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Chapter 6: Problem 40

Find the matrix representing each linear transformation \(T\) on \(\mathbf{R}^{3}\) relative to the usual basis of \(\mathbf{R}^{3}\) : (a) \(T(x, y, z)=(x, y, 0)\). (b) \(\quad T(x, y, z)=(z, y+z, x+y+z)\). (c) \(T(x, y, z)=(2 x-7 y-4 z, 3 x+y+4 z, 6 x-8 y+z)\).

Short Answer

Expert verified
The matrices representing the given linear transformations T are: (a) \[T = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}\] (b) \[T = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}\] (c) \[T = \begin{bmatrix} 2 & -7 & -4 \\ 3 & 1 & 4 \\ 6 & -8 & 1 \end{bmatrix}\]

Step by step solution

01

Apply Transformation to Basis Vectors

Apply the given transformation T(x, y, z) = (x, y, 0) to each of the basis vectors (1,0,0), (0,1,0), and (0,0,1).
02

Obtain Column Vectors

Find the resulting vectors: T(1,0,0) = (1,0,0) T(0,1,0) = (0,1,0) T(0,0,1) = (0,0,0) These vectors are the columns of the transformation matrix.
03

Form the Transformation Matrix

Combine the column vectors obtained in step 2 to form the matrix of the transformation T: \[T = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}\] #b) T(x, y, z) = (z, y+z, x+y+z)#
04

Apply Transformation to Basis Vectors

Apply the given transformation T(x, y, z) = (z, y+z, x+y+z) to each of the basis vectors (1,0,0), (0,1,0), and (0,0,1).
05

Obtain Column Vectors

Find the resulting vectors: T(1,0,0) = (0,0,1) T(0,1,0) = (0,1,1) T(0,0,1) = (1,1,1) These vectors are the columns of the transformation matrix.
06

Form the Transformation Matrix

Combine the column vectors obtained in step 2 to form the matrix of the transformation T: \[T = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}\] #c) T(x, y, z) = (2x-7y-4z, 3x+y+4z, 6x-8y+z)#
07

Apply Transformation to Basis Vectors

Apply the given transformation T(x, y, z) = (2x-7y-4z, 3x+y+4z, 6x-8y+z) to each of the basis vectors (1,0,0), (0,1,0), and (0,0,1).
08

Obtain Column Vectors

Find the resulting vectors: T(1,0,0) = (2,3,6) T(0,1,0) = (-7,1,-8) T(0,0,1) = (-4,4,1) These vectors are the columns of the transformation matrix.
09

Form the Transformation Matrix

Combine the column vectors obtained in step 2 to form the matrix of the transformation T: \[T = \begin{bmatrix} 2 & -7 & -4 \\ 3 & 1 & 4 \\ 6 & -8 & 1 \end{bmatrix}\] So, the matrices representing the linear transformations T for each case are: (a) \[T = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}\] (b) \[T = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}\] (c) \[T = \begin{bmatrix} 2 & -7 & -4 \\ 3 & 1 & 4 \\ 6 & -8 & 1 \end{bmatrix}\]

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Key Concepts

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

Basis Vectors
Basis vectors are the building blocks of vector space in linear algebra. They provide a reference framework for vector spaces, such as \( \mathbf{R}^{3} \) in our exercise example. Imagine them as the directional lines on a grid that we use to map out every possible point or vector in that space.

For \( \mathbf{R}^{3} \), the usual basis consists of three vectors, \( \mathbf{e}_1 = (1, 0, 0), \mathbf{e}_2 = (0, 1, 0), \mathbf{e}_3 = (0, 0, 1) \). These are akin to the X, Y, and Z axes in 3D space. What makes basis vectors special is that they are linearly independent and span the vector space. This means any vector in \( \mathbf{R}^{3} \) can be expressed as a unique combination of these basis vectors.
Transformation Matrix
A transformation matrix is a tool that represents how a linear transformation changes vectors in a vector space. It encapsulates the entire function of the transformation into a matrix format, which allows us to quickly and easily apply the transformation to any vector.

In the provided exercise, we see the transformation matrices for different linear transformations. Applying these matrices to any vector in \( \mathbf{R}^{3} \) gives us a new vector that is the result of the transformation. The columns of a transformation matrix are the images of the basis vectors under the transformation, which indicates how each basis vector is repositioned.
Column Vectors
Column vectors are the vertical arrays of numbers in matrices. In the context of linear transformations, each column vector in a transformation matrix represents what happens to each basis vector after the transformation is applied.

For example, in the exercise, after finding the images of the basis vectors through the transformation \( T \), those images get placed vertically in our new matrix as the column vectors. This process is crucial for understanding the nature of the linear transformation and visualizing how space, or in particular \( \mathbf{R}^{3} \), is being warped or shifted.
Linear Algebra
Linear algebra is the branch of mathematics concerning linear equations, linear functions, and their representations through matrices and vector spaces. It's essential in almost all areas of mathematics and is pivotal in applied sciences and engineering.

