Problem 5 - Advanced Linear Algebra

Question

Prove that a subspace \(S\) of a Hilbert space \(H\) is closed if and only if \(S=S^{\perp \perp}\).

Step-by-Step Solution

Answer

To prove that a subspace \(S\) of a Hilbert space \(H\) is closed if and only if \(S=S^{\perp \perp}\), we perform the following steps: 1. Prove that if \(S\) is closed, then \(S=S^{\perp \perp}\): (a) Show that \(S \subseteq S^{\perp \perp}\) by considering an arbitrary element \(x \in S\) and showing that \(x\) is also an element of \(S^{\perp \perp}\). (b) Show that \(S^{\perp \perp} \subseteq S\) by considering an arbitrary element \(x \in S^{\perp \perp}\) and showing that \(x\) belongs to \(S\). 2. Prove that if \(S=S^{\perp \perp}\), then \(S\) is closed. Since the orthogonal complement of \(S^\perp\), i.e., \(S^{\perp \perp}\), is always closed, it follows that \(S\) must also be closed. Together, these steps show that a subspace \(S\) of a Hilbert space \(H\) is closed if and only if \(S=S^{\perp \perp}\).

Step 1: Prove that if \(S\) is closed, then \(S=S^{\perp \perp}\)

We will first prove that if \(S\) is a closed subspace of the Hilbert space \(H\), then \(S=S^{\perp \perp}\). This will be done by proving the following two claims.

Step 1(a): Prove that \(S \subseteq S^{\perp \perp}\)

To prove that \(S \subseteq S^{\perp \perp}\), we have to show that for any element \(x \in S\), \(x\) is also an element of \(S^{\perp \perp}\). Consider any arbitrary element \(x \in S\). Let \(y \in S^\perp\). Since \(x\) and \(y\) belong to orthogonal subspaces, their inner product must be zero, i.e., \(\langle x, y \rangle = 0\). As \(y \in S^\perp\), it is orthogonal to every element of \(S\). Now, we know that for any \(y \in S^\perp\), \(\langle x, y \rangle = 0\). This implies that \(x \in S^{\perp \perp}\) since \(x\) is orthogonal to all elements in \(S^{\perp}\), and hence we establish that \(S \subseteq S^{\perp \perp}\).

Step 1(b): Prove that \(S^{\perp \perp} \subseteq S\)

To prove that \(S^{\perp \perp} \subseteq S\), we need to show that any element \(x \in S^{\perp \perp}\) must also belong to \(S\). Consider any arbitrary element \(x \in S^{\perp \perp}\). Using the property that the orthogonal complement of \(S^{\perp}\), i.e., \(S^{\perp \perp}\), is closed and applying the orthogonal decomposition theorem to \(x \in S^{\perp \perp}\), we can write \(x\) as the sum of its orthogonal projections onto \(S\) and \(S^\perp\): \(x = P_S (x) + P_{S^\perp}(x)\). Now, since \(x \in S^{\perp \perp}\), it is orthogonal to all elements in \(S^{\perp}\). Therefore, the projection of \(x\) onto \(S^{\perp}\) must be zero: \(P_{S^\perp}(x) = 0\). This implies that \(x = P_S (x) \in S\), hence proving that \(S^{\perp \perp} \subseteq S\). Together, Steps 1(a) and 1(b) prove that if \(S\) is closed, then \(S=S^{\perp \perp}\).

Step 2: Prove that if \(S=S^{\perp \perp}\), then \(S\) is closed

Now, suppose \(S=S^{\perp \perp}\). We want to prove that \(S\) is closed. Remember that the orthogonal complement of \(S^{\perp}\), i.e., \(S^{\perp \perp}\), is always closed. Since \(S = S^{\perp \perp}\), it follows that \(S\) is also closed. Thus, we have shown that if \(S=S^{\perp \perp}\), then \(S\) is a closed subspace of the Hilbert space \(H\). This completes the proof that a subspace \(S\) of a Hilbert space \(H\) is closed if and only if \(S=S^{\perp \perp}\).

Key Concepts

Closed Subspace

In a Hilbert space, a subspace is considered "closed" when it contains all its limit points from convergent sequences within the space. Think of it like completing a circle; you cannot have any gaps within the defined subspace. Mathematically, a subspace being closed also means that it includes the limit of every Cauchy sequence of its elements.

• A Hilbert space is essentially a vector space equipped with an inner product, helping us measure angles and lengths.
• Closed subspaces are important because they guarantee convergence; when working with functions or sequences, you know you won't "fall off" the space.
• An example is if you have a sequence of points \(x_n\) in a closed subspace that converges to a point \x\. This point \x\ must also be in the subspace.

Recognizing closed subspaces helps in understanding the geometry of the space and ensures consistent mathematical operations.

Orthogonal Complement

The concept of an orthogonal complement is central when exploring the structure of subspaces within a Hilbert space. The orthogonal complement of a subspace, denoted as \(S^\perp\), includes all elements in the Hilbert space that are perpendicular to every element in the subspace \(S\).

• Two elements are orthogonal if their inner product equals zero. Imagine two arrows pointing in completely different, non-overlapping directions.
• The orthogonal complement \(S^\perp\) forms a subspace itself when considering a closed subspace like \(S\).
• Every element in a Hilbert space can be uniquely expressed as a sum of two components: one in \(S\) and another in its \(S^\perp\). This is a result of the orthogonal decomposition theorem.

Understanding orthogonal complements allows for decomposing spaces into easily manageable parts—one within the subspace and the other absolutely disjoint from it.

Orthogonal Projections

Orthogonal projections within a Hilbert space involve "projecting" or "casting" elements onto a subspace. If you imagine shining light perpendicularly onto a surface, the orthogonal projection is like the shadow cast purely downward from the light onto that surface.

• An orthogonal projection \P_S(x)\ maps each element \x\ of the Hilbert space to the closest point in the subspace \(S\).
• It satisfies the property of minimal distance, meaning it finds the nearest point in \(S\) to \x\.
• In terms of calculations, to find the projection of \x\ onto \(S\), use the relationships defined by the inner product in the space.

Orthogonal projections are powerful tools; they simplify problems in geometry and functional analysis by breaking down vectors into components that are easier to work with inside the given space. These projections also facilitate the process of solving equations and understanding how functions behave within constrained spaces.