# Find a basis

1. Oct 17, 2005

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.
Find a basis for the following space of each of the following and specify its dimension
2x1+x2-x3+x4=0

B.

2. Oct 18, 2005

### HallsofIvy

Staff Emeritus
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.

3. Oct 18, 2005

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

4. Oct 18, 2005

### HallsofIvy

Staff Emeritus
.

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.