The power of linear algebra arises from its ability to concisely represent complex operations and solve systems of equations. It offers the tools, like matrices and vectors, to model real-world problems in a low-dimensional space that can be visualized and computed efficiently. The exercises we solve in linear algebra, like finding transformation matrices, are not just abstract number-crunching activities; they have real implications across various scientific domains.

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Most popular questions from this chapter

Prove Theorem 6.3: For any linear operators \(G, F \in A(V),[G \circ F]=[G][F]\). Using the notation in Problem 6.10, we have $$ \begin{aligned} (G \circ F)\left(u_{i}\right) &=G\left(F\left(u_{i}\right)\right)=G\left(\sum_{j=1}^{n} a_{i j} u_{j}\right)=\sum_{j=1}^{n} a_{i j} G\left(u_{j}\right) \\ &=\sum_{j=1}^{n} a_{i j}\left(\sum_{k=1}^{n} b_{j k} u_{k}\right)=\sum_{k=1}^{n}\left(\sum_{j=1}^{n} a_{i j} b_{j k}\right) u_{k} \end{aligned} $$ Recall that \(A B\) is the matrix \(A B=\left[c_{i k}\right]\), where \(c_{i k}=\sum_{j=1}^{n} a_{i j} b_{j k}\). Accordingly, $$ [G \circ F]=(A B)^{T}=B^{T} A^{T}=[G][F] $$ The theorem is proved.

Prove Theorem 6.2: Let \(S=\left\\{u_{1}, u_{2}, \ldots, u_{n}\right\\}\) be a basis for \(V\) over \(K\), and let \(\mathbf{M}\) be the algebra of \(n\) square matrices over \(K .\) Then the mapping \(m: A(V) \rightarrow \mathbf{M}\) defined by \(m(T)=[T]_{S}\) is a vector space isomorphism. That is, for any \(F, G \in A(V)\) and any \(k \in K\), we have (i) \([F+G]=[F]+[G]\) (ii) \([k F]=k[F]\), (iii) \(m\) is one-to-one and onto. (i) Suppose, for \(i=1, \ldots, n\), $$ F\left(u_{i}\right)=\sum_{j=1}^{n} a_{i j} u_{j} \quad \text { and } \quad G\left(u_{i}\right)=\sum_{j=1}^{n} b_{i j} u_{j} $$ Consider the matrices \(A=\left[a_{i j}\right]\) and \(B=\left[b_{i j}\right] .\) Then \([F]=A^{T}\) and \([G]=B^{T}\). We have, for \(i=1, \ldots, n\), $$ (F+G)\left(u_{i}\right)=F\left(u_{i}\right)+G\left(u_{i}\right)=\sum_{j=1}^{n}\left(a_{i j}+b_{i j}\right) u_{j} $$ Because \(A+B\) is the matrix \(\left(a_{i j}+b_{i j}\right)\), we have $$ [F+G]=(A+B)^{T}=\underline{A}^{T}+B^{T}=[F]+[G] $$ (ii) Also, for \(i=1, \ldots, n\), $$ (k F)\left(u_{i}\right)=k F\left(u_{i}\right)=k \sum_{j=1}^{n} a_{i j} u_{j}=\sum_{j=1}^{n}\left(k a_{i j}\right) u_{j} $$ Because \(k A\) is the matrix \(\left(k a_{i j}\right)\), we have $$ [k F]=(k A)^{T}=k A^{T}=k[F] $$ (iii) Finally, \(m\) is one-to-one, because a linear mapping is completely determined by its values on a basis. Also, \(m\) is onto, because matrix \(A=\left[a_{i j}\right]\) in \(\mathbf{M}\) is the image of the linear operator, $$ F\left(u_{i}\right)=\sum_{j=1}^{n} a_{i j} u_{j}, \quad i=1, \ldots, n $$ Thus, the theorem is proved.

Suppose \(V=U+W\), and suppose \(U\) and \(V\) are each invariant under a linear operator \(F: V \rightarrow V\). Also, suppose dim \(U=r\) and \(\operatorname{dim} W=S\). Show that \(F\) has a block diagonal matrix representation \(M=\left[\begin{array}{ll}A & 0 \\ 0 & B\end{array}\right]\) where \(A\) is an \(r \times r\) submatrix.

Let \(S\) and \(S^{\prime}\) be bases of \(V\), and let \(1_{V}\) be the identity mapping on \(V\). Show that the matrix \(A\) representing \(1_{V}\) relative to the bases \(S\) and \(S^{\prime}\) is the inverse of the change-of-basis matrix \(P\) from \(S\) to \(S^{\prime} ;\) that is, \(A=P^{-1}\).

Find the trace and determinant of each of the following linear maps on \(\mathbf{R}^{3}\) : (a) \(F(x, y, z)=(x+3 y, 3 x-2 z, x-4 y-3 z)\). (b) \(G(x, y, z)=(y+3 z, 2 x-4 z, 5 x+7 y)\).
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