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Chapter 5: Problem 49

Show that the following mappings are linear: (a) \(F: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}\) defined by \(F(x, y, z)=(x+2 y-3 z, 4 x-5 y+6 z)\). (b) \(F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\) defined by \(F(x, y)=(a x+b y, c x+d y)\), where \(a, b, c, d\) belong to \(\mathbf{R}\).

Short Answer

Expert verified
In conclusion, both mappings are linear as they satisfy the additivity and homogeneity conditions. For mapping (a), we showed that \(F(u+v) = F(u) + F(v)\) and \(F(\alpha u) = \alpha F(u)\) for any vectors \(u\) and \(v\) in \(\mathbf{R}^3\), and for mapping (b), we demonstrated the same properties for any vectors \(u\) and \(v\) in \(\mathbf{R}^2\).

Step by step solution

01

Check linearity for Mapping (a)

First, let's consider the mapping \(F: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}\) defined by \(F(x, y, z)=(x+2 y-3 z, 4 x-5 y+6 z)\). 1. Additivity: Let \(u = (u_1, u_2, u_3)\) and \(v = (v_1, v_2, v_3)\) be vectors in \(\mathbf{R}^3\). Then we have: $$F(u+v) = F((u_1+v_1, u_2+v_2, u_3+v_3)) = (u_1+v_1+2(u_2+v_2)-3(u_3+v_3), 4(u_1+v_1)-5(u_2+v_2)+6(u_3+v_3))$$ $$= (u_1+2 u_2-3 u_3+v_1+2 v_2-3 v_3, 4 u_1-5 u_2+6 u_3+4 v_1-5 v_2+6 v_3) = F(u) + F(v)$$ 2. Homogeneity: Let \(u = (u_1, u_2, u_3)\) be a vector in \(\mathbf{R}^3\) and let \(\alpha\) be a scalar. Then we have: $$F(\alpha u) = F(\alpha u_1, \alpha u_2, \alpha u_3) = (\alpha u_1+2 (\alpha u_2)-3 (\alpha u_3), 4 (\alpha u_1)-5 (\alpha u_2)+6 (\alpha u_3))$$ $$= \alpha(u_1+2 u_2-3 u_3, 4 u_1-5 u_2+6 u_3) = \alpha F(u)$$ Since both additivity and homogeneity properties are satisfied, the mapping (a) is linear.
02

Check linearity for Mapping (b)

Now, let's consider the mapping \(F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\) defined by \(F(x, y)=(a x+b y, c x+d y)\), where \(a, b, c, d\) belong to \(\mathbf{R}\). 1. Additivity: Let \(u = (u_1, u_2)\) and \(v = (v_1, v_2)\) be vectors in \(\mathbf{R}^2\). Then we have: $$F(u+v) = F((u_1+v_1, u_2+v_2)) = (a(u_1+v_1)+b(u_2+v_2), c(u_1+v_1)+d(u_2+v_2))$$ $$= (au_1+bu_2+av_1+bv_2, cu_1+du_2+cv_1+dv_2) = (au_1+bu_2, cu_1+du_2) + (av_1+bv_2, cv_1+dv_2) = F(u) + F(v)$$ 2. Homogeneity: Let \(u = (u_1, u_2)\) be a vector in \(\mathbf{R}^2\) and let \(\alpha\) be a scalar. Then we have: $$F(\alpha u) = F(\alpha u_1, \alpha u_2) = (a(\alpha u_1)+b(\alpha u_2), c(\alpha u_1)+d(\alpha u_2))$$ $$= \alpha(au_1+bu_2, cu_1+du_2) = \alpha F(u)$$ Since both additivity and homogeneity properties are satisfied, the mapping (b) is linear.

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Key Concepts

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

Linearity Properties
When we discuss linearity properties in mathematics, particularly in the context of linear mappings or transformations, we are referring to two fundamental characteristics that define linearity: additivity and homogeneity. Understanding these properties is crucial as they are foundational to the concept of linear transformations.

  • Additivity implies that if you take two vectors, say u and v, in a vector space, the result of applying a linear transformation F to the sum of u and v鈥攚ritten as F(u + v)鈥攕hould be the same as applying F to each vector individually and then adding the results, i.e., F(u) + F(v). This property ensures that the mapping respects vector addition.
  • Homogeneity, on the other hand, concerns scalar multiplication. It states that scaling a vector by some number 伪, and then applying the linear transformation鈥�贵(伪耻)鈥攕hould yield the same result as applying the transformation to the vector first and then scaling the result, denoted as 伪贵(耻).
It鈥檚 easy to see how these properties play out in practice. For example, in exercise (a) involving a mapping from 鈩澛� to 鈩澛�, the linearity properties are carefully checked through straightforward algebraic proofs. This rigorous approach instills in us the appreciation of the structure and predictability that linear mappings provide in higher mathematics.
Linear Transformation
A linear transformation is essentially a function between two vector spaces that preserves the operations of vector addition and scalar multiplication, as dictated by the linearity properties. This means that a linear transformation is compatible with the structure of the vector spaces involved.

