Intersection of vector subspaces

**dr-er** · March 30th 2009, 05:22 AM

Dear All,

I hope you can help me solving the following problem.

Given several subspaces $S_i$ of a linear $n$ -dimensional vector space, each subspace being determined by a set of $n_i$ $(n_i\le n)$ linearly independent vectors, find a set of vectors spanning the intersection of the subspaces.

It suffices to solve the problem just for two subspaces, as the rest can be done iteratively. Is there any simple method for determining the dimension of the intersection in question without finding the vectors?

Thank you for any hint.

**NonCommAlg** · March 30th 2009, 05:44 AM

Originally Posted by dr-er

Dear All,

I hope you can help me solving the following problem.

Given several subspaces $S_i$ of a linear $n$ -dimensional vector space, each subspace being determined by a set of $n_i$ $(n_i\le n)$ linearly independent vectors, find a set of vectors spanning the intersection of the subspaces.

It suffices to solve the problem just for two subspaces, as the rest can be done iteratively. Is there any simple method for determining the dimension of the intersection in question without finding the vectors?

Thank you for any hint.

$\dim (S_1 \cap S_2) = \dim S_1 + \dim S_2 - \dim (S_1 + S_2).$ so in order to find the dimension of the intersection of two vector spaces, besides the dimesnion of each vector space, you also need to know

the dimension of sum of those vector spaces.

**dr-er** · March 30th 2009, 05:55 AM

Originally Posted by NonCommAlg

$\dim (S_1 \cap S_2) = \dim S_1 + \dim S_2 - \dim (S_1 + S_2).$ so in order to find the dimension of the intersection of two vector spaces, besides the dimesnion of each vector space, you also need to know

the dimension of sum of those vector spaces.

This is clear. But the question about dimensionality concerned of course the case with more than two subspaces... The possibility to determine the dimension of sum (or union) of any subspaces is assumed.

**NonCommAlg** · March 30th 2009, 06:17 AM

Originally Posted by dr-er

This is clear. But the question about dimensionality concerned of course the case with more than two subspaces...

The possibility to determine the dimension of sum of any subspaces is assumed.

if you mean dimension of any subspace in the form $S_i + S_j$ is given, then that's not good enough for the general case. for example for 3 subspaces, we have:

$\dim(S_1 \cap S_2 \cap S_3)=\dim S_1 + \dim(S_2 \cap S_3) - \dim(S_1 + S_2 \cap S_3).$ so we also need to have $\dim(S_1 + S_2 \cap S_3).$

**dr-er** · March 30th 2009, 06:31 AM

Originally Posted by NonCommAlg

if you mean dimension of any subspace in the form $S_i + S_j$ is given, then that's not good enough for the general case. for example for 3 subspaces, we have:

$\dim(S_1 \cap S_2 \cap S_3)=\dim S_1 + \dim(S_2 \cap S_3) - \dim(S_1 + S_2 \cap S_3).$ so we also need to have $\dim(S_1 + S_2 \cap S_3).$

I meant that the dimension of any subspace in the form $\sum S_i$ , where the sum is running over any subset of subspace indices $i$ is known. This can be easily performed by computing the rank of the sum (union).

Your expression for 3 subspaces implies that one really needs to know $S_2 \cap S_3$ , so there is no possibility to avoid the solution of the main problem.

I see no other way to determine $\dim(S_1 + S_2 \cap S_3)$ .

**NonCommAlg** · March 30th 2009, 06:41 AM

Originally Posted by dr-er

I meant that the dimension of any subspace in the form $\sum S_i$ , where the sum is running over any subset of subspace indices $i$ is known. This can be easily performed by computing the rank of the sum (union).

From your expression for 3 subspaces it follows that one needs really to know $S_2 \cap S_3$ , so there is no possibility to avoid the solution of the main problem.

I see no other way to determine $\dim(S_1 + S_2 \cap S_3)$ .

as i've already mentioned $\dim(S_2 \cap S_3)=\dim S_2 + \dim S_3 - \dim(S_2 + S_3).$ so considering your assumptions, we know the value of $\dim(S_2 \cap S_3).$

by the way, the union of vector spaces need not be a vector space. so be careful when talking about "union of vector spaces".

**dr-er** · March 30th 2009, 06:47 AM

Originally Posted by NonCommAlg

as i've already mentioned $\dim(S_2 \cap S_3)=\dim S_2 + \dim S_3 - \dim(S_2 + S_3).$ so considering your assumptions, we know the value of $\dim(S_2 \cap S_3).$

I was speaking about $\dim(S_1 + S_2 \cap S_3)$ , not about $\dim(S_2 \cap S_3).$

by the way, the union of vector spaces need not be a vector space. so be careful when talking about "union of vector spaces".

Thank you for the remark!

**NonCommAlg** · March 30th 2009, 07:12 AM

Originally Posted by dr-er

I was speaking about $\dim(S_1 + S_2 \cap S_3)$ , not about $\dim(S_2 \cap S_3).$

really?! this is what you said:

Originally Posted by dr-er

Your expression for 3 subspaces implies that one really needs to know $S_2 \cap S_3$ , so there is no possibility to avoid the solution of the main problem.

**dr-er** · March 30th 2009, 07:47 AM

Originally Posted by NonCommAlg

really?! this is what you said:

You should have read further:

I see no other way to determine

.

what you have actually cited.

You could also note, that I have written $S_2\cap S_3$ , not $\dim(S_2\cap S_3)$ .

**dr-er** · March 30th 2009, 08:04 AM

Originally Posted by dr-er

Dear All,

I hope you can help me solving the following problem.

Given several subspaces $S_i$ of a linear $n$ -dimensional vector space, each subspace being determined by a set of $n_i$ $(n_i\le n)$ linearly independent vectors, find a set of vectors spanning the intersection of the subspaces.

It suffices to solve the problem just for two subspaces, as the rest can be done iteratively. Is there any simple method for determining the dimension of the intersection in question without finding the vectors?

Thank you for any hint.

Concerning the main problem it can be of course easily solved in the 3D case. Consider two subspaces of dimension 2 (if one of the subspaces has a smaller dimension the case is trivial). Than either the subspaces are equivalent and the intersection can be represented by any pair $(v^1_1,v^1_2), (v^2_1,v^2_2)$ or the intersection has dimension 1 and is represented by the vector

$v=(v^1_1\times v^1_2)\times(v^2_1\times v^2_2)$ .

Can it be somehow generalized to higher dimensions? I can imagine the following. Let $F(S)$ be an operation, which produces $n-\dim(S)$ vectors, orthogonal to the vectors of $S$ . Than the problem were solved by:

$S_1\cap S_2=F(F(S_1)+F(S_2))$

But what is the explicit construction of the operation $F(S)$ ? I would guess that the Gram-Schmidt orthogonalization should work... Any other suggestions?

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