# Proving that a subset is a subspace

## 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##.

## 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
Mentor

## 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##.

## 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.
Mr Davis 97 said:
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>##.
Mr Davis 97 said:
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?

Last edited:
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
Mentor
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.

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?