Finding intersection of vector subspaces

In summary: The conversation discusses finding the intersections and sums of various subspaces in R3. In summary, the intersections are the y-axis, a single point, the zero vector, and a line through the origin with direction vector <1, -1, 1>. The sums are R3, the plane through (1, 0, 0) and (0, 1, 1), and a three-dimensional space.
  • #1
tatianaiistb
47
0

Homework Statement



What are the intersections of the following pairs of subspaces?
(a) The x-y plane and the y-z plane in R'.
(b) The line through (1, 1, 1) and the plane through (1,0, 0) and (0, 1, 1).
(c) The zero vector and the whole space R'.
(d) The plane S perpendicular to (1, 1, 0) and perpendicular to (0, 1, I) in R3
What are the sums of those pairs of subspaces?

Homework Equations





The Attempt at a Solution



So I've been using logic, but I don't know if what I'm doing is right or makes sense...

a) I think the intersection is the y-axis (it's where the two planes I believe meet). And for the sum I have that R3= x-axis + y-axis+z-axis that can be written as a combination of a member of the xy- and yz-planes. So, R3=xy-plane_yz-plane

b) In 3D, a line is either parallel to a plane or intersects it in a single point. So, I'm thinking it should intersect at the single point (1,1,1) and that the sum should be the plane through (1,0,0) and (0,1,1), but I'm a bit at a loss here...

c) The intersection I think should be the lonely zero vector and their sum I think is the whole space R3.

d) This one I'm completely lost!

Can anyone please help me, particularly with parts B and D, and let me know if my logic seems right?

Thanks!
 
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  • #2
tatianaiistb said:

Homework Statement



What are the intersections of the following pairs of subspaces?
(a) The x-y plane and the y-z plane in R'.
What is R'? You seem to mean R3.
(b) The line through (1, 1, 1) and the plane through (1,0, 0) and (0, 1, 1).
(c) The zero vector and the whole space R'.
(d) The plane S perpendicular to (1, 1, 0) and perpendicular to (0, 1, I) in R3
What are the sums of those pairs of subspaces?

Homework Equations





The Attempt at a Solution



So I've been using logic, but I don't know if what I'm doing is right or makes sense...

a) I think the intersection is the y-axis (it's where the two planes I believe meet). And for the sum I have that R3= x-axis + y-axis+z-axis that can be written as a combination of a member of the xy- and yz-planes. So, R3=xy-plane_yz-plane
Yes, that is correct.

b) In 3D, a line is either parallel to a plane or intersects it in a single point.
No, that is not true. There are three possibilities. A line may per parallel to a plane (no points in common), intersect it in a single point (one point in common), or lie in the plane (an infinite number of points in common- all points on the line).

So, I'm thinking it should intersect at the single point (1,1,1) and that the sum should be the plane through (1,0,0) and (0,1,1), but I'm a bit at a loss here...
Look again. The origin, of course, and the point (1, 1, 1) are both in the plane containing (0, 0, 0), (1, 0, 0), and (0, 1, 1). (Remember that we are talking about subspaces so they must contain (0, 0, 0) and the plane must contain (1, 0, 0)+ (0, 1, 1).)

c) The intersection I think should be the lonely zero vector and their sum I think is the whole space R3.
Yes, that is correct.

d) This one I'm completely lost!
I don't understand what you mean by "The plane S perpendicular to (1, 1, 0) and perpendicular to (0, 1, I) in R3". I assume you meant (0, 1, 1) rather than (0, 1, I), but you only refer to a single space, S. Of course, a plane perpendicular to (1, 1, 0) cannot be also perpendicular to (0, 1, 1)! So I think you mean "The plane S perpendicular to (1, 1, 0) and the plane T perpendicular to (0, 1, 1)."
Recall, from Calculus, that a plane perpendicular to vector <A, B, C> containing [itex](x_0, y_0, z_0)[/itex] can be written as [itex]A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0[/itex]. That is, plane S (which must contain (0,0,0) to be a subspace) is x+ y= 0 while T is y+ z= 0. Those say y= -x and z= -y= x. Thus, taking x= t as parameter, the intersection of the two planes is the line x= t, y= -t, z= t. <x, y, z>= <t, -t, t>= t<1, -1, 1>.
The sum of two two distinct two dimensional subspaces is always a three dimensional space.