Returning to the exercise's examples, both (a) and (b) showcase functions that take vectors from one space and 'transform' them into vectors in another space. For instance, the function in (a) takes a three-dimensional vector and gives back a two-dimensional one while maintaining the structural integrity of the vector space operations. Hence, these transformations are termed 'linear' because they satisfy the conditions of additivity and homogeneity.

A fascinating aspect of linear transformations is their broad applicability across various fields such as physics, engineering, and economics. They provide a simplified but powerful way to model and analyze a wide range of phenomena that exhibit linear characteristics.
Vector Spaces
Vector spaces are a pivotal concept in linear algebra and a central stage where linear transformations perform their 'magic'. A vector space is a collection of objects, called vectors, where two operations are defined: vector addition and scalar multiplication. These operations must comply with certain axioms that guarantee the consistency and structure of the space.

In both exercises (a) and (b), we're looking at different instances of vector spaces: 鈩澛� and 鈩澛�. It's crucial to grasp that irrespective of the dimension, these spaces share the same fundamental rules that allow us to perform algebraic manipulations confidently. Understanding vector spaces is key to grasping how vectors can be added together or multiplied by scalars and how these operations relate to linear transformations. As we've seen, such transformations only 'make sense' within the context and rules provided by these remarkable mathematical constructs.
Homogeneity
Homogeneity in the context of linear mappings is one of the cornerstone properties that define a linear transformation. It ensures that transformations are 'scale-invariant' in a sense鈥攚hen a vector is scaled by a scalar, the transformation respects this scaling.

The demonstration of homogeneity in both exercises (a) and (b) involves taking a scalar multiple of a vector and showing that you can 'pull out' the scalar when applying the linear transformation. This feature is crucial when scaling objects in various applications, such as computer graphics and data analysis. It maintains proportional relationships and allows for the coherent magnification of vector quantities. The notion of homogeneity extends beyond mathematics, with applications in economics, physics, and other sciences when describing systems or phenomena that are consistent with respect to scale.

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Most popular questions from this chapter

Let \(V\) be of finite dimension and let \(T\) be a linear operator on \(V\) for which \(T R=I\), for some operator \(R\) on \(V .\) (We call \(R\) a right inverse of \(T\).) (a) Show that \(T\) is invertible. (b) Show that \(R=T^{-1}\). (c) Give an example showing that the above need not hold if \(V\) is of infinite dimension. (a) Let \(\operatorname{dim} V=n .\) By Theorem \(5.14, T\) is invertible if and only if \(T\) is onto; hence, \(T\) is invertible if and only if \(\operatorname{rank}(T)=n .\) We have \(n=\operatorname{rank}(I)=\operatorname{rank}(T R) \leq \operatorname{rank}(T) \leq n .\) Hence, \(\operatorname{rank}(T)=n\) and \(T\) is invertible. (b) \(T T^{-1}=T^{-1} T=I .\) Then \(R=I R=\left(T^{-1} T\right) R=T^{-1}(T R)=T^{-1} I=T^{-1}\). (c) Let \(V\) be the space of polynomials in \(t\) over \(K\); say, \(p(t)=a_{0}+a_{1} t+a_{2} t^{2}+\cdots+a_{s} t^{5}\). Let \(T\) and \(R\) be the operators on \(V\) defined by $$ T(p(t))=0+a_{1}+a_{2} t+\cdots+a_{s} t^{s-1} \quad \text { and } \quad R(p(t))=a_{0} t+a_{1} t^{2}+\cdots+a_{s} t^{s+1} $$ We have $$ (T R)(p(t))=T(R(p(t)))=T\left(a_{0} t+a_{1} t^{2}+\cdots+a_{s} t^{s+1}\right)=a_{0}+a_{1} t+\cdots+a_{5} t^{s}=p(t) $$ and so \(T R=I\), the identity mapping. On the other hand, if \(k \in K\) and \(k \neq 0\), then $$ (R T)(k)=R(T(k))=R(0)=0 \neq k $$ Accordingly, \(R T \neq I\)

Determine the number of different mappings from \((a)\\{1,2\\}\) into \(\\{1,2,3\\},(b)\\{1,2, \ldots, r\\}\) into \(\\{1,2, \ldots, s\\}\).

For any mapping \(f: A \rightarrow B\), show that \(\mathbf{1}_{B} \circ f=f=f \circ \mathbf{1}_{A}\).

Suppose the mapping \(F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\) is defined by \(F(x, y)=(x+y, x) .\) Show that \(F\) is linear.

Suppose \(F: V \rightarrow U\) is linear and \(k\) is a nonzero scalar. Prove that the maps \(F\) and \(k F\) have the same kemel and the same image.
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