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【转载】the span of a set of vectors  

2017-10-10 08:35:42|  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《the span of a set of vectors》

Span

Definition

We have seen in the last discussion that the span of vectors v1, v2, ... , vn is the set of linear combinations

        c1v1 + c2v2 + ... + cnvn  

and that this is a vector space.  

We now take this idea further.  If V is a vector space and S  =  {v1, v2, ... , vn) is a subset of V, then is Span(S) equal to V?

Let V be a vector space and let S  =  {v1, v2, ... , vn) be a subset of VWe say that S spans V if every vector v in V can be written as a linear combination of vectors in S.

        v  =  c1v1 + c2v2 + ... + cnvn  

Example

Show that the set 

        S =  {(0,1,1), (1,0,1), (1,1,0)} 

spans R3 and write the vector (2,4,8) as a linear combination of vectors in S.

 

Solution

A vector in R3 has the form 

        v  =  (x, y, z)

Hence we need to show that every such v can be written as

        (x,y,z)  =  c1(0, 1, 1) + c2(1, 0, 1) + c3(1, 1, 0)

        =  (c2 + c3, c1 + c3, c1 + c2)

This corresponds to the system of equations

              c2 + c3  =  x
        c1 +       c3  =  y
        c1 + c2         =  z

which can be written in matrix form

        span - jia_huiqiang - 平安甜橙的博客

We can write this as 

        Ac  =  b

Notice that 

        det(A)  =  2

Hence A is nonsingular and

        c  =  A-1b

So that a nontrivial solution exists.  To write (2,4,8) as a linear combination of vectors in S, we find that    

        span - jia_huiqiang - 平安甜橙的博客

so that 

        span - jia_huiqiang - 平安甜橙的博客

We have

        (2,4,8)  =  5(0,1,1) + 3(1,0,1) + (-1)(1,1,0)

 

Example

Show that if

        v1  =  t + 2        and         v2  =  t2 + 1

and  S  =  {v1, v2}

then 

        S does not span P2

 

Solution

A general element of P2 is of the form 

        v   =  at2 + bt + c

We set 

        v  =  c1v1 + c2v2 

or

        at2 + bt + c  =  c1(t + 2) + c2(t2 + 1)  =  c2t2 + c1t + c1 + c2 

Equating coefficients gives

        a  =  c2
   
      b  =  c1
   
      c  =  c1 + c2 

Notice that if 

        a  =  1        b  =  1        c  =  1

there is no solution to this.  Hence S does not span V.

 

Example

Let 

        span - jia_huiqiang - 平安甜橙的博客

Find a spanning set for the null space of A.

 

Solution

We want the set of all vectors x with 

        Ax  =  0

We find that the rref of A is

        span - jia_huiqiang - 平安甜橙的博客

The parametric equations are

        x1  =  7s + 6t
        x2  =  -4s - 5t
        x3  =  s
        x4  =  t

We can get the span in the following way.  We first let

        s  =  1        and         t  =  0

to get 

        v1  =  (7,-4,1,0)   

and let

        s  =  0        and     t  =  1

to get

        v2  =  (6,-5,0,1)

If we let S  =  {v1,v2} then S spans the null space of A.

 

Linear Independence

We now know how to find out if a collection of vectors span a vector space.  It should be clear that if S  =  {v1, v2, ... , vn) then Span(S) is spanned by S.  The question that we next ask is are there any redundancies.  That is, is there a smaller subset of S that also span Span(S).  If so, then one of the vectors can be written as a linear combination of the others. 

        vi  =  c1v1 + c2v2 + ... + ci -1vi -1 + ci+1vi+1 + ... + cnvn 

If this is the case then we call S a linearly dependent set.  Otherwise, we say that S is linearly independent.  There is another way of checking that a set of vectors are linearly dependent.

Theorem

Let S  =  {v1, v2, ... , vn) be a set of vectors, then S is linearly dependent if and only if 0 is a nontrivial linear combination of vectors in S.  That is, there are constants c1, ..., cn with at least one of the constants nonzero with

         c1v1 + c2v2 + ... + cnvn  =  0 

Proof

Suppose that S is linearly dependent, then

        vi  =  c1v1 + c2v2 + ... + ci -1vi -1 + ci+1vi+1 + ... + cnvn 

Subtracting vi from both sides, we get

         c1v1 + c2v2 + ... + ci -1vi -1 + vi  + ci+1vi+1 + ... + cnvn  =  0

In the above equation ci  =  1 which is nonzero, so that 0 is a nontrivial linear combination of vectors in S.  

Now let 

         c1v1 + c2v2 + ... + ci -1vi -1 + civi  + ci+1vi+1 + ... + cnvn  =  0

with ci nonzero.  Divide both sides of the equation by ci and let aj  =  -cj / ci to get

         -a1v1 - a2v2 - ... - ai -1vi -1 + vi - ai+1vi+1 - ... - anvn  =  0

finally move all the terms to the other right side of the equation to get

        vi  =  a1v1 + a2v2 + ... + ai -1vi -1 + ai+1vi+1 + ... + anvn 

Example

Show that the set of vectors 

        S  =  {(1, 1, 3, 4),  (0, 2, 3, 1),  (4, 0, 0, 2)}

are linearly independent.

 

Solution

We write

        c1(1, 1, 3, 4) + c2(0, 2, 3, 1) + c3(4, 0, 0, 2)  =  0

We get four equations

        c1 + 4c3  =  0
        c1 + 2c2  =  0
        3c1 + 3c2  =  0
        4c1 + c2 + 2c3  =  0

The matrix corresponding to this homogeneous system is 

        span - jia_huiqiang - 平安甜橙的博客

and

        span - jia_huiqiang - 平安甜橙的博客

Hence 

        c1  =  c2  =  c3  =  0

and we can conclude that the vectors are linearly independent.

Example

Let   

        S  =  {cos2 t, sin2 t, 4)

then S is a linearly dependent set of vectors since

        4  =  4cos2t + 4sin2t

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