# Dimensions of subspaces

1. Oct 30, 2008

### helix999

If V and W are 2-dimensional subspaces of R4 , what are the possible dimensions of the subspace V intersection W?

I am new to subspaces, so I have no clue to this question. Help guys!!!

Options: (A) 1 only (B) 2 only (C) 0 and 1 only (D) 0, 1, and 2 only (E) 0, 1, 2, 3, and 4

2. Oct 30, 2008

### Staff: Mentor

You don't need to be concerned about the fact that V and W are subspaces, nor that they are embedded in a four-dimensional space.

V and W are two-dimensional, so you know what they look like geometrically, don't you? Now, what are all of the possibilities for the intersection of V and W?

3. Oct 30, 2008

### helix999

you mean to say that V and W are 2x2 matrices and their intersection would also be a 2x2 matrix....so the required dimension is 2 only?

4. Oct 30, 2008

### Staff: Mentor

No, I didn't say that at all. What does any two-dimensional space look like geometrically?

5. Oct 30, 2008

### helix999

I am sorry but i am not able to grasp the idea u r trying to put.

6. Oct 30, 2008

### helix999

I think the possibility of dimensions would be 0,1, and 2.

7. Oct 30, 2008

### Staff: Mentor

Considering the real plane, what is the dimension of a single point? What is the dimension of a line?

8. Oct 30, 2008

### helix999

it is single dimension

9. Oct 30, 2008

### Dick

You do need to pay some attention to the dimension of the space they are embedded in. If they were embedded in three space a 0 dimensional intersection would be impossible.

10. Oct 30, 2008

### Staff: Mentor

11. Oct 30, 2008

### Staff: Mentor

Yes, but I wanted the OP to focus more on the subspaces and not get tangled up in trying to imagine a 4-D space.

12. Oct 31, 2008

### helix999

point is 0 dimension and line is single dimension

13. Oct 31, 2008

### HallsofIvy

Staff Emeritus
And a plane has 2 dimensions.

14. Oct 31, 2008

### HallsofIvy

Staff Emeritus
You also need to pay attention to the fact that they are subspaces. The planes (x, y, 0, 0) and (x, y, 0 , 1) have empty intersection. That is impossible for subspaces.

15. Oct 31, 2008

### helix999

I got the answer of my question, really appreciate ur posts for making it clear to me. Thnx a bunch!!

why is it impossible for subspaces?

16. Oct 31, 2008

### helix999

How come a point can exist as an intersection of subspaces in 4 space but not in 3 space?

17. Oct 31, 2008

### HallsofIvy

Staff Emeritus
The 2-dimensional subspaces of R4, {(x, y, 0, 0)} and {0, 0, u, v}, have only the single point (0, 0, 0, 0) in common. Two 2- dimensional subspaces in R2 must have a least a 1-dimensional subspace in common.

18. Oct 31, 2008

### Dick

If the intersection is the single point 0, then if you take union of a basis for one subspace with a basis of the other, the whole set must be linearly independent (otherwise you could construct another intersection point). In the case of two planes, that's 4 linearly independent vectors. Possible in 4 space, not in 3 space.

19. Oct 31, 2008

### helix999

those were the new concepts for me....thnx

How do I proceed for such type of questions:

Let V be the real vector space of all real 2 x 3 matrices, and let W be the real vector space of all real 4 x 1
column vectors. If T is a linear transformation from V onto W, what is the dimension of the subspace
{v belongs to V:T(v)=0}?

20. Oct 31, 2008

### HallsofIvy

Staff Emeritus
In other words, "what is the dimension of the kernel of T" or "what is the nullity of T".
Is that really the entire question? There are many possible answers. For example, T(v)= 0 for all v is a linear transformation and it's kernel is all of V. Since V is a 6 dimensional space (Taking Vij to be the matrix with 1 at "ith row, jth column" and 0 everywhere else, for i= 1 to 2, j= 1 to 3 gives a basis), so that in this example, the dimension of the kernel would be 6.

On the other hand the linear transformation that maps
$$\left[\begin{array}{ccc} a & b & c \\d & e & f\end{array}\right]\right)$$ to $$\left[\begin{array}{c} a \\ 0 \\ 0 \\ 0\end{array}\right]$$
has nullity 5: dimension of the kernel is 5.

Perhaps the problem is "what is the smallest possible dimension of the kernel of T".
Since the largest the image of T can be is 4 dimensional, the dimension of "the real vector space of all real 4 x 1 column vectors", the smallest the kernel can be is 6- 4= 2.