Question

Let \(U\) be a subspace of \(\mathbb{R}^{n} .\) If \(\mathbf{x}\) in \(\mathbb{R}^{n}\) can be written in any way at all as \(\mathbf{x}=\mathbf{p}+\mathbf{q}\) with \(\mathbf{p}\) in \(U\) and \(\mathbf{q}\) in \(U^{\perp}\), show that necessarily \(\mathbf{p}=\operatorname{proj}_{U} \mathbf{x}\).

Short answer

\( \mathbf{p} = \operatorname{proj}_U \mathbf{x} \) because \( \mathbf{x} - \mathbf{p} = \mathbf{q} \) and \( \mathbf{q} \in U^{\perp} \).

Step-by-step solution

Understanding the Problem

We need to show that when a vector \( \mathbf{x} \) from \( \mathbb{R}^n \) is decomposed into \( \mathbf{x} = \mathbf{p} + \mathbf{q} \) with \( \mathbf{p} \in U \) and \( \mathbf{q} \in U^{\perp} \), it implies that \( \mathbf{p} \) must be the orthogonal projection of \( \mathbf{x} \) onto \( U \).

Definition of Orthogonal Projection

Recall that the orthogonal projection of \( \mathbf{x} \) onto \( U \), denoted as \( \operatorname{proj}_U \mathbf{x} \), is a vector \( \mathbf{p} \in U \) such that the difference \( \mathbf{x} - \mathbf{p} \) is orthogonal to every vector in \( U \).

Using the Decomposition of \( \mathbf{x} \)

We have \( \mathbf{x} = \mathbf{p} + \mathbf{q} \) where \( \mathbf{p} \in U \) and \( \mathbf{q} \in U^{\perp} \). By definition of orthogonality, \( \mathbf{p} \) and \( \mathbf{q} \) are perpendicular, i.e., \( \langle \mathbf{p}, \mathbf{q} \rangle = 0 \).

Verification \( \mathbf{p} = \operatorname{proj}_U \mathbf{x} \)

Since \( \mathbf{q} \in U^{\perp} \), \( \mathbf{x} - \mathbf{p} = \mathbf{q} \), and \( \mathbf{p} \) is orthogonal to \( \mathbf{q} \), \( \mathbf{p} \) must equal \( \operatorname{proj}_U \mathbf{x} \) because it satisfies the condition that \( \mathbf{x} - \mathbf{p} \) is orthogonal to all vectors in \( U \). Thus, \( \mathbf{p} = \operatorname{proj}_U \mathbf{x} \).

Conclusion

The decomposition \( \mathbf{x} = \mathbf{p} + \mathbf{q} \), where \( \mathbf{p} \in U \) and \( \mathbf{q} \in U^{\perp} \), ensures \( \mathbf{p} \) is the orthogonal projection onto \( U \) because \( \mathbf{x} - \mathbf{p} = \mathbf{q} \) and \( \mathbf{q} \) is orthogonal to every vector in \( U \). Therefore, \( \mathbf{p} = \operatorname{proj}_U \mathbf{x} \).

Key concepts

Subspace

A subspace is a special kind of subset within a vector space. For it to be considered a subspace, it must satisfy the following conditions: it must include the zero vector, it must be closed under addition, and it must be closed under scalar multiplication. This means that if you take any two vectors from a subspace and add them, the resulting vector is also in the subspace. Similarly, if you take any vector from the subspace and multiply it by a scalar (which is just any number), the resulting vector is still within the subspace.
Understanding subspaces is key because they form the backbone of various vector space concepts. In the context of the exercise, the subspace is denoted by \( U \), and it consists of all vectors that fit certain criteria within \( \mathbb{R}^{n} \).

• Includes the zero vector
• Closed under addition
• Closed under scalar multiplication

These properties make subspaces predictable and reliable for all sorts of linear transformations and decompositions, like the ones we frequently see in linear algebra.

Orthogonality

Orthogonality is a core concept in linear algebra, and it refers to the idea of two vectors being perpendicular to each other. In simple terms, when two vectors are orthogonal, their dot product is zero, \( \langle \mathbf{v}, \mathbf{w} \rangle = 0 \). This property is incredibly useful, especially when dealing with projections and decompositions.
Orthogonality comes into play when you have a subspace \( U \) and its orthogonal complement \( U^{\perp} \). These two sets of vectors are related in a way that any vector \( \mathbf{q} \) from \( U^{\perp} \) is orthogonal to all vectors from \( U \). In the exercise we're discussing, this orthogonality allows us to perfectly decompose any vector \( \mathbf{x} \) from \( \mathbb{R}^{n} \) into two parts: one part that lies in \( U \) and one part that lies in \( U^{\perp} \).

Vector Decomposition

Vector decomposition is the method of breaking down a vector \( \mathbf{x} \) in a vector space into two specific components. In the context of orthogonal projections, this decomposition involves splitting \( \mathbf{x} \) into a sum of two vectors, \( \mathbf{p} \) and \( \mathbf{q} \). Here, \( \mathbf{p} \) is a vector that lies entirely in the subspace \( U \), while \( \mathbf{q} \) is a vector that lies entirely in the orthogonal complement \( U^{\perp} \) of that subspace.
This precise decomposition is what allows us to simplify complex vector calculations and make transformations more manageable. When \( \mathbf{x} \) is decomposed as \( \mathbf{x} = \mathbf{p} + \mathbf{q} \), it uniquely determines \( \mathbf{p} \) as the orthogonal projection of \( \mathbf{x} \) onto \( U \).

• Ensures \( \mathbf{p} \) is the closest vector in \( U \) to \( \mathbf{x} \)
• Maintains \( \mathbf{q} \) orthogonal to \( \mathbf{p} \)

Thus, vector decomposition not only aids in understanding the structure of vector spaces but also in performing computations that would otherwise be much more complex or non-intuitive.