Can anyone please help me, particularly with parts B and D, and let me know if my logic seems right?

Thanks!
 
  • #3
HallsofIvy said:
What is R'? You seem to mean R3.

Yes, that is correct.


No, that is not true. There are three possibilities. A line may per parallel to a plane (no points in common), intersect it in a single point (one point in common), or lie in the plane (an infinite number of points in common- all points on the line).


Look again. The origin, of course, and the point (1, 1, 1) are both in the plane containing (0, 0, 0), (1, 0, 0), and (0, 1, 1). (Remember that we are talking about subspaces so they must contain (0, 0, 0) and the plane must contain (1, 0, 0)+ (0, 1, 1).)


Yes, that is correct.


I don't understand what you mean by "The plane S perpendicular to (1, 1, 0) and perpendicular to (0, 1, I) in R3". I assume you meant (0, 1, 1) rather than (0, 1, I), but you only refer to a single space, S. Of course, a plane perpendicular to (1, 1, 0) cannot be also perpendicular to (0, 1, 1)! So I think you mean "The plane S perpendicular to (1, 1, 0) and the plane T perpendicular to (0, 1, 1)."
Recall, from Calculus, that a plane perpendicular to vector <A, B, C> containing [itex](x_0, y_0, z_0)[/itex] can be written as [itex]A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0[/itex]. That is, plane S (which must contain (0,0,0) to be a subspace) is x+ y= 0 while T is y+ z= 0. Those say y= -x and z= -y= x. Thus, taking x= t as parameter, the intersection of the two planes is the line x= t, y= -t, z= t. <x, y, z>= <t, -t, t>= t<1, -1, 1>.
The sum of two two distinct two dimensional subspaces is always a three dimensional space.

Yes I did mean R3. Thank you!

For part b:

I see now. So, because (1,0,0) + (0,1,1) = (1,1,1) can be written as a linear combination, it is obviously a subspace. And then the intersection would be the line, not the point, going through (1,1,1). And then the sum should be the plane through (1,0,0) and (0,1,1).

Am I thinking correctly now?


For part d:

This is why I am so confused. The problem states "The plane S perpendicular to (1,1,0) and perpendicular to (0,1,1) in R3. I too was thinking they should be two separate planes. I'm lost!
 
  • #4
So I'm thinking that for part d, it must refer to two separate planes because the main question asks for the intersection of the following PAIRS of subspaces. It's the only way I think makes sense.
 

Related to Finding intersection of vector subspaces

1. What is the definition of intersection of vector subspaces?

The intersection of two or more vector subspaces is the set of all vectors that belong to both subspaces. In other words, it is the common elements shared by all the subspaces.

2. How do you find the intersection of two vector subspaces?

To find the intersection of two vector subspaces, you can use the method of elimination. Set up a system of equations using the basis vectors of each subspace and solve for the variables. The solution will give you the coordinates of the vectors that are in the intersection.

3. Can the intersection of vector subspaces be empty?

Yes, it is possible for the intersection of vector subspaces to be empty. This means that there are no common elements between the subspaces and they are considered to be disjoint.

4. Is the intersection of vector subspaces always a subspace?

Yes, the intersection of vector subspaces is always a subspace. This is because it contains all the common elements of the subspaces, which means it satisfies all the properties of a subspace.

5. What is the significance of finding the intersection of vector subspaces?

Finding the intersection of vector subspaces allows us to understand the relationship between different subspaces. It can also help us determine if two subspaces are equal or if one subspace is contained within the other.

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