Proving that a subset is a subspace

[Mr Davis 97]

Homework Statement

Determine whether $W = \{(a_1, a_2, a_3) \in \mathbb{R}^3 : a_1 = 3a_3,~ a_3 = -a_2 \}$ is a subspace of $\mathbb{R}^3$.

Homework Equations

The Attempt at a Solution

To show that a subset of vector space is a subspace we need to show three things: 1) That the zero vector of R^3 is in W. 2) That W is closed under vector addition. 3) That W is closed under scalar multiplication.

1) if $a_2 = 0$ then $a_1 = 0,~a_3 = 0$, so the zero vector is in W.
2) I'm not exactly sure how to clearly show this one. Here is my attempt: $(a_1, a_2, a_3)+ (b_1, b_2, b_3) = (a_1 + b_1, a_2 + b_2, a_3 + b_3) = (3(a_2 + b_2), a_2 + b_2, -(a_2 + b_2))$, which is of the form of the vector defined in W.
3) Not exactly sure how to show this one either, but here is my attempt: $c(a_1, a_2, a_3) = (ca_1, ca_2, ca_3) = (3(ca_2), ca_2, -(ca_2))$, which is of the form of the vectors in W.

Thus, W is a subspace of R^3

Is this a correct proof? Am I doing 2) and 3) right or is there a better way?

 

[Mark44]

Homework Statement

Determine whether $W = \{(a_1, a_2, a_3) \in \mathbb{R}^3 : a_1 = 3a_3,~ a_3 = -a_2 \}$ is a subspace of $\mathbb{R}^3$.

Homework Equations

The Attempt at a Solution

To show that a subset of vector space is a subspace we need to show three things: 1) That the zero vector of R^3 is in W. 2) That W is closed under vector addition. 3) That W is closed under scalar multiplication.

1) if $a_2 = 0$ then $a_1 = 0,~a_3 = 0$, so the zero vector is in W.
2) I'm not exactly sure how to clearly show this one. Here is my attempt: $(a_1, a_2, a_3)+ (b_1, b_2, b_3) = (a_1 + b_1, a_2 + b_2, a_3 + b_3) = (3(a_2 + b_2), a_2 + b_2, -(a_2 + b_2))$, which is of the form of the vector defined in W.

Start with two vectors that clearly belong to W, such as $u = <-3u_2, u_2, -u_2>$ and $v = <-3v_2, v_2, -v_2>$.

Edit: From the problem statement, I determined that all three coordinates are directly or indirectly related to the second coordinate, so I wrote all three coordinates in terms of the second.

3) Not exactly sure how to show this one either, but here is my attempt: $c(a_1, a_2, a_3) = (ca_1, ca_2, ca_3) = (3(ca_2), ca_2, -(ca_2))$, which is of the form of the vectors in W.

Similar idea as above -- start with $u = <-3u_2, u_2, -u_2>$.

Thus, W is a subspace of R^3

Is this a correct proof? Am I doing 2) and 3) right or is there a better way?

 

[Mr Davis 97]

Start with two vectors that clearly belong to W, such as $u = <-3u_2, u_2, -u_2>$ and $v = <-3v_2, v_2, -v_2>$.

Edit: From the problem statement, I determined that all three coordinates are directly or indirectly related to the second coordinate, so I wrote all three coordinates in terms of the second.
Similar idea as above -- start with $u = <-3u_2, u_2, -u_2>$.

So what about for more complicated potential subspaces, such as $W = \{(a_1, a_2, a_3) \in \mathbb{R}^3 : 5a_1^2 - 3a_2^2 + 6a_3^2 = 0 \}$? Would adding specific solutions to see if it is closed be better than solving for $a_1$ and putting that into the tuple and adding that to another tuple of the same form to see if that's closed?

 

[Mark44]

So what about for more complicated potential subspaces, such as $W = \{(a_1, a_2, a_3) \in \mathbb{R}^3 : 5a_1^2 - 3a_2^2 + 6a_3^2 = 0 \}$? Would adding specific solutions to see if it is closed be better than solving for $a_1$ and putting that into the tuple and adding that to another tuple of the same form to see if that's closed?

You could solve for a₁ in terms of the other two variables.

$a_1 = \pm \sqrt{(3/5)a_2^2 - (6/5)a_3^2}$
$a_2 = a_2$
$a_3 = a_3$
Now if you have two such vectors in this set, is their sum in the set? Is a scalar multiple of this vector in the set? I didn't check for a zero vector, since there is obviously such a vector in the set.

 

[Mr Davis 97]

You could solve for a₁ in terms of the other two variables.

$a_1 = \pm \sqrt{(3/5)a_2^2 - (6/5)a_3^2}$
$a_2 = a_2$
$a_3 = a_3$
Now if you have two such vectors in this set, is their sum in the set? Is a scalar multiple of this vector in the set? I didn't check for a zero vector, since there is obviously such a vector in the set.

Well, $(\pm \sqrt{(3/5)a_2^2 - (6/5)a_3^2}) + (\pm \sqrt{(3/5)b_2^2 - (6/5)b_3^2}) \neq \pm \sqrt{(3/5)(a_2 + b_2)^2 - (6/5)(a_2 + b_2)^2}$, so is that enough to show that it is not closed under addition?