Is the Intersection of Two Subspaces Always a Subspace?

[lkh1986]

Homework Statement

R = { (a+1, b 0) | a, b are real numbers}
S = { (a+b, b, c) | a, b, c are real numbers)
T = R intersect S

I have shown that R and S are subspaces of R^3. Now I have to determine whether T is also a subspace of R^3.

The answer provided is that yes, T is also a subspace, because by theorem, if A and B are subspaces, then their intersection will also be a subspace.

Homework Equations

The 3 conditions for W to be a subspace.

(i) W is nonempty, or vector 0 inside W.
(ii) If u, v inside W, then u+ also inside W.
(iii) If u inside W and k is a scalar value, ku also inside W.

I have a question here for condition (i). Does it mean that if (0,0,0) inside W, then W is nonempty? I mean, W is nonempty because it has 3 elements in (0,0,0)? Take an example, let U = {(x,y,5)|x, y are real}, so U is not a subspace by condition (i) because the third entry or element will always be 5 and will never be 0. Am I right?

The Attempt at a Solution

To find the intersection of R and S, I equate a+1 = a+b, I will be b = 1.
Also, b=b and c=0.

So I will get T = {(a+1,1,0)| a is a real number}

Since (0,0,0) is not inside T, (because the second entry will always be 1 and never be 0) so T is NOT a subspace. However, I know from theorem that any intersection of 2 subspaces must also be a subspace. So there's a contradiction here.

The only thing that I can think of that could go wrong is that I cannot equate a + 1 = a+b and then b=1, because a, b are just dummy variables. They could be any real numbers. So, while both R and S use the unknown a in their description or notation, they somehow cannot be regarded as the "same" thing?

Thanks.

 

[HallsofIvy]

Homework Statement

R = { (a+1, b 0) | a, b are real numbers}
S = { (a+b, b, c) | a, b, c are real numbers)
T = R intersect S

I have shown that R and S are subspaces of R^3. Now I have to determine whether T is also a subspace of R^3.

The answer provided is that yes, T is also a subspace, because by theorem, if A and B are subspaces, then their intersection will also be a subspace.

Homework Equations

The 3 conditions for W to be a subspace.

(i) W is nonempty, or vector 0 inside W.
(ii) If u, v inside W, then u+ also inside W.
(iii) If u inside W and k is a scalar value, ku also inside W.

I have a question here for condition (i). Does it mean that if (0,0,0) inside W, then W is nonempty? I mean, W is nonempty because it has 3 elements in (0,0,0)? Take an example, let U = {(x,y,5)|x, y are real}, so U is not a subspace by condition (i) because the third entry or element will always be 5 and will never be 0. Am I right?

Yes, the fact that (0, 0, 0) is in W means it is nonempty. The fact that anything is in W means it is non-empty! But the point is not just that (0, 0, 0) "has 3 elements", it is because (0, 0, 0)= (-1+1, 0, 0), with a= -1, b= 0 is in R and (0, 0, 0)= (0+ 0, 0, 0) with a= b= c= 0 is in T and so is in their intersection.

If you were to look only at "(i) W is nonempty" then the fact that (0, 0, 5) is in a set proves it is nonempty! However, (ii) would fail. Once (0, 0, 5) is in the set (0, 0, 5)+ (0, 0, 5)= (0, 0, 10) whichi s NOT of the form (x, y, 5). The reason you can use either "W is nonempty" or "vector 0 is inside W" is because if v is in W then by (iii) so is -v and then, by (ii) v+ (-v)= 0 is in W.

Yes, if (0, 0, 0) is in W, then it is nonempty and if (ii) and (iii) are also true then T is a subspace of R^3.

The Attempt at a Solution

To find the intersection of R and S, I equate a+1 = a+b, I will be b = 1.
Also, b=b and c=0.

So I will get T = {(a+1,1,0)| a is a real number}

Since (0,0,0) is not inside T, (because the second entry will always be 1 and never be 0) so T is NOT a subspace. However, I know from theorem that any intersection of 2 subspaces must also be a subspace. So there's a contradiction here.

The only thing that I can think of that could go wrong is that I cannot equate a + 1 = a+b and then b=1, because a, b are just dummy variables. They could be any real numbers. So, while both R and S use the unknown a in their description or notation, they somehow cannot be regarded as the "same" thing?

Thanks.[/QUOTE]
I'm not sure what you mean by "regarded as the 'same' thing" but your error is as you say- to have a vector in both R and S, it does not have to be for the same a, b, and c. Taking a= -1, b= 0, c= 0, as I said above, gives (0, 0, 0). Indeed, saying R is a subspace implies that (0, 0, 0) is in the set. Similarly, taking a= b= c= 0 gives (0, 0, 0) in S. Since (0, 0, 0) is in both sets, it is in their intersection.

 

[lkh1986]

I'm not sure what you mean by "regarded as the 'same' thing" but your error is as you say- to have a vector in both R and S, it does not have to be for the same a, b, and c. Taking a= -1, b= 0, c= 0, as I said above, gives (0, 0, 0). Indeed, saying R is a subspace implies that (0, 0, 0) is in the set. Similarly, taking a= b= c= 0 gives (0, 0, 0) in S. Since (0, 0, 0) is in both sets, it is in their intersection.

The "regarded as the 'same' thing" means that the value for a is the same for both R and S. I think I understand the question now.

Thanks very much for the explanation, especially on how condition (iii) is related to condition (i), simply by choosing the scalar value k=-1. :)