vastness of vector spaces. A basis of a vector space is a set of vectors that are linearly independent and span the entire space, offering a reference frame to describe all other vectors.

In our exercise, the vectors in the basis E are like the cardinal points on a compass鈥攖hey鈥檙e the most straightforward directions one could follow. The vectors in basis S, on the other hand, are like paths meandering through the wilderness鈥攖hey also cover the entire field but in a more convoluted way. Any vector in our space can be described as a combination of these basis vectors, and changing the basis is akin to choosing a new pattern to weave through the field. Every linear transformation can be uniquely identified by how it interacts with the basis vectors鈥攁 fundamental aspect that is key to understanding the very nature of the linear mapping.
Linear Mapping
A linear mapping is akin to a magical spell in the realm of mathematics; it's a rule that takes vectors and transforms them, maintaining the linearity of the space. This means that if you imagine stretching, rotating, or flipping a shape, the transformation would keep straight lines unswervingly straight, and the origin fixed in place.

The effect of this enchantment is wholly captured by the linear transformation T in our exercise. By applying T to basis vectors, we explore its character鈥攄oes it flip the vectors over, stretch them out, or pivot them around the space? The matrices A and B are like the spellbook, encoding the essence of T in matrix form. These matrices answer the question 'How does T alter the vectors in our space?' While the matrices may appear to be simple grids of numbers, they hold the transformative power of T, enabling vectors to navigate through the multi-dimensional landscapes of linear algebra.

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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, \(P^{-1}[v]_{S}=[v]_{S^{\prime}}\). Suppose \(S=\left\\{u_{1}, \ldots, u_{n}\right\\}\) and \(S^{\prime}=\left\\{w_{1}, \ldots, w_{n}\right\\}\), and suppose, for \(i=1, \ldots, n\), $$ w_{i}=a_{i 1} u_{1}+a_{i 2} u_{2}+\cdots+a_{i n} u_{n}=\sum_{j=1}^{n} a_{i j} u_{j} $$ Then \(P\) is the \(n\)-square matrix whose \(j\) th row is $$ \left(a_{1 j}, a_{2 j}, \ldots, a_{n j}\right) $$ Also suppose \(v=k_{1} w_{1}+k_{2} w_{2}+\cdots+k_{n} w_{n}=\sum_{i=1}^{n} k_{i} w_{i}\). Then $$ [v]_{S^{\prime}}=\left[k_{1}, k_{2}, \ldots, k_{n}\right]^{T} $$ Substituting for \(w_{i}\) in the equation for \(v\), we obtain $$ \begin{aligned} v &=\sum_{i=1}^{n} k_{i} w_{i}=\sum_{i=1}^{n} k_{i}\left(\sum_{j=1}^{n} a_{i j} u_{j}\right)=\sum_{j=1}^{n}\left(\sum_{i=1}^{n} a_{i j} k_{i}\right) u_{j} \\\ &=\sum_{j=1}^{n}\left(a_{1 j} k_{1}+a_{2 j} k_{2}+\cdots+a_{n j} k_{n}\right) u_{j} \end{aligned} $$ Accordingly, \([v]_{S}\) 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 \(P[v]_{S^{\prime}}\) is obtained by multiplying the \(j\) th row of \(P\) by \([v]_{s^{\prime}}\)-that is, (1) by (2). However, the product of (1) and (2) is (3). Hence, \(P[v]_{S^{\prime}}\) and \([v]_{S}\) have the same entries. Thus, \(P[v]_{S}=[v]_{S}\), as claimed. Furthermore, multiplying the above by \(P^{-1}\) gives \(P^{-1}[v]_{S}=P^{-1} P[v]_{S^{\prime}}=[v]_{S^{\prime}}\).

Let \(F: \mathbf{R}^{n} \rightarrow \mathbf{R}^{m}\) be the linear mapping defined as follows: $$ F\left(x_{1}, x_{2}, \ldots, x_{n}\right)=\left(a_{11} x_{1}+\cdots+a_{1 n} x_{n}, a_{21} x_{1}+\cdots+a_{2 n} x_{n}, \ldots, a_{m 1} x_{1}+\cdots+a_{m n} x_{n}\right) $$ (a) Show that the rows of the matrix \([F]\) representing \(F\) relative to the usual bases of \(\mathbf{R}^{n}\) and \(\mathbf{R}^{m}\) are the coefficients of the \(x_{i}\) in the components of \(F\left(x_{1}, \ldots, x_{n}\right)\) (b) Find the matrix representation of each of the following linear mappings relative to the usual basis of \(\mathbf{R}^{n}\) : (i) \(F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{3}\) defined by \(F(x, y)=(3 x-y, \quad 2 x+4 y, \quad 5 x-6 y)\). (ii) \(F: \mathbf{R}^{4} \rightarrow \mathbf{R}^{2}\) defined by \(F(x, y, s, t)=(3 x-4 y+2 s-5 t, \quad 5 x+7 y-s-2 t)\). (iii) \(F: \mathbf{R}^{3} \rightarrow \mathbf{R}^{4}\) defined by \(F(x, y, z)=(2 x+3 y-8 z, \quad x+y+z, \quad 4 x-5 z, \quad 6 y)\). (a) We have (b) By part (a), we need only look at the coefficients of the unknown \(x, y, \ldots\) in \(F(x, y, \ldots)\). Thus, (i) \(\quad[F]=\left[\begin{array}{rr}3 & -1 \\ 2 & 4 \\ 5 & -6\end{array}\right]\) (ii) \([F]=\left[\begin{array}{rrrr}3 & -4 & 2 & -5 \\ 5 & 7 & -1 & -2\end{array}\right]\) (iii) \(\quad[F]=\left[\begin{array}{rrr}2 & 3 & -8 \\ 1 & 1 & 1 \\ 4 & 0 & -5 \\ 0 & 6 & 0\end{array}\right]\)

