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?