Spans and linear independance

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In summary, the conversation discusses solving problems related to spans and linear independence. It starts with a homework statement and then goes on to show how to prove different parts of the problem. The experts suggest using specific equations and the definition of "span" to solve the problem. They also provide a counterexample to prove that the equality in part b.) does not hold if the assumption of linear independence is dropped. The conversation ends with a final solution for part c.) and a further counterexample to support the solution.
  • #1
karnten07
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[SOLVED] Spans and linear independance

Homework Statement



Let V be a vector over a field F.

a.) Let x1,...,xn[tex]\in[/tex]V and y1,...,ym[tex]\in[/tex]V. Show that

Span(x1,...,xn,y1,...,ym) = Span(x1,...,xn) + Span(y1,...,ym)

B.) Let x1, x2, x3, x4 be four linearly independent vectors in V. Show hat

Span(x1, x2,x3) [tex]\cap[/tex] Span(x2, x3, x4) = Span(x2,x3)

c.) Show that the equality in part b.) does not hold if we drop the assumption that x1, x2, x3, x4 are linearly independent.

The Attempt at a Solution



a.) Does it suffice to show:
For a,b[tex]\in[/tex]R,

(a1x1+...+anxn+b1y1+...+bmym) = (a1x1+...+anxn)+(b1y1+...+bmym) ?

b.)
Does it suffice to show:
For a,b[tex]\in[/tex]R,

Span(x1, x2,x3) [tex]\cap[/tex] Span(x2, x3, x4)= (a1x2(1)+...+anx2n+b1x3(1)+...+bmx3m) = Span(x2, x3) ?

c.) If what i have doen so far is reasonably correct, the only part I'm unsure about is part c.) I will do some reading about it if i can, but any hints would be greatly appreciated. Thanks
 
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  • #2
karnten07 said:

Homework Statement



Let V be a vector over a field F.

a.) Let x1,...,xn[tex]\in[/tex]V and y1,...,ym[tex]\in[/tex]V. Show that

Span(x1,...,xn,y1,...,ym) = Span(x1,...,xn) + Span(y1,...,ym)

B.) Let x1, x2, x3, x4 be four linearly independent vectors in V. Show hat

Span(x1, x2,x3) [tex]\cap[/tex] Span(x2, x3, x4) = Span(x2,x3)

c.) Show that the equality in part b.) does not hold if we drop the assumption that x1, x2, x3, x4 are linearly independent.

The Attempt at a Solution



a.) Does it suffice to show:
For a,b[tex]\in[/tex]R,

(a1x1+...+anxn+b1y1+...+bmym) = (a1x1+...+anxn)+(b1y1+...+bmym) ?
Yes, that is exactly right.

b.)
Does it suffice to show:
For a,b[tex]\in[/tex]R,

Span(x1, x2,x3) [tex]\cap[/tex] Span(x2, x3, x4)= (a1x2(1)+...+anx2n+b1x3(1)+...+bmx3m) = Span(x2, x3) ?
By "x2(1)", "x2n", etc, you mean the components of vector x1? You shouldn't need to look at components. Just use the definition of "span". You will need to show why vectors x1 and x4 are not needed- and that has to do with the fact that the four vectors are independent.

c.) If what i have doen so far is reasonably correct, the only part I'm unsure about is part c.) I will do some reading about it if i can, but any hints would be greatly appreciated. Thanks
To prove that something is NOT true, you just need a counter-example. Take 4 vectors in, in R2, say, that are NOT idependent and see shy you can't do that. In fact, taking x1= x2= x3= x4 should work!
 
  • #3
HallsofIvy said:
Yes, that is exactly right.


By "x2(1)", "x2n", etc, you mean the components of vector x1? You shouldn't need to look at components. Just use the definition of "span". You will need to show why vectors x1 and x4 are not needed- and that has to do with the fact that the four vectors are independent.


To prove that something is NOT true, you just need a counter-example. Take 4 vectors in, in R2, say, that are NOT idependent and see shy you can't do that. In fact, taking x1= x2= x3= x4 should work!

Oh yes, so for b.)

