Sum and intersection of anihalator spaces

In summary, The statement is asking to prove that the orthogonal complement of the intersection of two finite-dimensional subspaces U and W is equal to the sum of the orthogonal complements of U and W. To prove this, it needs to be shown that any element in the orthogonal complement of the intersection is also in the sum of the orthogonal complements, and vice versa. However, the argument given in the conversation is faulty and does not prove the statement. A simpler and more direct approach can be taken to prove both directions of the statement.
  • #1
talolard
125
0

Homework Statement




prove that (U[tex]\bigcap[/tex]W)[tex]^{\circ}[/tex]=W[tex]^{\circ}[/tex]+U[tex]^{\circ}[/tex]


First prove That
(U[tex]\bigcap[/tex]W)[tex]^{\circ}[/tex][tex]\supseteq[/tex]W[tex]^{\circ}[/tex]+U[tex]^{\circ}[/tex]

Take any [tex]f\in (U\bigcap W)^{\circ}[/tex]
Then it is easy to see that for any [tex] f\in (U\bigcap W) f(v) =0 [/tex]

but since [tex] v\in U and v \in W [/tex] then [tex] f \in U^{\circ} and f \in W^{\circ}[/tex]



So we have

(U[tex]\bigcap[/tex]W)[tex]^{\circ}[/tex][tex]\supseteq[/tex]W[tex]^{\circ}[/tex]+U[tex]^{\circ}[/tex]

Next prove
(U[tex]\bigcap[/tex]W)[tex]^{\circ}[/tex][tex]\subseteq[/tex]W[tex]^{\circ}[/tex]+U[tex]^{\circ}[/tex]

[tex] W^{\circ} + U ^{\circ} = span(W)^{\circ} + span(U)^{\circ} [/tex]

because

[tex] S^{\circ} =Span(S)^{\circ} [/tex]

[tex] span(W)^{\circ} + span(U)^{\circ} = span (span(W)^{\circ} \cup span(u)^{\circ} [/tex]

By definition of addition of subspaces

[tex] span (span(W)^{\circ} \cup span(u)^{\circ}= span (W \cup U) ^{\circ} [/tex]

Which I am not sure of

And after all of that, we know that [tex] span (W \cap U) ^{\circ} \subseteq span (W \cup U) ^{\circ} [/tex]

Which proves it if I did not make a mistake? Am I correct?

Thanks
Tal
 
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  • #2
Need some more information here. What are [tex]U[/tex] and [tex]W[/tex]?
 
  • #3
Sorry Finite subspaces of the finite space V
 
  • #4
talolard said:
Sorry Finite subspaces of the finite space V

I'll assume you mean "finite-dimensional".

In the first half, your argument is both faulty and goes in the wrong direction. You say you intend to prove [tex](U \cap W)^\circ \supset U^\circ + W^\circ[/tex], and then you give an argument that begins with "take [tex]f \in (U \cap W)^\circ[/tex]" and concludes that "[tex]f \in U^\circ + W^\circ[/tex]". This argument, if correct, would prove [tex](U \cap W)^\circ \subset U^\circ + W^\circ[/tex], not [tex]\supset[/tex] : [tex]A \subset B[/tex] means that [tex]\alpha \in A[/tex] implies [tex]\alpha \in B[/tex].

However, the argument itself is not correct. If [tex]f \in (U \cap W)^\circ[/tex], and [tex]v \in U \cap W[/tex], then you are correct that [tex]f(v) = 0[/tex]. However, you cannot conclude from this that [tex]f \in U^\circ[/tex] on the grounds that [tex]f(v) = 0[/tex] and [tex]v \in U[/tex]. To conclude that [tex]f \in U^\circ[/tex] you would have to prove that [tex]f(v) = 0[/tex] for every [tex]v \in U[/tex], and this need not be true. What you actually want to do is give an equation [tex]f = g + h[/tex] where [tex]g \in U^\circ[/tex] and [tex]h \in W^\circ[/tex]; this proves that [tex]f \in U^\circ + W^\circ[/tex].

In the second half, I can't understand at all what you've written. This direction, [tex]U^\circ + W^\circ \subset (U \cap W)^\circ[/tex], actually has a simpler, direct argument like the above: take [tex]f \in U^\circ + W^\circ[/tex], and prove that [tex]f \in (U \cap W)^\circ[/tex].
 

Related to Sum and intersection of anihalator spaces

What is the definition of "sum of annihilator spaces"?

The sum of annihilator spaces is the set of all elements that are annihilated by at least one of the subspaces in a given collection of subspaces. In other words, it is the set of all elements that are mapped to zero by every linear functional in the collection.

What is the definition of "intersection of annihilator spaces"?

The intersection of annihilator spaces is the set of all elements that are annihilated by every linear functional in a given collection of subspaces. In other words, it is the set of all elements that are mapped to zero by every linear functional in the collection.

How are the sum and intersection of annihilator spaces related?

The sum and intersection of annihilator spaces are related by the fact that the sum of annihilator spaces is a subset of the intersection of annihilator spaces. This means that every element in the sum of annihilator spaces is also in the intersection of annihilator spaces, but not necessarily vice versa.

What is the significance of "sum of annihilator spaces" in linear algebra?

The sum of annihilator spaces is significant in linear algebra because it allows us to define a complementary subspace for a given subspace. This can be useful in solving linear systems and understanding the structure of vector spaces.

Can the sum and intersection of annihilator spaces be equal?

Yes, the sum and intersection of annihilator spaces can be equal in certain cases. This occurs when the given collection of subspaces is orthogonal, meaning that every pair of subspaces in the collection is perpendicular to each other. In this case, the sum and intersection of annihilator spaces will be equal and will be the orthogonal complement of the given collection of subspaces.

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