# Prove the W is the orthogonal complement of its orthogonal complement

• chipotleaway
In summary, the conversation discusses the proof of the property that the orthogonal complement of the orthogonal complement of a subspace W is equal to W. The attempt at a solution shows that all vectors in W must also be in (W^{\perp})^{\perp}, and then attempts to show that (W^{\perp})^{\perp} cannot contain anything not in W. However, it neglects the fact that this property only holds when W is closed. A standard proof can be found in Erwin Kreyszig's "Introduction to Functional Analysis" on page 149, lemma 3.3-6.
chipotleaway

## Homework Statement

Let W be a subspace of R^n. Show that the orthogonal complement of the orthogonal complement of W is W.

i.e. Show that $(W^{\perp})^{\perp}=W$

## The Attempt at a Solution

This is one of those 'obvious' properties that probably has a really simple proof but which continues to elude me. Here's my latest, rather ugly attempt. I still feel like I've overlooked something, if not, I think it definitely could be refined.

If $W^{\perp}$ is the orthogonal complement of $W$, then $w.u=0, \forall w\in W, \forall u \in W^{\perp}$.
Similarly, $x.u=0, \forall u\in W^{\perp}$ and $\forall x\in (W^{\perp})^{\perp}$.
All vectors $w \in W$ must also be in $(W^{\perp})^{\perp}$ since $W$ consists of only vectors to perpendicular to those in $W^{\perp}$. Now we need to show that $(W^{\perp})^{\perp}$ cannot contain anything not in $W$. Suppose there is a vector $x' \in (W^{\perp})^{\perp}$ but not in $W$. Then $x'.u=0$, but it must also have some property that prevents it from being in $W$, namely $x'$ would also have to be orthogonal to $W$ (if not, then it must be in $W$). But this would imply it is in $W^{\perp}$ and contradict the property $W^{\perp} \cap (W^{\perp})^{\perp}=0$

chipotleaway said:

## Homework Statement

Let W be a subspace of R^n. Show that the orthogonal complement of the orthogonal complement of W is W.

i.e. Show that $(W^{\perp})^{\perp}=W$

## The Attempt at a Solution

This is one of those 'obvious' properties that probably has a really simple proof but which continues to elude me. Here's my latest, rather ugly attempt. I still feel like I've overlooked something, if not, I think it definitely could be refined.

If $W^{\perp}$ is the orthogonal complement of $W$, then $w.u=0, \forall w\in W, \forall u \in W^{\perp}$.
Similarly, $x.u=0, \forall u\in W^{\perp}$ and $\forall x\in (W^{\perp})^{\perp}$.
All vectors $w \in W$ must also be in $(W^{\perp})^{\perp}$ since $W$ consists of only vectors to perpendicular to those in $W^{\perp}$.
Is it okay to simply assert that or should you explicitly show it? One way to show equality of two sets A and B is to show that A is a subset of B and B is a subset of A, which is essentially what you seem to be doing. But if you can simply make the claim you did, can't you do the same thing with the two sets swapped and be finished?

Now we need to show that $(W^{\perp})^{\perp}$ cannot contain anything not in $W$. Suppose there is a vector $x' \in (W^{\perp})^{\perp}$ but not in $W$. Then $x'.u=0$, but it must also have some property that prevents it from being in $W$, namely $x'$ would also have to be orthogonal to $W$ (if not, then it must be in $W$). But this would imply it is in $W^{\perp}$ and contradict the property $W^{\perp} \cap (W^{\perp})^{\perp}=0$

1 person
You have neglected the fact that your claim is true only when W is closed. You can check Erwin Kreyszig - Introduction to Functional analysis.. pg 149, lemma 3.3-6 for a standard proof.
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## 1. What is an orthogonal complement?

An orthogonal complement is a vector space that consists of all vectors that are perpendicular to a given vector space. In other words, it is the set of all vectors that have a dot product of zero with all vectors in the original space.

## 2. Why is the orthogonal complement important in linear algebra?

The orthogonal complement plays a crucial role in linear algebra because it allows us to decompose a vector space into two mutually perpendicular subspaces, making it easier to understand and solve problems in higher dimensions.

## 3. How do you prove that W is the orthogonal complement of its orthogonal complement?

To prove that W is the orthogonal complement of its orthogonal complement, we must show that any vector in W is perpendicular to any vector in the orthogonal complement of W, and vice versa. This can be done by using the properties of dot product and the definition of orthogonal complement.

## 4. Can the orthogonal complement of a vector space be empty?

Yes, the orthogonal complement of a vector space can be empty if the vector space itself is the zero vector or if it contains only the zero vector. In this case, all vectors would have a dot product of zero and would therefore be in the orthogonal complement, resulting in an empty set.

## 5. How does the concept of orthogonal complement relate to orthogonality?

The concept of orthogonal complement is closely related to orthogonality because a vector space and its orthogonal complement are mutually perpendicular. This means that any vector in the original space is perpendicular to any vector in its orthogonal complement, and vice versa.

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