# Vector subspaces

1. Mar 6, 2014

### Panphobia

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

(x,y,z) where 2x + 2y + z = 1

Is this set a subspace of R^3?

3. The attempt at a solution

I am thinking it is not since it does not contain the origin since
2(0)+2(0) + 0 = 1
0 != 1
(!= means not equal)
Am I right? I am kind of having trouble with this part of linear algebra since the textbook is so vague about this.

2. Mar 7, 2014

### Staff: Mentor

Yes, this is enough to show that this set is not a subspace of R3. BTW, the set is a plane in R3.

3. Mar 7, 2014

### Panphobia

Yea I know that, but how can I prove that something is a subspace of R^3? I know that proof against is easier, since you only need one tuple to disprove. But to do you need to do anything extra?

4. Mar 7, 2014

### Staff: Mentor

To prove that a subset of a vector space (such as R3) is actually a subspace of that vector space, you need to show three things:
1. The zero vector is an element of the set.
2. If u and v are arbitrary elements of the set, then u + v is also an element of the set. (This shows that the set is closed under vector addition.)
3. If u is an arbitrary element of the set, and k is a scalar, then ku is also an element of the set. (This shows that the set is closed under scalar multiplication.)
That's it.

It's a lot more work to show that a set together with two operations (vector addition and scalar multiplication) is actually a vector space - you have to verify that all 10 or so vector space axioms are satisfied. If the set already belongs to a vector space, then all you need to verify are the three I listed above.

5. Mar 7, 2014

### Panphobia

Thanks, I figured it out, and kind of on the same subject. If you are given u = (1,2,-1,1), v = (2,-1,1,0), and w = (-1,2,0,3) write x = (2,-3,-2,-8) as a linear combination of u,v, and w. I really don't know where to start, the text book gives examples of 3 3D vectors, not 3 4D vectors so how would I go ()u + ()v + ()w if you have 4 values for x?

6. Mar 7, 2014

### kduna

You can use good old row reduction to do this. Set up an augmented matrix whose first three columns are $u, v, w$. Then make the last column $x$. Then row reduce this.

Note: You could also do it by inspection, although that is harder and may require a lot of playing around/ getting lucky.

Last edited: Mar 7, 2014
7. Mar 7, 2014

### Staff: Mentor

$$\begin{bmatrix} 2 \\ -3 \\ -2 \\ -8\end{bmatrix} = c_1\begin{bmatrix}1 \\ 2 \\ -1 \\ 1 \end{bmatrix} + c_2\begin{bmatrix}2 \\ -1 \\ 1 \\0 \end{bmatrix} + c_3\begin{bmatrix}-1 \\ 2 \\ 0 \\ 3 \end{bmatrix}$$