# Homework Help: Linear Algebra - Finding different flats

1. Sep 30, 2011

### Throwback

1. The problem statement, all variables and given/known data

I'm given the following equations in R^4 which make up set A:

x + y = 1
x + y + z + w = 2
x + y - z - w = 0
z + w = 1

After some row reduction to row echelon form, I get the following in vector form:

[x y z w] = [1 0 1 0] + s[-1 1 0 0] + t[0 0 -1 1]

Questions:
1 Identify a one flat in set A
2 Identify a one flat not in set A
3 Identify a two flat that intersects A in a point
4 Identify a two flat that intersects A in a line
5 Identify a two flat that does not intersect A

2. Relevant equations

x + y = 1
x + y + z + w = 2
x + y - z - w = 0
z + w = 1

x = 1 - s
y = s
z = 1 - t
w = t

[x y z w] = [1 0 1 0] + s[-1 1 0 0] + t[0 0 -1 1]

3. The attempt at a solution

1

I'm assuming the equation in vector form would work here, but first I have to remove a parameter to make it a one flat:

[x y z w] = [1 0 1 0] + s[-1 1 0 0]

Would changing the parameter s (and thus changing the flat by some multiple) also work?

2

Again using the following in vector form

[x y z w] = [1 0 1 0] + s[-1 1 0 0] + t[0 0 -1 1]

if i change the position vector [1 0 1 0] to [0 0 0 0], it will be parallel to the original, and thus not in set A?

I'm also thinking this might work also because if we look at the original equations and the parametric equations

x + y = 1
x + y + z + w = 2
x + y - z - w = 0
z + w = 1

x = 1 - s
y = s
z = 1 - t
w = t

we can see that by setting the position vector to zero would result in

x = -s
y = s
z = -t
w = t

and now plugging these into the original equations

x + y = 1 becomes -s + s = 0 which is a contradiction and hence by setting the position vector to zero, the new set of equations are not found in A

And now because it must be a one flat, I'll remove a parameter, leaving me with the following

[x y z w] = [0 0 0 0] + s[-1 1 0 0]

3

I'm really not sure.

4

[x y z w] = [1 0 1 0] + s[1 1 0 0] + t[0 0 -1 1] ???

5

Find a contradiction like before? I'll use the original equation without the position vector

[x y z w] = s[-1 1 0 0] + t[0 0 -1 1]

I'll change s[-1 1 0 0] and t[0 0 -1 1] to s[1 1 0 0] and t[0 0 -1 1]. respectively. Now plugging these new vectors in the following parametric equations from before, I'll have the following

so now x = y

from

x = 1 - s
y = s

we have

1 -s = s, a contradiction