Suppose V and W are finite-dimensional. Then L.V; W / is finitedimensional and dimL(V,W) = D (dim V)( dim W)

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
icon
Related questions
Question

Given vector spaces VV and WW over FF. Let L(V,W)L(V,W) be the set of all linear maps from VV to WW. With addition and scalar multiplication as given in Definition 3.6 of the textbook, show that L(V,W)L(V,W) satisfies the distributive properties of a vector space.

 

Suppose V and W are finite-dimensional. Then L.V; W / is finitedimensional and
dimL(V,W) = D (dim V)( dim W)

84
CHAPTER 3 Linear Maps
Linear Maps Thought of as Matrix Multiplication
Previously we defined the matrix of a linear map. Now we define the matrix
of a vector.
3.62 Definition matrix of a vector, M(v)
Suppose v e V and v1,... Vn is a basis of V. The matrix of v with
respect to this basis is the n-by-1 matrix
M(v) =
Cn
where c1,..,Cn are the scalars such that
v = c1V1 +.+ CnVn-
The matrix M(v) of a vector v e V depends on the basis v1,.., Vn of V,
as well as on v. However, the basis should be clear from the context and thus
it is not included in the notation.
3.63 Example matrix of a vector
• The matrix of 2– 7x +5x³ with respect to the standard basis of P3(R)
is
()
2
-7
5
• The matrix of a vector x e F" with respect to the standard basis is
obtained by writing the coordinates of x as the entries in an n-by-1
matrix. In other words, if x = (x1,..., Xn) € F", then
()
X1
M(x) =
Xn
Occasionally we want to think of elements of V as relabeled to be n-by-1
matrices. Once a basis v1,..., Vn is chosen, the function M that takes v e V
to M(v) is an isomorphism of V onto F",1 that implements this relabeling.
Transcribed Image Text:84 CHAPTER 3 Linear Maps Linear Maps Thought of as Matrix Multiplication Previously we defined the matrix of a linear map. Now we define the matrix of a vector. 3.62 Definition matrix of a vector, M(v) Suppose v e V and v1,... Vn is a basis of V. The matrix of v with respect to this basis is the n-by-1 matrix M(v) = Cn where c1,..,Cn are the scalars such that v = c1V1 +.+ CnVn- The matrix M(v) of a vector v e V depends on the basis v1,.., Vn of V, as well as on v. However, the basis should be clear from the context and thus it is not included in the notation. 3.63 Example matrix of a vector • The matrix of 2– 7x +5x³ with respect to the standard basis of P3(R) is () 2 -7 5 • The matrix of a vector x e F" with respect to the standard basis is obtained by writing the coordinates of x as the entries in an n-by-1 matrix. In other words, if x = (x1,..., Xn) € F", then () X1 M(x) = Xn Occasionally we want to think of elements of V as relabeled to be n-by-1 matrices. Once a basis v1,..., Vn is chosen, the function M that takes v e V to M(v) is an isomorphism of V onto F",1 that implements this relabeling.
86
CHAPTER 3 Linear Maps
Because the result above allows us to think (via isomorphisms) of each
linear map as multiplication on F".1 by some matrix A, keep in mind that the
specific matrix A depends not only on the linear map but also on the choice
of bases. One of the themes of many of the most important results in later
chapters will be the choice of a basis that makes the matrix A as simple as
possible.
In this book, we concentrate on linear maps rather than on matrices. How-
ever, sometimes thinking of linear maps as matrices (or thinking of matrices
as linear maps) gives important insights that we will find useful.
Operators
Linear maps from a vector space to itself are so important that they get a
special name and special notation.
3.67 Definition operator, L(V)
A linear map from a vector space to itself is called an operator.
The notation L(V) denotes the set of all operators on V. In other
words, L(V) = L(V, V).
The deepest and most important
parts of linear algebra, as well as
most of the rest of this book, deal
with operators.
A linear map is invertible if it is
injective and surjective. For an op-
erator, you might wonder whether in-
jectivity alone, or surjectivity alone,
is enough to imply invertibility. On
infinite-dimensional vector spaces, neither condition alone implies invert-
ibility, as illustrated by the next example, which uses two familiar operators
from Example 3.4.
3.68 Example neither injectivity nor surjectivity implies invertibility
• The multiplication by x² operator on P(R) is injective but not surjective.
The backward shift operator on F is surjective but not injective.
In view of the example above, the next result is remarkable-it states
that for operators on a finite-dimensional vector space, either injectivity or
surjectivity alone implies the other condition. Often it is easier to check that
an operator on a finite-dimensional vector space is injective, and then we get
surjectivity for free.
Transcribed Image Text:86 CHAPTER 3 Linear Maps Because the result above allows us to think (via isomorphisms) of each linear map as multiplication on F".1 by some matrix A, keep in mind that the specific matrix A depends not only on the linear map but also on the choice of bases. One of the themes of many of the most important results in later chapters will be the choice of a basis that makes the matrix A as simple as possible. In this book, we concentrate on linear maps rather than on matrices. How- ever, sometimes thinking of linear maps as matrices (or thinking of matrices as linear maps) gives important insights that we will find useful. Operators Linear maps from a vector space to itself are so important that they get a special name and special notation. 3.67 Definition operator, L(V) A linear map from a vector space to itself is called an operator. The notation L(V) denotes the set of all operators on V. In other words, L(V) = L(V, V). The deepest and most important parts of linear algebra, as well as most of the rest of this book, deal with operators. A linear map is invertible if it is injective and surjective. For an op- erator, you might wonder whether in- jectivity alone, or surjectivity alone, is enough to imply invertibility. On infinite-dimensional vector spaces, neither condition alone implies invert- ibility, as illustrated by the next example, which uses two familiar operators from Example 3.4. 3.68 Example neither injectivity nor surjectivity implies invertibility • The multiplication by x² operator on P(R) is injective but not surjective. The backward shift operator on F is surjective but not injective. In view of the example above, the next result is remarkable-it states that for operators on a finite-dimensional vector space, either injectivity or surjectivity alone implies the other condition. Often it is easier to check that an operator on a finite-dimensional vector space is injective, and then we get surjectivity for free.
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 4 steps with 4 images

Blurred answer
Similar questions
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
Basic Technical Mathematics
Basic Technical Mathematics
Advanced Math
ISBN:
9780134437705
Author:
Washington
Publisher:
PEARSON
Topology
Topology
Advanced Math
ISBN:
9780134689517
Author:
Munkres, James R.
Publisher:
Pearson,