The vectors \(u_{1}=(1,1,0), u_{2}=(0,1,1), u_{3}=(1,2,2)\) form a basis \(S\) of \(\mathbf{R}^{3}\). Find the coordinates of an arbitrary vector \(v=(a, b, c)\) relative to the basis \(S\). Method 1. Express \(v\) as a linear combination of \(u_{1}, u_{2}, u_{3}\) using unknowns \(x, y, z\). We have $$ (a, b, c)=x(1,1,0)+y(0,1,1)+z(1,2,2)=(x+z, x+y+2 z, y+2 z) $$ this yields the system $$ \begin{array}{rlrlrl} x+\quad z & =a & x+\quad z & =a & x+\quad z & =a \\ x+y+2 z & =b & \text { or } & y+z & =-a+b & \text { or } & y+z & =-a+b \\ y+2 z & =c & y+2 z & =c & \quad z & =a-b+c \end{array} $$ Solving by back-substitution yields \(x=b-c, y=-2 a+2 b-c, \quad z=a-b+c\). Thus, $$ [v]_{S}=[b-c,-2 a+2 b-c, a-b+c]^{T} $$ Method 2. Find \(P^{-1}\) by row reducing \(M=[P, I]\) to the form \(\left[I, P^{-1}\right]\), where \(P\) is the change-of-basis matrix from the usual basis \(E\) to \(S\) or, in other words, the matrix whose columns are the basis vectors of \(S\). We have $$ \begin{aligned} M &=\left[\begin{array}{llllll} 1 & 0 & 1 & 1 & 0 & 0 \\ 1 & 1 & 2 & 0 & 1 & 0 \\ 0 & 1 & 2 & 0 & 0 & 1 \end{array}\right] \sim\left[\begin{array}{rrrrrr} 1 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & -1 & 1 & 0 \\ 0 & 1 & 2 & 0 & 0 & 1 \end{array}\right] \\ & \sim\left[\begin{array}{llllll} 1 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & -1 & 1 & 0 \\ 0 & 0 & 1 & 1 & -1 & 1 \end{array}\right] \sim\left[\begin{array}{rrrrrr} 1 & 0 & 0 & 0 & 1 & -1 \\ 0 & 1 & 0 & -2 & 2 & -1 \\ 0 & 0 & 1 & 1 & -1 & 1 \end{array}\right]=\left[I, P^{-1}\right] \end{aligned} $$ $$ \text { Thus, } \quad P^{-1}=\left[\begin{array}{rrr} 0 & 1 & -1 \\ -2 & 2 & -1 \\ 1 & -1 & 1 \end{array}\right] \text { and }[v]_{S}=P^{-1}[v]_{E}=\left[\begin{array}{rrr} 0 & 1 & -1 \\ -2 & 2 & -1 \\ 1 & -1 & 1 \end{array}\right]\left[\begin{array}{l} a \\ b \\ c \end{array}\right]=\left[\begin{array}{c} b-c \\ -2 a+2 b-c \\ a-b+c \end{array}\right] $$

Consider the following \(3 \times 3\) matrix \(A\) and basis \(S\) of \(\mathbf{R}^{3}\) : \\[A=\left[\begin{array}{rrr}1 & -2 & 1 \\\3 & -1 & 0 \\\1 & 4 & -2 \end{array}\right] \quad \text { and } \quad S=\left\\{u_{1}, u_{2}, u_{3}\right\\}=\left\\{\left[\begin{array}{l}1 \\\1 \\\1 \end{array}\right], \quad\left[\begin{array}{l}0 \\\1 \\ 1\end{array}\right], \quad\left[\begin{array}{l}1 \\\2 \\\3\end{array}\right]\right\\}\\]

Find the trace and determinant of each of the following linear maps on \(\mathbf{R}^{2}\) : (a) \(F(x, y)=(2 x-3 y, 5 x+4 y)\). (b) \(G(x, y)=(a x+b y, c x+d y)\).
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