What is the basis and dimension of a specific space?

[brad sue]

Hi please can you help me in checking my work?
Find a basis for the following space of each of the following and specify its dimension.
2x₁+x₂-x₃+x₄=0
x₂+x₃-x₄=0
x₂+x₄=0
I found in solving the system that x₄ can be an arbitrary and I let x₄=t
The solution is:
x₁=t
x₂=-t
x₃=2t
x₄=t
The dimension is one.
But I am confused about this one:
Find a basis for the following space of each of the following and specify its dimension
2x₁+x₂-x₃+x₄=0

Thank you for your help

B.

 

[HallsofIvy]

Hi please can you help me in checking my work?
Find a basis for the following space of each of the following and specify its dimension.
2x₁+x₂-x₃+x₄=0
x₂+x₃-x₄=0
x₂+x₄=0

Are you sure that's exactly what the problem said? It doesn't make a lot of sense to me! For one thing I wonder what the "following space" is! Also, you appear to be thinking of these as simultaneous equations so it doensn't make sense to look at "each".

If the problem has said "find a basis for the solution set of the following" that would make sense.
To do that do exactly what you did:
Reduce the equations to the point where you can say x₄ is arbitrary, then let x₄= t and write
x1=t ,x2=-t, x3=2t, x4=t.

As you said the dimension is 1 and a basis would consist of a single vector derived from that by taking t to be whatever you want. In particular,
taking t= 1, {<1,-1, 2, 1>} is a basis.

 

[brad sue]

Are you sure that's exactly what the problem said? It doesn't make a lot of sense to me! For one thing I wonder what the "following space" is! Also, you appear to be thinking of these as simultaneous equations so it doensn't make sense to look at "each".
If the problem has said "find a basis for the solution set of the following" that would make sense.
To do that do exactly what you did:
Reduce the equations to the point where you can say x₄ is arbitrary, then let x₄= t and write
x1=t ,x2=-t, x3=2t, x4=t.
As you said the dimension is 1 and a basis would consist of a single vector derived from that by taking t to be whatever you want. In particular,
taking t= 1, {<1,-1, 2, 1>} is a basis.

Yes your are right I was mixing two problem in my head.
Thank you very much.
But what about the second problem?
Find a basis for the solution set of the following:
2x1 + x2 - x3 + x4 = 0
Since the only solution would be x1=x2=x3=x4=0, we can say that there is no basis, can't we?

Thank you

bertrand

 

[HallsofIvy]

Find a basis for the solution set of the following:
2x1 + x2 - x3 + x4 = 0
Since the only solution would be x1=x2=x3=x4=0, we can say that there is no basis, can't we?

.

But that's clearly NOT the only solution! There are a (triply) infinite number of solutions.

If x1, x2, x3 are any numbers, choose x4= -2x1- x2- x3 and the equation is clearly satisfied. Since you can choose 3 numbers arbitrarily, this has dimension 3 and any basis contains 3 vectors.. A simple way of finding a basis is to choose x1= 1, x2= x3= 0, then x1=0, x2= 1, x3= 0, then x1= x2= 0, x3= 1.