Homework Help: Simple linear algebra questions

1. Jun 1, 2006

Simfish

So as for the basis of a vector space, I can express it in terms of 2 variables, say, x3 and x4, or x1 and x2, etc... Does it matter which two variables I express the components in the form of? I would think that there is no difference, but just making sure...

Also, aren't the null space and range complementary? Such that you need to do one to form the other?

Last edited: Jun 1, 2006
2. Jun 1, 2006

Simfish

The question though, is how can we prove that the two spans, expressed in terms of different variables, are the same?

3. Jun 1, 2006

matt grime

I think you ought to post an example to illustrate what you mean for the first part. A matrix does not have basis. Vector spaces have bases. A matrix is not a vector space.

The kernel (null space) and range, assuming you're speaking of a linear map (matrix) from V to V, are not complementary. They can even coincide: exrcise, find an example in 2x2 matrices.

4. Jun 1, 2006

matt grime

If you're asking how can we tell if two sets of vectors span the same subspace, then put them in two matrices (as the rows) and put the matrices in reduced row echelon form.

5. Jun 1, 2006

Simfish

Ok. So I have to find the basis of -x1 + 2x2 - x4 = 0 and x2 + x3 = 0. Now the official solution expresses them in terms of x3 and x4. I expressed them in terms of x1 and x3. So then essentially, I have to put them in matrices and then simplify to reduced echelon form to see if they are row equivalent or not, right?

edit: okay, it's easy to see. Same thing. And we can see that it forms a basis once again. Just in different variables - but we re-arrange them to see that it doesn't really matter.

Last edited: Jun 1, 2006
6. Jun 1, 2006

matt grime

What are x1, x2 and so on?

7. Jun 1, 2006

Simfish

Just components of x, which is a subspace of R4

8. Jun 1, 2006

HallsofIvy

Please state problems clearly! What you mean is that you have to find a basis for the subspace of R4 of vectors of the for <x1, x2, x3, x4> that satisfy those equations: -x1 + 2x2 - x4 = 0 and x2 + x3 = 0.

Here's one way to do that: since there are two equations in four unknowns, we can expect to be able to solve for two of them in terms of the other 2. Here it is obvious, from the second equation, that x3= -x2. Since x3 doesn't appear in the first equation at all, we can just solve it for, say, x1 in terms of x2 and x4: x1[\sub]= 2x2- x4.
Now we know that 3= -x2 and x1[\sub]= 2x2- x4 for any vectors in this subspace. If we take x2= 1, x4= 0, then x3= -1 and x1= 2(1)- 0= 2 so the vector is <2, 1, -1, 0>. If we take x2= 0, x4= 0 then we have x3= 0 and x1= 2(0)- 1= -1 so the vector is <-1, 0, 0, 1>. Those two vectors form a basis for the subspace.