Describing the intersection of two subspaces

[Mr Davis 97]

Homework Statement

$W_1 = \{(a_1, a_2, a_3) \in \mathbb{R}^3 : a_1 = 3a_3,~ a_3 = -a_2 \}$
$W_2 = \{(a_1, a_2, a_3) \in \mathbb{R}^3 : 2a_1 - 7a_2 + a_3 = 0 \}$
Given that these are two subspaces of $\mathbb{R}^3$, describe the intersection of the two, i.e. $W_1 \cap W_2$ and show that it is a subspace.

Homework Equations

The Attempt at a Solution

Would it be sufficient just to say that $W_1 \cap W_2 = \{(a_1, a_2, a_3) \in \mathbb{R}^3 : 2a_1 - 7a_2 + a_3 = 0,~a_1 = 3a_3,~ a_3 = -a_2 \}$ and proceed to show that it is a subspace?

 

[FactChecker]

You should look at those equations and you can draw some strong conclusions.

 

[Mr Davis 97]

You should look at those equations and you can draw some strong conclusions.

Ah, so would it be more correct to say that $W_1 \cap W_2 = \{ (0,0,0) \}$? If I leave it in the form $W_1 \cap W_2 = \{(a_1, a_2, a_3) \in \mathbb{R}^3 : 2a_1 - 7a_2 + a_3 = 0,~a_1 = 3a_3,~ a_3 = -a_2 \}$ is that incorrect?

 

[FactChecker]

Ah, so would it be more correct to say that $W_1 \cap W_2 = \{ (0,0,0) \}$?

Yes. W₁ is a line through (0,0,0) and W₂ is a plane through (0,0,0). There are only two ways those can intersect. Either the W₁ line is on the W₂ plane or (0,0,0) is the only intersection point. In this case it is the later.

If I leave it in the form $W_1 \cap W_2 = \{(a_1, a_2, a_3) \in \mathbb{R}^3 : 2a_1 - 7a_2 + a_3 = 0,~a_1 = 3a_3,~ a_3 = -a_2 \}$ is that incorrect?

It is technically correct but I assume that more is desired.

 

[Mr Davis 97]

Yes. W₁ is a line through (0,0,0) and W₂ is a plane through (0,0,0). There are only two ways those can intersect. Either the W₁ line is on the W₂ plane or (0,0,0) is the only intersection point. In this case it is the later.
It is technically correct but I assume that more is desired.

So about in the case when we have
$W_1 = \{(a_1, a_2, a_3) \in \mathbb{R}^3 : a_1 = 3a_3,~ a_3 = -a_2 \}$
$W_2 = \{(a_1, a_2, a_3) \in \mathbb{R}^3 : a_1 - 4a_2 - a_3 = 0 \}$
If we want to find $W_1 \cap W_2$, would this just be $W_1$ since the intersection of a line and a three-dimensional space is just the line?

 

[Mark44]

So about in the case when we have
$W_1 = \{(a_1, a_2, a_3) \in \mathbb{R}^3 : a_1 = 3a_3,~ a_3 = -a_2 \}$
$W_2 = \{(a_1, a_2, a_3) \in \mathbb{R}^3 : a_1 - 4a_2 - a_3 = 0 \}$
If we want to find $W_1 \cap W_2$, would this just be $W_1$ since the intersection of a line and a three-dimensional space is just the line?

Your reasoning is wrong here. Neither of these subspaces is the entire R³. Both of them are planes, and both include the point (0, 0, 0).

 

[Mr Davis 97]

Your reasoning is wrong here. Neither of these subspaces is the entire R³. Both of them are planes, and both include the point (0, 0, 0).

So in that case can I only leave $W_1 \cap W_2$ like $\{(a_1, a_2, a_3) \in \mathbb{R}^3 : a_1 = 3a_3,~ a_3 = -a_2,~a_1 - 4a_2 + a_3 = 0 \}$

 

[Mark44]

So in that case can I only leave $W_1 \cap W_2$ like $\{(a_1, a_2, a_3) \in \mathbb{R}^3 : a_1 = 3a_3,~ a_3 = -a_2,~a_1 - 4a_2 + a_3 = 0 \}$

It depends on what they're asking for, which could be what sort of geometric object is represented by this intersection, what's its equation, what is a basis for this intersection, among others.

 

[FactChecker]

So about in the case when we have
$W_1 = \{(a_1, a_2, a_3) \in \mathbb{R}^3 : a_1 = 3a_3,~ a_3 = -a_2 \}$
$W_2 = \{(a_1, a_2, a_3) \in \mathbb{R}^3 : a_1 - 4a_2 - a_3 = 0 \}$
If we want to find $W_1 \cap W_2$, would this just be $W_1$ since the intersection of a line and a three-dimensional space is just the line?

I think you calculated that wrong. If
$W_2 = \{(a_1, a_2, a_3) \in \mathbb{R}^3 : a_1 + 4a_2 + a_3 = 0 \}$
then I would agree that $W_1 \cap W_2$ = $W_1$ since $3a_3 + 4(-a_3) + a_3 = 0$ so all the points on W₁ are on the plane W₂.