Question

If \(U\) is a subspace of \(\mathbb{R}^{n},\) show that \(\mathbf{x}=\operatorname{proj}_{U} \mathbf{x}+\operatorname{proj}_{U^{\perp}} \mathbf{x}\) for all \(\mathbf{x}\) in \(\mathbb{R}^{n}\)

Short answer

Any vector \( \mathbf{x} \) in \( \mathbb{R}^n \) can be decomposed as \( \mathbf{x} = \operatorname{proj}_{U} \mathbf{x} + \operatorname{proj}_{U^{\perp}} \mathbf{x} \).

Step-by-step solution

Understand the Problem

We need to show that any vector \( \mathbf{x} \) in \( \mathbb{R}^n \) can be decomposed into the sum of its projection onto a subspace \( U \) and the orthogonal complement of \( U \). This relies on the properties of orthogonal projections.

Recall Definitions

Recall that the projection of \( \mathbf{x} \) onto a subspace \( U \), denoted \( \operatorname{proj}_{U} \mathbf{x} \), is the vector in \( U \) closest to \( \mathbf{x} \). The orthogonal complement \( U^{\perp} \) consists of all vectors in \( \mathbb{R}^n \) orthogonal to \( U \). The projection onto \( U^{\perp} \) is \( \operatorname{proj}_{U^{\perp}} \mathbf{x} \).

Use Orthogonality

The fundamental property of \( U \) and \( U^{\perp} \) is that any vector \( \mathbf{x} \) in \( \mathbb{R}^n \) can be uniquely expressed as the sum of a vector in \( U \) and a vector in \( U^{\perp} \). This means \( \operatorname{proj}_{U} \mathbf{x} \) and \( \operatorname{proj}_{U^{\perp}} \mathbf{x} \) are perpendicular, and their sum equals \( \mathbf{x} \).

Write the Expression

According to the definitions and orthogonality condition, we can express \( \mathbf{x} \) as: \[ \mathbf{x} = \operatorname{proj}_{U} \mathbf{x} + \operatorname{proj}_{U^{\perp}} \mathbf{x} \] This is because both projections together span the whole space \( \mathbb{R}^n \) through their direct sum, capturing the totality of \( \mathbf{x} \).

Provide Conclusion

Hence, we have shown that for any vector \( \mathbf{x} \) in \( \mathbb{R}^n \), the following holds true: \[ \mathbf{x} = \operatorname{proj}_{U} \mathbf{x} + \operatorname{proj}_{U^{\perp}} \mathbf{x} \] This confirms the ability of vectors in \( \mathbb{R}^n \) to be decomposed into orthogonal components related to \( U \) and \( U^{\perp} \).

Key concepts

Subspace

In the realm of linear algebra, a subspace is a fundamental concept that helps to bring order and simplicity to the study of vector spaces. A subspace is a special type of subset of a vector space that retains the core properties of vector addition and scalar multiplication within its confines. This means if you take any two vectors within a subspace and add them, or multiply them by a scalar, the result will still be within the same subspace.To better understand, think about a flat plane living inside a three-dimensional space like our regular world. This plane can be considered a subspace of the entire three-dimensional space. Any point or vector you take on this plane will remain on this plane when linearly manipulated. A line through the origin is also a subspace of the plane and the original space, illustrating how subspaces can nest within each other.Subspaces are crucial because they define the space onto which we project vectors, which is essential for simplifying complex calculations. Within the context of the exercise, understanding subspaces helps in realizing how part of a vector in \( \mathbb{R}^n \) essentially belongs to a subspace \( U \), helping in its orthogonal decomposition.

Orthogonal Complement

The orthogonal complement is an intriguing concept that allows us to understand what lies outside a given subspace but still contained within the original vector space. Given a subspace \( U \) in a vector space \( \mathbb{R}^n \), the orthogonal complement \( U^{\perp} \) comprises all vectors in \( \mathbb{R}^n \) that are perpendicular, or orthogonal, to every vector in \( U \).Imagine a shadow cast by a tall building. The shadow is entirely separate from the building's structure, yet it interacts with it. Similarly, \( U^{\perp} \) exists independently of \( U \), yet complements it such that together, both spaces capture all directions within \( \mathbb{R}^n \). The dimension of \( U^{\perp} \) is such that when added to the dimension of \( U \), the result is the dimension of the entire space \( \mathbb{R}^n \).A significant property of the orthogonal complement used in the exercise is that any vector in \( \mathbb{R}^n \) can be split into two parts: one that lies in \( U \), and one in \( U^{\perp} \). These parts are distinct yet when summed, they reconstruct the original vector \( \mathbf{x} \). This unique way of slicing vectors is central to understanding projections.

Vector Decomposition

Vector decomposition is like breaking down a jigsaw puzzle into individual pieces that fit together to form a complete picture. It is the process of breaking a vector into parts, usually in such a way that those parts are easier to work with based on the problem at hand.One of the practical and profound techniques in linear algebra is decomposing a vector \( \mathbf{x} \) from \( \mathbb{R}^n \) into two components: the orthogonal projection onto a subspace \( U \) and the projection onto the orthogonal complement of \( U \), \( U^{\perp} \). In simple terms, it can be thought of as resolving a vector into two directions: one pointing along a plane or line you are interested in (\( U \)), and the other perpendicular to it (\( U^{\perp} \)).The expression \( \mathbf{x} = \operatorname{proj}_{U} \mathbf{x} + \operatorname{proj}_{U^{\perp}} \mathbf{x} \) sums up vector decomposition in this context perfectly. Here, \( \operatorname{proj}_{U} \mathbf{x} \) is the shadow of \( \mathbf{x} \) in \( U \), and \( \operatorname{proj}_{U^{\perp}} \mathbf{x} \) is the reminder component that makes \( \mathbf{x} \) whole again by adding the orthogonal part back.This decomposition is incredibly useful, especially in practical applications like data compression, computer graphics, and solving multivariable equations, where separating components simplifies complex issues into manageable parts.