Question

We often write vectors in \(\mathbb{R}^{n}\) as rows. Show that every proper subspace \(U\) of \(\mathbb{R}^{2}\) is a line through the origin. [Hint: If \(\mathbf{d}\) is a nonzero vector in \(U,\) let \(L=\mathbb{R} \mathbf{d}=\\{r \mathbf{d} \mid r\) in \(\mathbb{R}\\}\) denote the line with direction vector \(\mathbf{d}\). If \(\mathbf{u}\) is in \(U\) but not in \(L\), argue geometrically that every vector \(\mathbf{v}\) in \(\mathbb{R}^{2}\) is a linear combination of \(\mathbf{u}\) and \(\mathbf{d}\).

Short answer

Every proper subspace of \( \mathbb{R}^2 \) is a line through the origin.

Step-by-step solution

Identify the Components of the Subspace

First, recognize that a proper subspace of \( \mathbb{R}^2 \) cannot be the entire space itself. Identify that a subspace must pass through the origin and, since we are considering \( \mathbb{R}^2 \), the subspace can either be the origin itself or a line through the origin.

Build a Line Using a Nonzero Vector

Let \( \mathbf{d} \) be a nonzero vector in \( U \). The line \( L \) defined by \( L = \{ r \mathbf{d} \mid r \in \mathbb{R} \} \) is a line through the origin that passes through \( \mathbf{d} \). Thus, every vector in \( L \) can be expressed as a scalar multiple of \( \mathbf{d} \).

Consider a Vector Not in the Line

Assume there is a vector \( \mathbf{u} \) in \( U \) that is not in \( L \). The vector \( \mathbf{u} \) is linearly independent from \( \mathbf{d} \), meaning it cannot be expressed as a scalar multiple of \( \mathbf{d} \) alone.

Form Basis for \( \mathbb{R}^2 \)

If \( \mathbf{u} \) is not in \( L \), \( \{ \mathbf{u}, \mathbf{d} \} \) forms a basis for \( \mathbb{R}^2 \), which means any vector \( \mathbf{v} \) in \( \mathbb{R}^2 \) can be written as a linear combination of \( \mathbf{u} \) and \( \mathbf{d} \).

Conclude That \( U \) Must Be a Line

Since \( U \) is a proper subspace of \( \mathbb{R}^2 \), but if \( \mathbf{u} \) existed it would make \( U \) the whole space \( \mathbb{R}^2 \), we conclude that such \( \mathbf{u} \) cannot exist. Hence, \( U \), a proper subspace, must consist only of all scalar multiples of \( \mathbf{d} \), forming a line through the origin.

Key concepts

Linearity

Linearity is a crucial concept when discussing subspaces, especially in the context of vector spaces like \(\mathbb{R}^2\). In this space, vectors can be added together or multiplied by scalars to form new vectors. A linear operation, therefore, respects these two operations: addition and scalar multiplication.
In simple terms, a function or operation is linear if combining vectors and scaling them leads to predictable, additive outcomes. For instance, if you take a vector \( \mathbf{v} \) and multiply it by a scalar \( r \), you'll stretch or shrink its size in a consistent manner.
This holds true for any vector in a proper subspace of \( \mathbb{R}^2 \). Every action you take following these rules keeps you within that subspace.

Vector Spaces

A vector space is a set of vectors that can be scaled and added together, staying within the same set. In the realm of \( \mathbb{R}^2 \), where vectors are written as pairs of numbers, the vector space includes all possible linear combinations of these pairs.
To classify something as a vector space, it must satisfy a few essential properties:

• Closure under addition: The sum of any two vectors in the space remains in the space.
• Closure under scalar multiplication: Multiplying a vector by a scalar keeps it in the system.
• Contains the zero vector: Every vector space must have a zero vector, which is essentially a vector of all zeros.

Hence, any proper subspace of \( \mathbb{R}^2 \) will include the origin, i.e., the zero vector, and adheres to the principles of vector spaces. Importantly, these subspaces could either be a single point or extend as a line through the origin.

Linear Combination

A linear combination refers to the sum formed by multiplying each vector in a given set by a corresponding scalar and then adding the results together. For example, consider vectors \( \mathbf{u} \) and \( \mathbf{d} \) in \( \mathbb{R}^2 \). A vector \( \mathbf{v} \) can be expressed as \( r \mathbf{u} + s \mathbf{d} \), where \( r \) and \( s \) are scalars.
This approach is pivotal in establishing whether a set of vectors spans a subspace such as \( \mathbb{R}^2 \). In the context given, if both vectors form a basis for the space, then any vector can be written as a linear combination of these basis vectors.
When discussing subspaces, if you find a non-zero vector \( \mathbf{d} \) and another vector \( \mathbf{u} \) not lying along the line \( L \) generated by \( \mathbf{d} \), their linear combination can represent any vector in \( \mathbb{R}^2 \). This is only possible if both vectors are independent and span the entire space, demonstrating why a proper subspace would only consist of multiples of one vector, forming a line through the origin.