# 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

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.