Show that {u+v,v-w,w-u} is a basis of R^3 if {u,v,w} is

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In summary, the conversation discusses proving that {u+v,v-w,w-u} is a basis for (R3) given that {u,v,w} is also a basis for (R3). The conversation includes a discussion of proving linear independence and the use of simultaneous equations to do so. It is also mentioned that in this case, it is possible to prove the basis by inspection.
  • #1
Hydroxide
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Homework Statement



Suppose {u,v,w} is a basis for (R3). Show that {u+v,v-w,w-u} is also a basis for (R3).

The Attempt at a Solution



By definition, u,v and w are linearly independent and they span (R3).
I really don't know where to go from here though.

Thanks
 
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  • #2
Can you prove that u+v , v-w , w-v is also linearly independent? If so, do you know why that's enough to make it a base?
 
  • #3
I think there is an error in what is written. That set of three vectors is obvisouly linearly dependent: w-v = -(v-w).
 
  • #4
matt grime said:
I think there is an error in what is written. That set of three vectors is obvisouly linearly dependent: w-v = -(v-w).

My bad, its fixed now
 
  • #5
Was post 2 sufficient to help you? There is an easier way, in this case. It is perfectly possible to do it by inspection. Just adding some of those vectors together yields that u,v,w are all in the span of the three new ones, hence they form a basis (assuming char=/=2).
 
  • #6
daniel_i_l said:
Can you prove that u+v , v-w , w-v is also linearly independent? If so, do you know why that's enough to make it a base?
How would I show that they are Linearly independent.

matt grime said:
Was post 2 sufficient to help you? There is an easier way, in this case. It is perfectly possible to do it by inspection. Just adding some of those vectors together yields that u,v,w are all in the span of the three new ones, hence they form a basis (assuming char=/=2).
But part (ii) is :
Suppose {u,v,w,x} is a basis for R4. Show that {u+v,v+w,w+x,x+u} is NOT a basis for R4 So I can't really do it by inspection.

Thanks guys
 
  • #7
Yes, you can. Because it is easy to see by inspection that those 4 elements are linearly dependent. It is also good to spot easy things. And it will work in general, since that new set of vectors is a basis if and only if you can expess u,v,w as combinations of the three new ones, and it is easy to see how to do that by eye. In general, you need to set up and solve some simultaneous equations. In this case you can see the solution without doing and linear algebra.

As for your other question: you know how to solve simultaneous equations, so do it. Find the a,b,c such that a(u+v)+b(v-w)+c(w-u)=0, and show that the only solution is for a=b=c=0. Note you have to rearrange to write things in terms of u,v,w which are known to be a basis. More succintly, note this implies (a-c)u+(a+b)v+(c-b)w=0, so since, u,v,w are a basis, this implies a-c=a+b=c-b=0, whence a=b=c=0.
 
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  • #8
How do you show things are linearly independent in general? Remember that lin. independence comes from

[tex]c_1 \mathbf{v_1} + c_2 \mathbf{v_2} + ... + c_n \mathbf{v_n} = 0[/tex]

Is there a way to pick the constants in the linear combinations such that you get one of the other vectors?
 
  • #9
thanks everyone and especially matt grime for your patience
I've done it now.
 
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FAQ: Show that {u+v,v-w,w-u} is a basis of R^3 if {u,v,w} is

What is a basis in linear algebra?

A basis in linear algebra is a set of linearly independent vectors that can be used to represent any other vector in a vector space. It provides a framework for understanding the structure and operations of the vector space.

How do you prove that a set of vectors is a basis?

To prove that a set of vectors is a basis, you need to show that the vectors are linearly independent and span the entire vector space. This means that none of the vectors can be expressed as a linear combination of the others, and every vector in the vector space can be written as a linear combination of the basis vectors.

What does it mean for {u+v,v-w,w-u} to be a basis of R^3?

This means that the three vectors {u+v,v-w,w-u} form a basis for the vector space R^3, which consists of all 3-dimensional vectors. Any vector in R^3 can be written as a linear combination of these three vectors, and they are linearly independent.

How does the fact that {u,v,w} is a basis of R^3 relate to {u+v,v-w,w-u} being a basis?

Since {u,v,w} is a basis of R^3, it means that these three vectors are linearly independent and span the entire vector space. By combining these three vectors in different ways (i.e. adding and subtracting them), we can form a new basis {u+v,v-w,w-u} for R^3 that also has these properties.

Can you provide an example of how to show that {u+v,v-w,w-u} is a basis of R^3 if {u,v,w} is?

Yes, for example, if we have the vectors u = (1,0,0), v = (0,1,0), and w = (0,0,1), we can show that they form a basis for R^3. Then, by adding and subtracting these vectors, we can create a new basis {u+v,v-w,w-u} which would be (1,1,0), (-1,1,0), and (1,-1,-1). We can then verify that these three vectors are linearly independent and span R^3, making them a basis for the vector space.

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