Proving Direct Sum Decomposition: V=(C*1) \oplus W

[xitoa]

Homework Statement

Let V=Cⁿ and 1 be all ones vector 1 = e₁ + ... +e_n. Let W be the subspace of V spanned by those vectors of the form \lambdae₁ + \lambdae₂ + ... + \lambdae_n such that \lambda₁ + ... + \lambda_n= 0 \in C. Prove that there is a direct sum decomposition

V=(C*1) \oplus W

as complex vector spaces.

Homework Equations

V=(C*1) \oplus W



How should I approach this problem?

 

[micromass]

What did you already try?

 

[xitoa]

Okay, so I have a few questions:

1) i think i could say that W is linearly dependent? since \lambda₁ + \lambda₂ + ... + \lambda_n = 0 \in C

2) I'm not quite sure about what it means that 1=e₁ + ... + e_n
Does it mean that each e_n is something like <1,0,0...,0>, <0,1,0,...,0>, etc?

3) Also, I am still foggy on how direct sums work...say: two subspaces U and W exist in V(that are out of the context of this problem).

U and V are a direct sum if (is it iff?)

1. U\wedgeW = {0} (zero vector)
2. U+W=V

So, if they satisfy those two conditions, its a direct sum?

4) To help prove that V is a direct sum, could I say that Cⁿ is the zero vector?

 

[ystael]

Some of the things you say suggest that you're confused about really basic things.

1) i think i could say that W is linearly dependent? since \lambda₁ + \lambda₂ + ... + \lambda_n = 0 \in C

One doesn't say that a subspace is "linearly dependent". A list of vectors (finite list, when doing finite-dimensional linear algebra, which you are) is "linearly independent" or "linearly dependent". In any case, the fact that the equation defining W is "the sum of the coordinates is zero" has nothing to do with any kind of linear dependence property; a list of vectors is linearly dependent when there is a linear combination of vectors in the list that comes out to the zero vector.

In any case this has nothing to do with your problem.

[According to the definitions, every nontrivial subspace of a vector space is a "linearly dependent set" of vectors; that's why this isn't useful and we don't say it.]

2) I'm not quite sure about what it means that 1=e₁ + ... + e_n
Does it mean that each e_n is something like <1,0,0...,0>, <0,1,0,...,0>, etc?

\mathbf{\mathrm{e}}_j is conventionally defined to be the vector with a 1 in the jth place and 0 elsewhere. This equation is meant to tell you what \mathbf{1} is (that it's the vector with 1 in every coordinate), not what the \mathbf{\mathrm{e}}_j are.

3) Also, I am still foggy on how direct sums work...say: two subspaces U and W exist in V(that are out of the context of this problem).

U and Ware a direct sum if (is it iff?)

1. U\capW = {0} (zero vector)
2. U+W=V

So, if they satisfy those two conditions, its a direct sum?

Yes, this is a correct definition for "V is the (internal) direct sum of U and W."

4) To help prove that V is a direct sum, could I say that Cⁿ is the zero vector?

This makes no sense whatsoever. \mathbb{C}^n is the vector space in which all of this problem takes place. It is an n-dimensional complex vector space and there is no way in which you can "say that it is the zero vector".

You need to prove two things, taken from the definition you quote in your third point:

(a) that \mathbb{C}\mathbf{1} \cap W = \{0\}, and

(b) that \mathbb{C}\mathbf{1} + W = \mathbb{C}^n. This second point means to prove that every vector of \mathbb{C}^n can be expressed as the sum of a vector in \mathbb{C}\mathbf{1} and a vector in W.

 

[xitoa]

Thanks for replying, I think i meant to ask if \mathbb{C}^n contains the zero vector. ie the 0 vector is an element of \mathbb{C}\mathbf{1} ?

 

[ystael]

Thanks for replying, I think i meant to ask if \mathbb{C}^n contains the zero vector. ie the 0 vector is an element of \mathbb{C}\mathbf{1} ?

Yes, \mathbb{C}^n contains the zero vector (as every vector space must), and yes, \mathbb{C}\mathbf{1} contains the zero vector, because \mathbf{0} = 0\mathbf{1}. However, your use of "i.e." indicates an implication which does not hold: the (true) fact that \mathbf{0} \in \mathbb{C}^n does not directly imply (or indeed have anything to do with) the (true) fact that \mathbf{0} \in \mathbb{C}\mathbf{1}.

