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【转载】Linear Transformations on Rn  

2017-10-10 08:34:44|  分类: 线性代数 |  标签: |举报 |字号 订阅

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本文转载自jia_huiqiang《Linear Transformations on Rn》

Definition of a Linear Transformation

In your travels throughout your mathematical career there has been one theme that persists in every course.  That theme is functions.  Recall that a function is a rule that assigns every element from a domain set to a unique element of a range set.  If the domain and range are both the real numbers, then a function is the familiar real valued function.  If the domain is a real number and the range is Rn then the function is a vector valued function or a parametrically defined curve.  If the domain is Rn and the range is the real numbers, then the function is a function of several variables.  In linear algebra we are interested in special functions where the domain is Rn and the range is Rm

Definition

Let

        L:  Rn  --->  Rm

be a function such that the following two properties hold:

  1. L(u + v)  =  L(u) + L(v)
  2. L(cu)  =  cL(u)

Then the function is called a linear transformation.

This definition calls for some examples. 

 

Example

Let

        L:  R2  --->  R3

be defined by

        L(x,y)  =  (y, x, x + y)

Show that L is a linear transformation.

 

Solution

First, we prove the first property.  Suppose that

        u  =  (u1,u2)        and       v  =  (v1,v2)

then

        L(u + v)  =  L(u1 + v1, u2 + v2)

        =  (u2 + v2, u1 + v1, u1 + v1 + u2 + v2)

and

        L(u) + L(v)  =  (u2, u1, u1 + u2) + (v2, v1, v1 + v2)

         =  (u2 + v2, u1 + v1, u1 + u2 +v1 + v2)

Hence

        L(u + v)  =  L(u) + L(v)

Now for the second property.  We have

        L(cu)  =  L(c(u1, u2))  =  L(cu1, cu2

        =  (cu2, cu1, cu1 + cu2)  =  c(u2, u1, u1 + u2)  =  cL(u)

Since properties 1 and 2 hold, we can conclude that L is a linear transformation.

Example

Show that the function

        f:  R3 --->  R2

defined by

        f(x, y, z)  =  (xy, yz)

is not a linear transformation

 

Solution

To show a function is not a linear transformation, we just need to find an example that demonstrates the failure of one of the properties.  We have

        f(2(3, 4, 5))  =  f(6, 8, 10)  =  (48, 80)

and

        2(f(3, 4, 5))  =  2(12, 20)  =  (24, 40)

since these are not equal, we can conclude that f is not a linear transformation.

 

It is a simple consequence to the two properties that if L is a linear transformation then

        L(c1v1 + c2v2 + ... + ckvk)  =  c1L(v1) + c2L(v2) + ... + ckL(vk)

or in sigma notation

        L(Scivi)  =  S L(civi)

The Matrix of a Linear Transformation

 

Example/Theorem

Let A be an m x n matrix and let

        L:  Rn  --->  Rm

be defined by

        L(u)  =  AuT

Then L is a linear transformation.

 

Proof

The proof is just a matter of stating the corresponding properties of matrices. We have

        L(u + v)  =  A(u + v)T  =  A(uT + vT)  

                       =  AuT + AvT  =  L(u) + L(v)

and 

        L(cu)  =  A(cu)T  =  A(cuT)  

                   =  cAuT  =  cL(u)

  

The converse is also true.

 

Theorem

Let 

        L:  Rm  --->  Rn 

be a linear transformation.  Then there is a unique matrix A such that 

        L(u)  =  AuT

 

Proof

Recall that the vector ei is the vector with ith component equal to 1 and all others zero.  Any vector 

        v  =  (v1,v2, ... ,vm)  =  v1e1 + v2e2 + ... + vmem 

We let the ith column of A be the vector

        L(ei)

Notice that 

        Aei  =  the ith column of A 

so that

        L(ei)  =  Aei 

Then 

        L(v)  =  L(v1, v2, ... , vm)  =  L(v1e1 + v2e2 + ... + vmem)  

        =   v1L(e1) + v2L(e2) + ... + vmL(em)  =  v1Ae1 + v2Ae2 + ... + vmAem 

        =  A(v1e1 + v2e2 + ... + vmem)  =   AvT 

To show that the matrix is unique, we just notice that if B is a matrix with 

        L(v)  =  BvT

then 

        Bei  =  L(ei)  =  Aei 

so that the columns of A and B are the same.

 

Example

For the linear transformation from the first example,

        L(x, y)  =  (y, x, x + y)

We have

        L(1, 0)  =  (0, 1, 1)        L(0, 1)  =  (1, 0, 1)

so that the matrix A that represents the linear transformation is

        Linear Transformations on Rn - jia_huiqiang - 平安甜橙的博客

Properties of Linear Transformations

There are a few notable properties of linear transformation that are especially useful.  They are the following.

  1. L(0)  =  0
  2. L(u - v)  =  L(u) - L(v)

Notice that in the first property, the 0's on the left and right hand side are different.  The left hand 0 is the zero vector in Rm and the right hand 0 is the zero vector in Rn.  The proofs of these can be done in two ways.  One way is to use the definition of a linear transformation.  A quicker way is to note that a linear transformation can be represented as a matrix.  These two properties are just properties of matrices.

reference:http://ltcconline.net/greenl/courses/203/Vectors/linearTransRn.htm

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