Prove that V is the internal direct sum of two subspaces

In summary: It could be 1-dimensional, 2-dimensional, etc. But it must have a non-zero vector in it to overlap with U1 or U2.
  • #1
Austin Chang
38
0
Let V be a vector space. If U 1 and U2 are subspaces of V s.t. U1+U2 = V and U1 and U1∩U2 = {0V}, then we say that V is the internal direct sum of U1 and U2. In this case we write V = U1⊕U2. Show that V is internal direct sum of U1 and U2if and only if every vector in V may be written uniquely in the form v1+v2 with v1∈U1 and v2∈ U2.

What does it mean to be unique? Does it matter if it is unique?
 
Physics news on Phys.org
  • #2
Austin Chang said:
What does it mean to be unique?
It means that for any vector ##u\in V##, if ##w_1+w_2=u=v_1+v_2##, for ##w_1,v_1\in U_1## and ##w_2,v_2\in U_2##, then ##v_1=w_1## and ##v_2=w_2##.

Yes, the uniqueness matters. If the two subspaces overlap, there is more than one way of writing a vector in ##V## as a sum of two vectors one from each subspace. And vice versa, if the representation as a sum is not unique, the two subspaces must overlap.
 
  • #3
andrewkirk said:
It means that for any vector ##u\in V##, if ##w_1+w_2=u=v_1+v_2##, for ##w_1,v_1\in U_1## and ##w_2,v_2\in U_2##, then ##v_1=w_1## and ##v_2=w_2##.

Yes, the uniqueness matters. If the two subspaces overlap, there is more than one way of writing a vector in ##V## as a sum of two vectors one from each subspace. And vice versa, if the representation as a sum is not unique, the two subspaces must overlap.
So for your example if it was not unique U2 = U1 and you can write w1 in U2 and w2 in U1? therefore w2+w1 is not unique anymore because the same things came from different subspaces?
 
  • #4
Austin Chang said:
Let V be a vector space. If U 1 and U2 are subspaces of V s.t. U1+U2 = V and U1 and U1∩U2 = {0V}, then we say that V is the internal direct sum of U1 and U2. In this case we write V = U1⊕U2. Show that V is internal direct sum of U1 and U2if and only if every vector in V may be written uniquely in the form v1+v2 with v1∈U1 and v2∈ U2.

What does it mean to be unique? Does it matter if it is unique?

Perhaps an example of non-uniqueness will make this clear. You can represent 12 as the product of two factors ##12 = x_1 x_2## but the two factors are not unique since they and be chosen to have various different values - e.g. (12)(1), (4)(3), (2)(6) etc.

As whether uniqueness matters - that's a subjective question. Uniqueness is often a convenient property to have. For example, the rigorous approach to defining "a" zero z_1 in arithmetic would be to define it by the property that for all numbers x, z1 + x = x. It is convenient that there is a unique number, denoted by "0", that has this property. If there were several unequal zeroes, arithmetic would be more complicated.
 
  • Like
Likes Austin Chang
  • #5
Austin Chang said:
So for your example if it was not unique U2 = U1
No, that would be going too far. U2 and U1 will overlap nontrivially, but that doesn't mean they are identical. All it means that their intersection is a vector space of dimension at least 1.
 
  • #6
Stephen Tashi said:
Perhaps an example of non-uniqueness will make this clear. You can represent 12 as the product of two factors ##12 = x_1 x_2## but the two factors are not unique since they and be chosen to have various different values - e.g. (12)(1), (4)(3), (2)(6) etc.

As whether uniqueness matters - that's a subjective question. Uniqueness is often a convenient property to have. For example, the rigorous approach to defining "a" zero z_1 in arithmetic would be to define it by the property that for all numbers x, z1 + x = x. It is convenient that there is a unique number, denoted by "0", that has this property. If there were several unequal zeroes, arithmetic would be more complicated.
Thanks! That was a good example.
 
  • #7
andrewkirk said:
No, that would be going too far. U2 and U1 will overlap nontrivially, but that doesn't mean they are identical. All it means that their intersection is a vector space of dimension at least 1.
What do you mean by at least a vector space of dimension 1? I don't think I've gotten that far to understand it completely. Could you elaborate?
 
  • #8
Austin Chang said:
What do you mean by a vector space of dimension at least 1?
A vector space that is not just ##\{0_V\}##.
 

Related to Prove that V is the internal direct sum of two subspaces

What does it mean for V to be the internal direct sum of two subspaces?

When we say that V is the internal direct sum of two subspaces, it means that V can be written as the direct sum of two subspaces, say W and U, such that V = W + U and W ∩ U = {0}, where {0} is the zero vector. In other words, every vector in V can be uniquely expressed as the sum of a vector in W and a vector in U.

How do you prove that V is the internal direct sum of two subspaces?

To prove that V is the internal direct sum of two subspaces, we need to show that V = W + U and W ∩ U = {0}. This can be done by showing that every vector in V can be written as the sum of a vector in W and a vector in U, and also by showing that the only vector in W ∩ U is the zero vector.

Can V be the internal direct sum of more than two subspaces?

Yes, V can be the internal direct sum of more than two subspaces. In fact, V can be the internal direct sum of any number of subspaces, as long as the intersection of all those subspaces is only the zero vector.

Is the internal direct sum of two subspaces unique?

Yes, the internal direct sum of two subspaces is unique. This means that if V can be written as the direct sum of two subspaces, say W and U, then there are no other subspaces that can also be used to express V as a direct sum.

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

The internal direct sum of two subspaces is closely related to the concept of linear independence. If V is the internal direct sum of two subspaces, then the vectors in W and U are linearly independent, which means that none of the vectors in W can be written as a linear combination of the vectors in U, and vice versa.

Similar threads

  • Mechanical Engineering
Replies
2
Views
1K
  • Linear and Abstract Algebra
Replies
10
Views
1K
  • Linear and Abstract Algebra
Replies
3
Views
3K
  • Linear and Abstract Algebra
Replies
4
Views
1K
  • Linear and Abstract Algebra
2
Replies
59
Views
8K
  • Linear and Abstract Algebra
Replies
2
Views
2K
  • Linear and Abstract Algebra
Replies
6
Views
981
Replies
4
Views
3K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Linear and Abstract Algebra
Replies
3
Views
1K
Back
Top