Question

Show that the intersection of two subspaces is also a subspace.

Short answer

Yes, the intersection of two subspaces is also a subspace. This is because the intersection contains the zero vector, and is closed under vector addition and scalar multiplication, which are the primary conditions for any set to be a subspace.

Step-by-step solution

Definition of Subspaces

It is important to first recall the definition of subspace. A subspace \(V\) of a vector space \(W\) over a field \(F\) (commonly \(\mathbb{R}\) or \(\mathbb{C}\)) is a subset of \(W\) that is itself a vector space over \(F\). In other words, for a subset \(V\) to be a subspace of \(W\), it must be closed under vector addition and scalar multiplication, and should contain the zero vector from \(W\).

Considering the intersection of two subspaces

Let's denote two subspaces of a given vector space \(W\) as \(V_1\) and \(V_2\). The intersection of these subspaces, denoted as \(V_1 \cap V_2\), contains all vectors that are present in both \(V_1\) and \(V_2\). We need to show that this intersection \(V_1 \cap V_2\) also satisfies the properties of a subspace.

Zero Vector Verification

First, as the zero vector is included in all subspaces, it will also be included in \(V_1 \cap V_2\). Therefore, the intersection contains the zero vector.

Closure under Addition Verification

Consider two vectors \(x\) and \(y\) that belong to \(V_1 \cap V_2\). By definition of the intersection, both these vectors are also in \(V_1\) and \(V_2\). As both \(V_1\) and \(V_2\) are subspaces, they are closed under addition. Therefore \(x + y\) is in \(V_1\) and in \(V_2\), implying that \(x + y\) belongs to \(V_1 \cap V_2\). This shows the closure under addition for \(V_1 \cap V_2\).

Closure under Scalar multiplication Verification

Take a scalar \(c\) in F and a vector \(x\) in \(V_1 \cap V_2\). Again by definition, \(x\) is both in \(V_1\) and \(V_2\). Since these are subspaces and are closed under scalar multiplication, it follows that \(cx\) is in \(V_1\) and in \(V_2\). So \(cx\) must also be part of \(V_1 \cap V_2\). This verifies the closure under scalar multiplication for \(V_1 \cap V_2\).

Conclusion of Proof

Hence, we have shown that \(V_1 \cap V_2\) includes the zero vector from \(W\), and is closed under addition and scalar multiplication. Therefore, by definition, the intersection of any two subspaces is also a subspace.

Key concepts

Vector Space

A vector space is a collection of vectors that satisfy certain mathematical properties. These properties involve vector addition and scalar multiplication, which allow the vectors to be scaled and combined in specific ways.
For a set to qualify as a vector space, it must include a zero vector (the vector that does nothing when added to another vector) and

• be closed under addition; This means that adding two vectors from this space results in a vector that is still in the space.
• be closed under scalar multiplication; This means multiplying a vector by a scalar results in a vector that is still in the space.

Common examples of vector spaces include spaces of real numbers (denoted by \( \mathbb{R}^n \)) or spaces of complex numbers (denoted by \( \mathbb{C}^n \)). Understanding these fundamental properties is crucial when examining subsets and subspaces within larger vector spaces.

Intersection

The intersection in the context of vector spaces refers to the common elements shared by two or more sets. When discussing subspaces, the intersection represents vectors that belong to all considered subspaces.
For subspaces \( V_1 \) and \( V_2 \) within a vector space \( W \), their intersection is denoted as \( V_1 \cap V_2 \). By definition, this intersection contains only those vectors that are present in both \( V_1 \) and \( V_2 \).
To show that \( V_1 \cap V_2 \) is also a subspace, we need to confirm that it holds all the properties of a vector space: i.e., it includes the zero vector and is closed under both vector addition and scalar multiplication. This concept establishes how different subspaces can overlap and still retain the characteristics of a unified subspace.

Scalar Multiplication

Scalar multiplication is a fundamental operation in vector spaces that involves multiplying a vector by a scalar (a single number from the field over which the vector space is defined). This operation must satisfy specific properties to maintain the structure of the vector space.
When you multiply a vector in a vector space by a scalar, the resulting vector should still be within that space. For instance, if \( v \) is in \( V \) (a vector space), then multiplying \( v \) by any scalar \( c \) results in \( cv \), which must also belong to \( V \).
This property ensures consistency across operations and highlights that scalar multiplication does not push the vector out of its original space. Verifying this property within intersecting subspaces \( V_1 \cap V_2 \) is crucial for proving that an intersection is itself a subspace.

Closure under Addition

Closure under addition is a critical property for any subset within a vector space to be classified as a subspace. This property ensures that the sum of any two vectors from the subset remains within the subset.
Consider vectors \( x \) and \( y \) that are elements of a subspace \( V \). If adding them results in a new vector \( x + y \) that is still a member of the same subspace \( V \), then \( V \) is closed under addition.
This property is vital when proving that the intersection of two subspaces \( V_1 \cap V_2 \) is itself a subspace. In the intersection, if any two vectors from \( V_1 \cap V_2 \) also belong to the individual subspaces \( V_1 \) and \( V_2 \), their sum must lie in both \( V_1 \) and \( V_2 \), thus in \( V_1 \cap V_2 \), guaranteeing closure under addition.