Is This Proof of the Annihilator of a Set Correct?

In summary, an inner product space is a vector space equipped with an inner product function that has properties such as linearity, symmetry, and positive definiteness. The annihilator of a set in an inner product space is the set of all vectors that are orthogonal to every vector in the original set. Garling, Proposition 11.3.5 - 1 is a theorem that states the linear independence of a set of vectors and its orthogonal complement. The annihilator of a set is equivalent to the orthogonal complement of the subspace spanned by that set. Inner product spaces and their properties can be extended to infinite-dimensional spaces, but some modifications may be necessary.
  • #1
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I am reading D. J. H. Garling's book: "A Course in Mathematical Analysis: Volume II: Metric and Topological Spaces, Functions of a Vector Variable" ... ...

I am focused on Chapter 11: Metric Spaces and Normed Spaces ... ...

I need some help to confirm my thinking on Proposition 11.3.5 - 1 ...

Garling's statement and proof of Proposition 11.3.5 reads as follows:
View attachment 8963
In Proposition 11.3.5 - 1, given the definition of \(\displaystyle A^{ \bot }\), we are required to prove that

\(\displaystyle A^{ \bot } = \{ x \in V \ : \ \langle a, x \rangle = 0\) for all \(\displaystyle a \in A \}\)

We need to show

\(\displaystyle A^{ \bot } = \{ x \in V \ : \ \langle x, a \rangle = 0\) for all \(\displaystyle a \in A \}\)
Proof:

By definition \(\displaystyle A^{ \bot } = \{ x \in V \ : \ \langle x, a \rangle = 0\) for all \(\displaystyle a \in A \}\)

\(\displaystyle \Longrightarrow A^{ \bot } = \{ x \in V \ : \ \overline{ \langle x, a \rangle } = 0\) for all \(\displaystyle a \in A \}\)
Now suppose \(\displaystyle \langle a, x \rangle = u + i v\)

Then \(\displaystyle \overline{ \langle x, a \rangle } = \langle a, x \rangle = u + i v\) by definition of an inner product ...

and so \(\displaystyle \langle x, a \rangle = u - i v\) Thus if \(\displaystyle \langle a, x \rangle = 0\) then \(\displaystyle u = v = 0\)

... which implies \(\displaystyle \langle x, a \rangle = 0\) Thus if

\(\displaystyle A^{ \bot } = \{ x \in V \ : \ \langle a, x \rangle = 0 for all a \in A \}\)

then

\(\displaystyle A^{ \bot } = \{ x \in V \ : \ \langle x, a \rangle = 0 for all a \in A \} \)

Is the above proof correct? Is there a better way to prove the above ...?Peter
 

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  • #2
Hi Peter.

You could just say
$$\langle a,x\rangle=0\ \iff\ \langle x,a\rangle=\overline{\langle a,x\rangle}=\overline 0=0.$$
 
  • #3
Hi Peter,

Your proof looks correct to me. Another way to prove this proposition is by using the fact that the orthogonal complement is the set of all vectors that are perpendicular to the given set. So, we can show that A^{\bot} = \{x \in V : \langle x, a \rangle = 0 \text{ for all } a \in A\} by showing that every vector in A^{\bot} is perpendicular to every vector in A, and vice versa.

To show that every vector in A^{\bot} is perpendicular to every vector in A, we can use the definition of the inner product and the fact that the inner product is linear. Then, for any x \in A^{\bot} and a \in A, we have:

\langle x, a \rangle = \langle x, \sum_{i=1}^n \lambda_i a_i \rangle = \sum_{i=1}^n \lambda_i \langle x, a_i \rangle = \sum_{i=1}^n \lambda_i \cdot 0 = 0

where \lambda_i are scalars and a_i are vectors in A. This shows that x is perpendicular to every vector in A, and thus x \in A^{\bot}.

Similarly, to show that every vector in A^{\bot} is perpendicular to every vector in A, we can use the same argument in reverse. For any x \in A^{\bot} and a \in A, we have:

\langle a, x \rangle = \langle \sum_{i=1}^n \lambda_i a_i, x \rangle = \sum_{i=1}^n \lambda_i \langle a_i, x \rangle = \sum_{i=1}^n \lambda_i \cdot 0 = 0

This shows that a is perpendicular to every vector in A^{\bot}, and thus a \in A^{\bot}.

Therefore, we have shown that A^{\bot} = \{x \in V : \langle x, a \rangle = 0 \text{ for all } a \in A\}, and thus the proposition is proven.

I hope this helps! Let me know if you have any further questions or if you would like me to clarify anything. Happy studying!
 

FAQ: Is This Proof of the Annihilator of a Set Correct?

1. What is an inner product space?

An inner product space is a mathematical concept that involves a vector space equipped with an inner product, which is a function that takes in two vectors and returns a scalar value. This inner product satisfies certain properties, such as linearity and positive definiteness, and allows for the calculation of angles and lengths within the vector space.

2. What is the annihilator of a set in an inner product space?

The annihilator of a set in an inner product space is the set of all linear functionals that map the set to the zero scalar. In other words, it is the set of all vectors that are orthogonal to every vector in the given set. This concept is useful in linear algebra and functional analysis, particularly in the study of dual spaces.

3. What is Garling, Proposition 11.3.5 - 1?

Garling, Proposition 11.3.5 - 1 is a theorem in functional analysis that states that the annihilator of a closed subspace in an inner product space is equal to the orthogonal complement of that subspace. In other words, the set of all vectors that are orthogonal to a closed subspace is the same as the set of all linear functionals that map the subspace to the zero scalar.

4. How is Garling, Proposition 11.3.5 - 1 useful in mathematics?

This proposition is useful in proving other theorems and results in functional analysis, as it provides a relationship between two important concepts: the annihilator of a set and the orthogonal complement of a subspace. It also has applications in linear algebra and can aid in the understanding of dual spaces and orthogonal projections.

5. Can Garling, Proposition 11.3.5 - 1 be extended to other mathematical concepts?

Yes, this proposition can be extended to other mathematical concepts, such as Hilbert spaces and Banach spaces. In these contexts, the annihilator and orthogonal complement have similar definitions and properties, making this proposition applicable in a broader range of mathematical settings.

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