What Are the Possible Dimensions of V Intersection W in R4?

[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

 

[Mark44]

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?

 

[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?

 

[Mark44]

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?

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

 

[helix999]

What does any two-dimensional space look like geometrically?

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

 

[helix999]

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

 

[Mark44]

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

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

 

[helix999]

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

it is single dimension

 

[Dick]

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?

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.

 

[Mark44]

it is single dimension

I asked two questions. Your answer above is the answer to which question?

 

[Mark44]

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.

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.

 

[helix999]

I asked two questions. Your answer above is the answer to which question?

point is 0 dimension and line is single dimension

 

[HallsofIvy]

And a plane has 2 dimensions.

 

[HallsofIvy]

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.

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.

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.

 

[helix999]

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

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.

why is it impossible for subspaces?

 

[helix999]

If they were embedded in three space a 0 dimensional intersection would be impossible.

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

 

[HallsofIvy]

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.

 

[Dick]

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

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.

 

[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}?

 

[HallsofIvy]

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}?

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 V_ij 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.

 

[helix999]

That was the complete question and i guess i understood the explanation.

We solve the integrals like sqrt(1+x^4) dx (limit 0 to 1) by substituting x^2 as tan theta but how do we proceed for integrals like sqrt(1-x^4) dx (limit 0 to 1)?

 

[HallsofIvy]

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?

Any subspace must contain the 0 vector so the intersection of two subspaces must contain at least the 0 vector. It is impossible for the intersection to be empty.

 

[HallsofIvy]

That was the complete question and i guess i understood the explanation.

We solve the integrals like sqrt(1+x^4) dx (limit 0 to 1) by substituting x^2 as tan theta but how do we proceed for integrals like sqrt(1-x^4) dx (limit 0 to 1)?

No, we do NOT substitute x^2 as tan theta. If we did then we would have 2 xdx= sec² theta dtheta- and there is no "x" outside the square root to use with dx.

 

[helix999]

then how do we solve such integrals

 

[kimhs]

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 V_ij 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.

There is only one answer because T is a linear transformation from V onto W.
Therefore the rank of T (or range of T) has to be 4.
Rank(T) + Nullity (T) = dimension of the domain of T (Where V supplies the domain with a dimension of 6).
And then you get Nullity (T) = 2 = dimension of Ker (T) => answer