Correct reasoning about direct sums proof?

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In summary: But these are the ones that come to mind first.In summary, the theorem states that if V is the direct sum of two subspaces U and W, then Z cannot equal the direct sums of (i) the intersection of Z and U and (ii) the intersection of Z and W unless Z is a subspace of U or W.
  • #1
quasar_4
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Howdy everyone. I'm not very good at writing proofs, so I am wondering if someone can tell me if I'm even on the right page with this. I am not sure if I understand the idea correctly. The theorem goes as follows: Suppose V is a finite dim. vector space with subspaces U and W such that V is the direct sum of U and W. Let Z also be a subspace of V. Then Z cannot equal the direct sums of (i) the intersection of Z and U and (ii) the intersection of Z and W unless Z is a subspace of U or W.

Here's my thoughts:

Let "C" denote the direct sum of the intersections (i) and (ii) above. Then two conditions must hold:
1) The intersection of the intersections (i) and (ii) must be the zero vector; and
2) The sum of the intersections must equal V.

The elements of C are given as {x|x is in Z and x is in U and x is in W}={0v}. Thus every x in Z that intersects U and W is in C. But we have that C={0v} since it is a direct sum. The only way that {0v}=(ZintersectU)intersect(ZintersectW) is if Z contains only the zero vector. Thus Z is the zero subspace, and it follows that Z must be a subspace of U and W.

Yeah, let me know if that's right... I'm not sure that I am correct in asserting that Z contains only the zero vector. :rolleyes:
 
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quasar_4 said:
The elements of C are given as {x|x is in Z and x is in U and x is in W}={0v}.
Why?
 
  • #3
well it sounded reasonable to me... if C is the set of {(Z intersects U) intersects (Z intersects W)}. So for (Z intersects U) we have the set containing {x|x is in Z and x is in U} and similarly for (Z intersects W) we have the set {x|x is in Z and x is in W}.
The direct sum would have all of these x too... wouldn't it? Although looking above it seems that I mixed up an intersection and direct sum. But even so, the intersection of the two intersections has to be the zero vector.
 
  • #4
quasar_4 said:
Although looking above it seems that I mixed up an intersection and direct sum. But even so, the intersection of the two intersections has to be the zero vector.
Yes, which means your second sentence is irrelevant! So what if the intersection of those two things is {0}? C (the direct sum) will certainly contain this, but it definitely does not have to equal it.
 
  • #5
quasar_4 said:
Howdy everyone. I'm not very good at writing proofs, so I am wondering if someone can tell me if I'm even on the right page with this. I am not sure if I understand the idea correctly. The theorem goes as follows: Suppose V is a finite dim. vector space with subspaces U and W such that V is the direct sum of U and W. Let Z also be a subspace of V. Then Z cannot equal the direct sums of (i) the intersection of Z and U and (ii) the intersection of Z and W unless Z is a subspace of U or W.

Here's my thoughts:

Let "C" denote the direct sum of the intersections (i) and (ii) above. Then two conditions must hold:
1) The intersection of the intersections (i) and (ii) must be the zero vector; and
2) The sum of the intersections must equal V.

The elements of C are given as {x|x is in Z and x is in U and x is in W}={0v}. Thus every x in Z that intersects U and W is in C. But we have that C={0v} since it is a direct sum. The only way that {0v}=(ZintersectU)intersect(ZintersectW) is if Z contains only the zero vector. Thus Z is the zero subspace, and it follows that Z must be a subspace of U and W.

Yeah, let me know if that's right... I'm not sure that I am correct in asserting that Z contains only the zero vector. :rolleyes:

Maybe you've already proved it. If not, here's a suggestion.

Put the statement into a form that'll shed some more light on the different options you have for proving it.
The first thing I'd do is get rid of the "unless"; then you'll have a positive statement. It'll be an implication.
Move its antecedent into your list of hypotheses (giving you more stuff to work with).
Notice the antecedent's a disjunction (which suggests one proof method).
Of course, there are other options too.
 

Related to Correct reasoning about direct sums proof?

1. What is a direct sum?

A direct sum is a mathematical concept that involves combining two or more objects or structures to create a new object or structure. In the context of linear algebra, a direct sum of vector spaces is the combination of two or more vector spaces to create a new vector space. This new vector space contains all the elements of the individual vector spaces and has certain properties that make it different from the original spaces.

2. How is a direct sum different from a direct product?

A direct sum is different from a direct product in that the elements of a direct sum are combinations of the individual elements from the original spaces, while the elements of a direct product are ordered pairs of elements from the original spaces. Additionally, a direct sum has a unique decomposition into its component spaces, while a direct product does not.

3. What is the importance of direct sums in mathematics?

Direct sums are important in mathematics because they allow us to combine and study objects in a structured way. They also provide a way to understand and analyze the properties of complex systems by breaking them down into simpler components. In addition, direct sums have applications in various fields such as linear algebra, group theory, and category theory.

4. How are direct sums used in proofs?

Direct sums are often used in proofs to demonstrate the structure and properties of a mathematical object. In particular, they are useful in proving the existence and uniqueness of certain structures, as well as in showing the relationship between different objects. Direct sums can also be used to simplify complicated expressions and equations, making them easier to work with in a proof.

5. Can you provide an example of a proof involving direct sums?

Sure, here is a simple proof involving direct sums:
Suppose V and W are vector spaces over a field F. Then the direct sum V⊕W is a vector space over F.
Proof:
First, we need to show that V⊕W is closed under vector addition. Let (v,w) and (v',w') be elements of V⊕W. Then (v+w, w+w') is also an element of V⊕W, since V and W are both closed under addition.
Next, we need to show that V⊕W is closed under scalar multiplication. Let (v,w) be an element of V⊕W and let c be an element of F. Then (cv, cw) is also an element of V⊕W, since V and W are both closed under scalar multiplication.
Finally, we need to show that V⊕W satisfies the vector space axioms. This can be easily verified using the properties of V and W. Therefore, V⊕W is a vector space over F.
QED.

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