Span(x1, x2,x3)[tex]\cap[/tex] Span(x2, x3, x4)= (ax2+bx3) = Span(x2, x3)
(cx1, dx4) [tex]\neq[/tex] (ax2, bx3) and so x1 and x4 are not of Span(x2, x3)

I doubt this is enough to show what you are asking, how do i show x1 and x4 are not needed.
 
  • #4
Don't talk in generalities- be specific and write specific equations:

If v is in span{x1, x2, x3}, what does that mean? v= what?

If v is in span{x2, x3, x4}, what does that mean? v= what?

If v is in the intersection of the two spans, then both of those are true. Set them equal, shift everything to one side of the equation and use the fact that the vectors are independent.
 
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  • #5
HallsofIvy said:
Don't talk in generalities- be specific and write specific equations:

If v is in span{x1, x2, x3}, what does that mean? v= what?

If v is in span{x2, x3, x4}, what does that mean? v= what?

If v is in the intersection of the two spans, then both of those are true. Set them equal, shift everything to one side of the equation and use the fact that the vectors are independent.

So for v[tex]\in[/tex]R,

Span(x1, x2,x3) [tex]\cap[/tex]Span(x2, x3, x4)= (v1x1+v2x2+v3x3)[tex]\cap[/tex](v2x2+v3x3+v4x4)= v2x2+v3x3= Span(x2,x3)

Because the vectors are linearlry independent there is only one trivial solution to the equation:

v1x1+v2x2+v3x3=v2x2+v3x3+v4x4=0

Which is v1=v2=v3=v4=0.Therefore x1 and x4 are not in the given intersection asked for in the question.

Is this right?? Thanks for all you help
 
  • #6
For part c.)

If x1, x2, x3, x4 are not linearly independent then

v1x1+v2x2+v3x3=v2x2+v3x3+v4x4=0

has a solution other than the trivial solution, that is v1, v2, v3, v4 are not all zero.

Therefore,

Span(x1, x2,x3)[tex]\in[/tex] Span(x2, x3, x4)= (v1x1+v2x2+v3x3)[tex]\cap[/tex](v2x2+v3x3+v4x4)=v1x1+v2x2+v3x3+v4x4= Span(x1, x2, x3, x4)

If v1, v2, v3, v4 [tex]\neq[/tex]0

Is this right? I am not sure if i need to say that all the vectors aren't zero or if just not all of them are zero?

Edit: Oh yes, i can do this one by counterexample,

So if x1, x2, x3, x4 are not linearly independent,

Span(x1, x2,x3)[tex]\in[/tex] Span(x2, x3, x4)[tex]\neq[/tex]Span(x2,x3)

By counterexample:

If x1=x2=x3=x4 and For v [tex]\in[/tex]R,

Span(x1, x2,x3)[tex]\in[/tex] Span(x2, x3, x4)= (v1x1+v2x2+v3x3)[tex]\cap[/tex](v2x2+v3x3+v4x4)=v1x1+v2x2+v3x3+v4x4= Span(x1, x2, x3, x4)[tex]\neq[/tex]Span(x2,x3)

Is this okay? Also was my other answer sufficient, if not it would be helpful to know why. Thanks
 
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  • #7
karnten07 said:
So for v[tex]\in[/tex]R,

Span(x1, x2,x3) [tex]\cap[/tex]Span(x2, x3, x4)= (v1x1+v2x2+v3x3)[tex]\cap[/tex](v2x2+v3x3+v4x4)= v2x2+v3x3= Span(x2,x3)

Because the vectors are linearlry independent there is only one trivial solution to the equation:

v1x1+v2x2+v3x3=v2x2+v3x3+v4x4=0
You are assuming here that the coefficents of x2 and x3, in the two linear combinations, are the same! That is not necessarily true.

I find the use of v1, etc. confusing, since "v" is often used to represent a vector. If v is in both spans then
v= ax1+ bx2+ cx3= px2+ qx3+ rx4 for some numbers, a, b, c, p, q, r.

Which is v1=v2=v3=v4=0.Therefore x1 and x4 are not in the given intersection asked for in the question.

Is this right?? Thanks for all you help
If all four are 0, then NONE of the vectors are in the intersection!

