Sum of two subspaces - question.

In summary: Since a basis for W_1+W_2 has 4 vectors, then dim(W_1+W_2) = 4. If dim(V) is something other than 4, there's no way that W_1+W_2 = V.In summary, the sum of subspaces W_1 = Sp{(1,0,0)} and W_3 = Sp{(0,1,-1), (0,0,1)} is not possible. Their direct sum, W_1 \oplus W_3, is the set of vectors v such that v = w1 + w3, where w1 ##\in## W1 and w3 ##\in## W3. Additionally, the sum
  • #1
peripatein
880
0

Homework Statement


Is it possible to add the following subspaces: W_1 = Sp{(1,0,0)} and W_3 = Sp{(0,1,-1), (0,0,1)}?


Homework Equations





The Attempt at a Solution


Will their sum be: Sp{(1,1,-1),(1,0,1)}?
 
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  • #2
peripatein said:

Homework Statement


Is it possible to add the following subspaces: W_1 = Sp{(1,0,0)} and W_3 = Sp{(0,1,-1), (0,0,1)}?


Homework Equations





The Attempt at a Solution


Will their sum be: Sp{(1,1,-1),(1,0,1)}?
No.

It would help to think about the geometry here. What does the W1 subspace look like? Can you describe it geometrically?

What does the W3 subspace look like? Can you describe it geometrically?

Are any vectors in W1 also in W3? Are any vectors in W3 also in W1?

The "sum" of the two sets (actually the direct sum, W1[itex] \oplus[/itex] W3) is the set of vectors v such that v = w1 + w3, where w1 ##\in## W1 and w3 ##\in## W3.
 
  • #3
I understand that their sum is a direct sum as the intersection is null, but I am not sure how that helps me to find Sp(W_1+W_3).
 
  • #4
I was also asked to prove that (W_1 intersection W_2) + (W_1 intersection W_3) = W_1 intersection (W_2+W_3), where W_1, W_2, W_3 are subspaces in vector space V.

Attempt at solution:
Let W_1=Sp(K), W_2=Sp(U), W_3=Sp(L)
Hence, (W_1 intersection W_2) = Sp(K) intersection Sp(U) = Sp(K intersection U)
Hence, (W_1 intersection W_3) = Sp(K) intersection Sp(L) = Sp(K intersection L)
Hence, (W_1 intersection W_2) + (W_1 intersection W_3) = Sp(K intersection U) + Sp(K intersection L) = Sp(K intersection U + K intersection L) = Sp(K intersection (U+L))
I am really not sure this is correct. Is it? Is this really how I should try proving it?
 
  • #5
Mark44, their geometrical representation will be:
W_3 = y-(-z) plane and z axis, W_1 = x axis
 
  • #6
peripatein said:
Mark44, their geometrical representation will be:
W_3 = y-(-z) plane and z axis, W_1 = x axis

So W1 contains all vectors of the form (x,0,0), and W3 contains all vectors of the form (0,y,z).

What vectors can be expressed as a sum of one vector from W1 and one from W3?
 
  • #7
R^3.
How may I check whether W_1+W_2=V, where V is a vector space and W_1 and W_2 are two subspaces in V, and W_1=(t,s,t-2s,2t) where t,s belong to R, and W_2=(x,y,z,-x-y-z)?
I first tried to write W_1 as {(1,0,1,2), (0,1,-2,0)} and W_2 as {(1,0,0,-1), (0,1,0,-1),(0,0,1,-1)} and then concluded that since dim(W_1+W_2) was not equal to dim(V) they could not possibly be equal. But I am really not sure that's correct. Could anyone please advise?
 
  • #8
peripatein said:
R^3.
How may I check whether W_1+W_2=V, where V is a vector space and W_1 and W_2 are two subspaces in V, and W_1=(t,s,t-2s,2t) where t,s belong to R, and W_2=(x,y,z,-x-y-z)?
I first tried to write W_1 as {(1,0,1,2), (0,1,-2,0)} and W_2 as {(1,0,0,-1), (0,1,0,-1),(0,0,1,-1)} and then concluded that since dim(W_1+W_2) was not equal to dim(V) they could not possibly be equal. But I am really not sure that's correct. Could anyone please advise?

There's a systematic way to do these problems. Set up a matrix where the rows are your vectors and row reduce it. When you are done the remaining nonzero vectors will be a basis for W_1+W_2.
 
Last edited:

Related to Sum of two subspaces - question.

What is the definition of "sum of two subspaces"?

The sum of two subspaces is the set of all vectors that can be written as the sum of a vector from one subspace and a vector from another subspace. It is denoted as W = U + V.

How is the sum of two subspaces calculated?

The sum of two subspaces is calculated by adding all possible combinations of vectors from each subspace. This means taking a vector from the first subspace and adding it to a vector from the second subspace, and repeating this for all possible combinations. The result is the sum of the two subspaces.

What is the relationship between the dimension of the sum of two subspaces and the dimensions of the individual subspaces?

The dimension of the sum of two subspaces is equal to the sum of the dimensions of the individual subspaces minus the dimension of their intersection. In other words, if U and V are subspaces of a vector space W, then dim(U + V) = dim(U) + dim(V) - dim(U ∩ V).

Can the sum of two subspaces ever be equal to one of the individual subspaces?

Yes, it is possible for the sum of two subspaces to be equal to one of the individual subspaces. This can happen when one subspace is a subset of the other subspace, or when the two subspaces are identical.

How is the sum of two subspaces related to linear independence?

If the two subspaces U and V are linearly independent, then the sum of the two subspaces will be the direct sum of U and V, meaning that there is no overlap between the two subspaces. However, if U and V are linearly dependent, then the sum of the two subspaces will have some shared elements. This relationship between linear independence and the sum of two subspaces can be useful in solving problems involving linear combinations of subspaces.

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