 

[xitoa]

So, in my proof, I should state \mathbb{C}\mathbf{1} contains the zero vector because \mathbf{0} = 0\mathbf{1}, so that V is a direct sum of C1 and W?

 

[ystael]

That's only the smallest part of the proof: to prove that \mathbb{C}\mathbf{1} \cap W = \{\mathbf{0}\}, it is not enough to prove that \mathbf{0} \in \mathbb{C}\mathbf{1} and \mathbf{0} \in W! And then you still need to show that \mathbb{C}\mathbf{1} + W = \mathbb{C}^n.

 

[xitoa]

W and C1 are both subspaces of V (which is \mathbb{C}^n)

So doesn't that mean that\mathbb{C}\mathbf{1} + W = \mathbb{C}^n is true?

 

[sheepover]

I'm also working on this problem. To me it seems like the only way V=(C*1) \oplus W can be true is if all coordinates in W=0. I'm not sure how to show this, or why it is even true. Can anyone please help?

 

[ystael]

W and C1 are both subspaces of V (which is \mathbb{C}^n)

So doesn't that mean that\mathbb{C}\mathbf{1} + W = \mathbb{C}^n is true?

I'll answer your questions as best I can, but you are misunderstanding some really basic ideas, and I think you would be better served by in-person tutoring.

To prove that two sets A, B are equal, you prove that they have the same elements, or equivalently that each is contained in the other: "A = B" is equivalent to "A \subset B and A \supset B" and also to "x\in A if and only if x\in B".

Here you are trying to prove that \mathbb{C}\mathbf{1} + W = \mathbb{C}^n. You are correct to observe that \mathbb{C}\mathbf{1} and W are both subspaces of \mathbb{C}^n. Since \mathbb{C}\mathbf{1} + W is the set \{v + w \mid v \in \mathbb{C}\mathbf{1}, w \in W\} and \mathbb{C}^n is closed under addition (being a vector space), this proves that \mathbb{C}\mathbf{1} + W \subset \mathbb{C}^n.

You still need to prove the other direction, that \mathbb{C}\mathbf{1} + W \supset \mathbb{C}^n; in words, that every vector in \mathbb{C}^n lies in \mathbb{C}\mathbf{1} + W, that is, every vector in \mathbb{C}^n is the sum of a vector in \mathbb{C}\mathbf{1} and a vector in W.

 

[ystael]

I'm also working on this problem. To me it seems like the only way V=(C*1) \oplus W can be true is if all coordinates in W=0. I'm not sure how to show this, or why it is even true. Can anyone please help?

You can't show this because it is false. To see why it can't be true, think about the dimensions. What are the dimensions of V = \mathbb{C}^n and \mathbb{C}\mathbf{1}? How does the dimension of a direct sum relate to the dimensions of the summands? What does that tell you about the dimension of W?

 

[sheepover]

You can't show this because it is false. To see why it can't be true, think about the dimensions. What are the dimensions of V = \mathbb{C}^n and \mathbb{C}\mathbf{1}? How does the dimension of a direct sum relate to the dimensions of the summands? What does that tell you about the dimension of W?

C^n is a dimension of order n, so does that mean W must be a dimension of order n as well? I'm not exactly sure what this means.

This is what I was trying to do:
If we want to prove C*1 (intersect) W = {0}

pick (a₁, ..., a_n) such that (a₁,..., a_n) is an element of C*1 (intersect) W.
because (a₁,..., a_n) is an element of W --> (a₁+...+a_n) = 0 (which is an element of C)

Now, my teacher wrote something like this:

because (a₁, ..., a_n) is an element of C*1, all coordinates are equal.
then, if a₁ = ... = a_n, it follows that a₁ = ... = a_n=0
therefore, C*1 (intersect) W = {0}

But I can't figure out how he got there.

 

[sheepover]

You still need to prove the other direction, that \mathbb{C}\mathbf{1} + W \supset \mathbb{C}^n; in words, that every vector in \mathbb{C}^n lies in \mathbb{C}\mathbf{1} + W, that is, every vector in \mathbb{C}^n is the sum of a vector in \mathbb{C}\mathbf{1} and a vector in W.

Also, I can't figure out a way to do this I'm really struggling with this problem.