Hard to tell since you haven't said that what "v1", etc. are! They must be members of the field in order that "v1x1", etc. make sense. In that case, "Span(x1, x2,x3) [tex]\cap[/tex]Span(x2, x3, x4)= (v1x1+v2x2+v3x3)[tex]\cap[/tex](v2x2+v3x3+v4x4)= v2x2+v3x3= Span(x2,x3)" makes no sense because "Span(x1, x2,x3) [tex]\cap[/tex]Span(x2, x3, x4)" is a set of vectors while (v1x1+v2x2+v3x3) is a single vector.

Again, you are writing generalities. Answer the questions I asked: if v is in Span(x1, x2, x3, x4), v= what? If v is in Span(x2, x3, x4), v= what? Write done the equations for v.
 
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  • #8
karnten07 said:
For part c.)

If x1, x2, x3, x4 are not linearly independent then

v1x1+v2x2+v3x3=v2x2+v3x3+v4x4=0
Again, you are assuming that the one combination has the same coefficients as the other.

has a solution other than the trivial solution, that is v1, v2, v3, v4 are not all zero.

Therefore,

Span(x1, x2,x3)[tex]\in[/tex] Span(x2, x3, x4)= (v1x1+v2x2+v3x3)[tex]\cap[/tex](v2x2+v3x3+v4x4)=v1x1+v2x2+v3x3+v4x4= Span(x1, x2, x3, x4)

If v1, v2, v3, v4 [tex]\neq[/tex]0

Is this right? I am not sure if i need to say that all the vectors aren't zero or if just not all of them are zero?

Edit: Oh yes, i can do this one by counterexample,

So if x1, x2, x3, x4 are not linearly independent,

Span(x1, x2,x3)[tex]\in[/tex] Span(x2, x3, x4)[tex]\neq[/tex]Span(x2,x3)

By counterexample:

If x1=x2=x3=x4 and For v [tex]\in[/tex]R,

Span(x1, x2,x3)[tex]\in[/tex] Span(x2, x3, x4)= (v1x1+v2x2+v3x3)[tex]\cap[/tex](v2x2+v3x3+v4x4)=v1x1+v2x2+v3x3+v4x4= Span(x1, x2, x3, x4)[tex]\neq[/tex]Span(x2,x3)

Is this okay? Also was my other answer sufficient, if not it would be helpful to know why. Thanks
I'm not sure I like this counter example. If x1= x2= x3= x4, the saying you need all of them is really the same as saying you only need x2 and x3.

Consider this: x1= <1, 1, 0>, x2= <0, 1, 0>, x3= <0, 0, 1>, x4= <1, 0, 1>.

What does a vector in their intersection look like? Can you write it using only x2 and x3?
 
  • #9
Thats great hallsofivy, i think i see why the counterexample wasnt good and the other example is better. I think I've got it now. Thanks
 
  • #10


Suppose that u, v and w are vectors in Rn. Show that if w E span{u,v} then span{u,v} = span{u,v,w}
 
  • #11


How do I tackle this problem?
 

What is a span?

A span is a set of all possible linear combinations of a given set of vectors. In other words, it is the space that can be formed by multiplying each vector in the set by a constant and adding them together.

What is linear independence?

Linear independence is a property of a set of vectors where no vector can be written as a linear combination of the other vectors in the set. In simpler terms, it means that none of the vectors in the set can be "created" by adding or multiplying the other vectors together.

Why is linear independence important?

Linear independence is important because it allows us to represent a set of vectors in a more concise and efficient way. It also helps us to understand the relationships between vectors and to solve problems related to vector spaces.

How can I determine if a set of vectors is linearly independent?

To determine if a set of vectors is linearly independent, you can use the determinant method or the coefficient method. The determinant method involves calculating the determinant of a matrix formed by the vectors, and if the determinant is non-zero, the vectors are linearly independent. The coefficient method involves setting up a system of equations and solving for the constants, and if there is only one unique solution, the vectors are linearly independent.

What is the difference between linear independence and orthogonality?

Linear independence and orthogonality are two different properties of vectors. Linear independence refers to the ability to create a set of vectors that cannot be "created" by adding or multiplying other vectors together. Orthogonality, on the other hand, refers to the perpendicular relationship between two vectors, meaning they form a right angle. While a set of linearly independent vectors can also be orthogonal, it is not always the case, and vice versa.

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