# Homework Help: Confused about subspaces!

1. Dec 3, 2011

### kr0z3n

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

Determine whether the following sets form subspaces of ℝ2:
(a) {(x1, x2)T | x1 + x2 = 0}

(b) {(x1, x2)T | x1 * x2 = 0}

2. Relevant equations

3. The attempt at a solution
I know that a is a subspace and b is not, but I would like to know why.
For part A, I let x=[c, -c]T
∂[c,-c]= [∂c, -∂c]
[c, -c] + [ ∂, -∂] = [c+∂, -c-∂]
Thus S is closed under scalar multiplication and addition.
But what if I let x=[1, -1]? Wouldn't that break the conditions since
∂[1,-1]=[∂,-∂] and [1,-1] + [1, -1]= [2,-2]?

And for part B the book states "No, this is not a subspace. Every element of S has at
least one component equal to 0. The set is closed under scalar multiplication, but
not under addition. For example, both (1, 0)T and (0,1)T are elements of S, but their sum is not."

But can't I let [x1 and x2] be the zero vectors and S would be a subspace?

I am confused about how sometimes I can multiply or add using variables and other times I have to use constants. Can someone please explain to me. Thanks

Last edited: Dec 3, 2011
2. Dec 3, 2011

### Quinzio

well, x1=2, x2=-2, so x1+x2=0, so it's still a member of the subspace.

3. Dec 3, 2011

### kru_

In order for a set to form a subspace, it must meet the three criteria. It must contain the identity, and be closed under addition and scalar multiplication.

Closure under an operation means that when the operation is performed on any two elements from a set, it will produce another element from the set.

For example, for any two real numbers a and b, a + b is still a real number. Thus a + b is still in the set of real numbers. Hence, the real numbers are closed under addition.

The key fact is that the operation must produce an element from the set for any two elements. Every possible combination of elements must produce another element in the set when the operation is applied on them. So if you can find two real numbers that, when added together, produce a complex number, then you would have proven that the real numbers are not closed under addition.

For part B you want to take any two vectors from R2, such that both vectors are in the set, but their sum is not. In order for a vector to be an element of set B, it needs to have components x1, x2 such that x1*x2=0.

Your answer sheet gives you two elements that fit the definition of set B, but when you add them together, the resulting vector has components that do not fit the definition. Hence the result of addition on the set B is not contained in B. So the set B is not closed under addition.

So your thought to let x1 and x2 be zero is insufficient to show closure, since closure demands the operation holds for every element of the set.

4. Dec 4, 2011

### HallsofIvy

It is not sufficient that there exist vectors of the set whose sum is in the set, it must be true that the sum of any two vectors in the set is in the set.

I have no idea what you mean by this. What "variables" or "constants"? Could you give an example?

5. Dec 5, 2011

### kr0z3n

After reading all the comments, I understand it now! Thanks to all who replied!

6. Jan 16, 2012

### Ernie

I have read through these posts and still don't seem to understand how to determine whether the following sets form subspaces of ℝ$^{2}$ :
(a) {(x$_{1}$ , x$_{2}$ )$^{T}$ | x$_{1}$ + x$_{2}$ = 0}
(b) {(x$_{1}$ , x$_{2}$ )$^{T}$ | x$_{1}$x$_{2}$ = 0}