# Math Help - Which subsets of R^3 are subspaces?

1. ## Which subsets of R^3 are subspaces?

SUBSPACES !

Hi, I don't understand subspaces! I came across this question in homework tasks, and I'm guessing it's very important to the course?!

Which of the following subsets of R^3 are subspaces? (Justify answers)

(a) A = {(s, 0, 2s) | s Є R }
(b) B = {(x, y, z) | y ≥ z}
(c) C = {(t+1, t, 3t) | t Є R }
(d) D = {(x, y, z) | 4x + y - z = 0}

2. Originally Posted by iExcavate
SUBSPACES !

Hi, I don't understand subspaces! I came across this question in homework tasks, and I'm guessing it's very important to the course?!

Which of the following subsets of R^3 are subspaces? (Justify answers)

(a) A = {(s, 0, 2s) | s Є R }
(b) B = {(x, y, z) | y ≥ z}
(c) C = {(t+1, t, 3t) | t Є R }
(d) D = {(x, y, z) | 4x + y - z = 0}

A subspace is a vector space which is contained in the bigger vector space.

If you look at your notes you will see that you only have two things to prove - closure under addition and closure under scalar multiplication. This is because we get the other axioms for free from the big space.

That is, if you are wanting to prove that for $U \subset V$ that $U$ is a subspace of $V$ you need to check that $\alpha u \in U$ for all $\alpha$ in the field, $u \in U$, and that for all $u_1, u_2 \in U$, $u_1+u_2 \in U$.

I believe two of the four are subspaces, the other two are not closed under scalar multiplication.

3. Originally Posted by iExcavate
SUBSPACES !

Hi, I don't understand subspaces! I came across this question in homework tasks, and I'm guessing it's very important to the course?!

Which of the following subsets of R^3 are subspaces? (Justify answers)
To be a subspace, we must have closure under addtion and scalar multiplication- and we can put those together- A is a subspace if and only if, for all u and v in A and scalars a and b, au+ bv is also in A.

(a) A = {(s, 0, 2s) | s Є R }
Let u= (s, 0, 2s) and v= (t, 0, 2t). au+ bv= (as, 0, 2as)+ (bt, 0, 2bt)= (as+ bt, 0, 2(as+ bt)). Is that also in A?

(b) B = {(x, y, z) | y ≥ z}
Is -1(x, y, z) also in B?

(c) C = {(t+1, t, 3t) | t Є R }
Is 0(t+1, t, 3t) in C? (The 0 vector, of course, is in every subspace.)

(d) D = {(x, y, z) | 4x + y - z = 0}
Let u= (x, y, z) and v= (i, j, k) where 4x+ y- z= 0 and 4i+ j- k= 0 so that they are in D. au+ bv= (ax, ay, az)+ (bi, bj, bk)= (ax+ bi, ay+ bj, az+ bk). For that to be in D, we must have 4(ax+ bi)+ (ay+ bj)- (az+ bk)= 0. Rewrite that as (4ax+ ay- az)+ (4bi+ bj- bk= a(4x+ y- z)+ b(4i+ j- k).

4. Originally Posted by HallsofIvy

Is -1(x, y, z) also in B?

I don't know? Is it? Wouldn't it just be (-x, -y, -z) with the rule saying that y is greater than or equal to z?

Also, I found A is a subspace, B not sure yet, C isn't a subspace, and I'm guessing D is a subspace?

5. Originally Posted by iExcavate
I don't know? Is it? Wouldn't it just be (-x, -y, -z) with the rule saying that y is greater than or equal to z?

Also, I found A is a subspace, B not sure yet, C isn't a subspace, and I'm guessing D is a subspace?
If $y \geq z$ then is $-y \geq -z$? Take an example, for instance $4 \geq 3$. Is $-4 \geq -3$?

Part D does give you a subspace. What you need to do is check that the equation holds when you multiply by a scalar and when you add two vectors together. That is, if $4x_1 + y_1 - z_1 = 0$ and if $4x_2 + y_2 - z_2 = 0$ then does

$4\alpha x_2 + \alpha y_2 + \alpha z_2 = 0$ hold? (As $(\alpha x_1, \alpha y_1, \alpha z_1)$ must be of the form required).

$4(x_1 + x_2) + (y_1 + y_2) - (z_1 + z_2) = 0$ hold? (As $(x_1 + x_2, y_1 + y_2, z_1 + z_2)$ must be of the form required).

Originally Posted by HallsofIvy
To be a subspace, we must have closure under addtion and scalar multiplication- and we can put those together- A is a subspace if and only if, for all u and v in A and scalars a and b, au+ bv is also in A.

Let u= (s, 0, 2s) and v= (t, 0, 2t). au+ bv= (as, 0, 2as)+ (bt, 0, 2bt)= (as+ bt, 0, 2(as+ bt)). Is that also in A?

Is -1(x, y, z) also in B?

Is 0(t+1, t, 3t) in C? (The 0 vector, of course, is in every subspace.)

Let u= (x, y, z) and v= (i, j, k) where 4x+ y- z= 0 and 4i+ j- k= 0 so that they are in D. au+ bv= (ax, ay, az)+ (bi, bj, bk)= (ax+ bi, ay+ bj, az+ bk). For that to be in D, we must have 4(ax+ bi)+ (ay+ bj)- (az+ bk)= 0. Rewrite that as (4ax+ ay- az)+ (4bi+ bj- bk= a(4x+ y- z)+ b(4i+ j- k).
What was wrong with my first post, and whatever happened to letting the OP think for themselves?