Question

Let \(U\) and \(W\) denote subspaces of \(\mathbb{R}^{n}\), and assume that \(U \subseteq W\). If \(\operatorname{dim} U=n-1\), show that either \(W=U\) or \(W=\mathbb{R}^{n}\)

Short answer

Either \(W = U\) or \(W = \mathbb{R}^{n}\).

Step-by-step solution

Understanding the Given Information

We are given two subspaces, \(U\) and \(W\) of \(\mathbb{R}^{n}\), with \(U \subseteq W\). Additionally, we know that \(\text{dim}(U) = n-1\). We need to demonstrate that either \(W = U\) or \(W = \mathbb{R}^{n}\).

Possible Dimensions of W

Since \(U\) is a subspace of \(W\), the dimension of \(W\) must be at least that of \(U\), so \(\text{dim}(W) \geq n-1\). Consequently, \(\text{dim}(W)\) can be either \(n-1\) or \(n\) because the maximum dimension of any subspace in \(\mathbb{R}^{n}\) is \(n\).

Case 1: W = U

If \(\text{dim}(W) = n-1\), then \(W\) and \(U\) have the same dimension. Given that \(U\) is a subset of \(W\) and both have dimension \(n-1\), it follows that \(W = U\).

Case 2: W is the Whole Space

If \(\text{dim}(W) = n\), then \(W\) is not just equal to \(U\) but encompasses additional dimensions, making \(W = \mathbb{R}^{n}\) since it maximizes the dimension.

Conclusion from the Cases

The two possible cases for \(W\) have been considered: either \(W = U\) when the dimension is \(n-1\) or \(W = \mathbb{R}^{n}\) if the dimension is \(n\). Therefore, the statement is proven.

Key concepts

Subspaces

In linear algebra, a subspace is a set within a vector space that itself behaves like a vector space. Imagine having a large storage box, which represents a vector space, and inside it, smaller boxes that represent subspaces. For a set to qualify as a subspace, there are some important characteristics it must have:

• It must include the zero vector from the overarching vector space.
• It must be closed under addition. This means if two vectors are inside the subspace, their sum must also be inside the subspace.
• It must be closed under scalar multiplication. If you take any vector within the subspace and multiply it by a scalar, it should still reside in the subspace.

Understanding subspaces is crucial when dealing with larger structures in linear algebra. They help solve complex problems by breaking them down into manageable parts.

Vector Spaces

A vector space is a fundamental concept in linear algebra that consists of vectors and consists of operations of vector addition and scalar multiplication. These operations must adhere to specific rules, such as associativity, commutativity, and the existence of an additive identity and inverses.
To visualize, consider a plane as a two-dimensional vector space where infinitely many lines can exist across any direction. If you pluck any two vectors and add them, or if you multiply them by any scalar (number), they'd still lie within the same plane.
Vector spaces are more than just planes; they can exist in any number of dimensions. The "dimension" of a vector space tells us the minimum number of independent vectors needed to span the entire space. In essence, these vector spaces form the backbone of solutions in linear algebra.

Dimension Theory

Dimension theory deals with how we measure the "size" or "capacity" of vector spaces and subspaces. Dimension is essentially the count of the least number of vectors required to cover or "span" a space completely without leaving any gaps.
For instance, consider egin{align} U ext{ and } W, ext{ where } W ext{ is in } ' ext{ and contains } U ext{ with dimension . } U =n-1. ext{ In dimension theory we realize that ext{ if the dimension of } U ext{ is } n-1, ext{ then } W ext{ has to either equal } U ext { or go up to dimension } n, ext{ representing the entire space.} egin{endalign} This forms the crux of understanding how vector spaces with different dimensions relate to one another, guiding us in drawing conclusions about their relationships, as demonstrated in the exercise. Therefore, dimension theory provides a framework to evaluate and compare the roles and sizes of subspaces within vector spaces.