Homework Help: Finding a basis

1. Aug 29, 2011

Warpenguin

I'm moving on to my next section of work and i come across this example:
Consider the homogeneous system
x + 2y − z + u + 2v = 0
x + y + 2z − 3u + v = 0

It asks for a basis to be found for the solution space S of this system. And also what is the dimension of S.

I know this might be basic but I can't remember how to form a matrix from that system. I believe I will know how to answer the question once i have the matrix. Can anyone help?

2. Aug 29, 2011

micromass

Let me do an example. From the system

$$\left\{\begin{array}{c} 2x-y+z=0\\ 3x-2y+z=0\\ 5x+9y-z=0\end{array}\right.$$

yields the matrix

$$\left(\begin{array}{cccc} 2 & -1 & 1 & 0\\ 3 & -2 & 1 & 0\\ 5 & 9 & -1 & 0 \end{array}\right)$$

Did you see what I did?? Can you do the same thing?

3. Aug 29, 2011

HallsofIvy

Another way of looking at it:
If we subtract the second equation from the first, we get y- 3z+ 4u+ v= 0. Solving for y,
y= 3z- 4u- v. Any vector, (x, y, z, u, v) satisfying those equations can be written as (x, 3z-4u-v, z, u, v)= (x, 0, 0, 0)+ (0, 3z, z, 0, 0)+ (0, -4u, 0, u, 0)+ (0, -v, 0, 0, v). Can you get the basis for the solution space from that? What is the dimension?

4. Aug 30, 2011

Warpenguin

Well I got the matrix
(1 2 -1 1 2 0)
(1 1 2 -3 1 0)

Then I used Gaussian elimination and I'm not sure if I did it correctly but I did end up with
(1 0)
(0 1)

which is my basis and the dimension is 2?

5. Aug 30, 2011

Warpenguin

I think what I did was wrong. If I follow what HallsofIvy said; is this how I'm supposed to do it:
(x, 3z-4u-v, z, u, v)= (x, 0, 0, 0)+ (0, 3z, z, 0, 0)+ (0, -4u, 0, u, 0)+ (0, -v, 0, 0, v)
= x(1, 0, 0, 0) + z(0, 3, 1, 0, 0) + u(0, -4, 0, 1, 0) + v(0, -1, 0, 0, 1)
i.e v1=(1, 0, 0, 0) v2=(0, 3, 1, 0, 0) v3=(0, -4, 0, 1, 0) v4=(0, -1, 0, 0, 1)
which are linearly independent, thus S={v1, v2, v3, v4} is the basis and the dimension=4

6. Aug 30, 2011

Warpenguin

double post

7. Aug 30, 2011

Char. Limit

I'm pretty sure that you did it incorrectly, as Gaussian elimination doesn't usually take a 5x2 matrix and make it 2x2.

8. Aug 30, 2011

Warpenguin

Char.Limit can you check if my post after that is correct?

9. Aug 30, 2011

Char. Limit

Yeah, that looks right to me.

10. Aug 30, 2011

stallionx

I would like to pose a question here that i am not so clear about, if someone would please help.

All Basis Vectors should be orthogonal to each other . Is that right ?

Thank You.

11. Aug 30, 2011

HallsofIvy

No, that is not correct. All we require of a basis is that it span the space and its vectors be independent.

Of course, it's nice if the basis is orthogonal- for any vector v, the coefficient in its expansion in an orthogonal basis $\{v_i\}$ is just $<v, v_i>/<v_i, v_i>$, where < , > is the inner product. It's even better if the basis is orthonormal. In that case the coefficient is just $<v, v_i>$.

Note, by the way, that in order to talk about "orthogonal" and/or "normal", a vector space must have an inner product and a vector space, in general, does not have to have an inner product.