Subspaces

Before we proceed to discuss the idea of subspaces, I’ll give some background on the idea of a set, as it is a concept that will appear in Linear Algebra often.

A set can be defined as a collection of unique objects. When we discuss the real numbers, for instance, we are really discussing the set of real numbers. The set of real numbers contains all of the numbers which are real, such as integers, rational numbers, and irrational numbers. We represent the set of real numbers with a special symbol, \mathbb{R}.

Let’s take a look at a simple example of a set to get more familiar with the ideas and notations associated with them. Suppose we have a set S, which contains three elements. The three elements that are in the set S are the numbers: 1,2 and 3. If we want to formally define this set, we define it as S = {1,2,3}. This tells us we have a set S which has three elements, and those three elements are the values 1,2, and 3. 

If we want to state that an element is inside the set S, we use the \in symbol. For instance, we can say that 1 \in S, which reads “1 is an element of S” or “1 is in the set S”. If we want to say that an element is not in S, then we can use the \notin symbol. For instance, we can say that 4 \notin S, which reads “4 is not an element of S” or “4 is not in the set S”. One final note is the idea of a subset. A subset is some set that contains elements from another set. For example, A = {1} is a subset of S, since A is composed of elements that exist in S, and nothing that doesn’t exist in S.

We discussed in the last article that we often refer to the 2 component vectors as \mathbb{R}^2, the three component vectors as \mathbb{R}^3, and the n component vectors as \mathbb{R}^n. These are examples of sets, where for example \mathbb{R}^2 is the set of all vectors with two real components. When we dicuss ideas like subspaces, we are going to use the idea of “the set of all vectors with n components”, which is represented by \mathbb{R}^n. This type of notation will be referenced throughout Linear Algebra, so it is important to know and understand.

With that in mind let’s proceed to the idea of subspaces. The set of vectors \mathbb{R}^n is often referred to as a space. This means a set of objects which have an operation for addition and multiplication defined for them. In terms of vectors, the object is a vector, which we have defined vector addition and scalar multiplication for. Any subset of the set \mathbb{R}^n is considered a subspace of \mathbb{R}^n if it satisfies three conditions defined below.

Subspace: A non-empty subset S of \mathbb{R}^n is called a subspace if for every vector in S, the following conditions hold:

  1. \vec{x} + \vec{y} \in S (closed under addition)
  2. t\vec{x} \in S (closed under scalar multiplication)
  3. It contains the zero vector, \vec{0}

To put this in plain English, a subspae is any subset of \mathbb{R}^n that is not empty, where we can add any two elements, do scalar multiplication on any element, and still receive an end result that is inside of the set S. To better understand this, let’s look at an example.

Example: Show that the following set is a subspace of \mathbb{R}^3. S = \left\{\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \mid  x_1 - x_2 + x_3 = 0\right\}

First, let’s understand what it means for some vector to be an element in the set S. The set S requires a vector to have three components, and those components must be defined such that the first element minus the second element plus the third element is equal to 0.

Understanding this, we can now move on to showing that the three properties of a subspace are true for this particular subset.

We will start by showing that this set is closed under addition. This means that any two vectors added together should give us a vector still in the set S.

Suppose we have two vectors, a = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} and b = \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix}. If we were to add these two vectors, we would add component by component, and get \begin{bmatrix} x_1 + y_1 \\ x_2 + y_2 \\ x_3 + y_3 \end{bmatrix}

From here, we just need to show that this new vector still exists in the set S. In order for this to be true, we need to check that x_1 - x_2 + x_3 = 0

Applying this formula, we get (x_1 + y_1) - (x_2 + y_2) + (x_3 + y_3). If we rearrange the brackets, we get (x_1 - x_2 + x_3) + (y_1 - y_2 + y_3). From the definitions of \vec{x} and \vec{y}, we already know that x_1-x_2+x_3 = 0 and y_1-y_2+y_3 = 0, therefore we have 0 + 0 = 0, meaning that \vec{x} + \vec{y} \in S.

Showing that scalar multiplication is closed is even easier. Suppose we have a vector \vec{x} \in S. If we were to compute t \vec{x}, we would get \begin{bmatrix} t x_1 \\ t x_2 \\ t x_3 \end{bmatrix}. Since we know that x_1 - x_2 + x_3 = 0, it follows that t(x_1 - x_2 + x_3) = t*0 = 0. Therefore, t\vec{x} \in S

Finally, we just need to show that \vec{0} \in S. For the zero vector to be in S, we need to show that it follows x_1 - x_2 + x_3 = 0. If all components are 0, then of course 0 - 0 + 0 = 0, therefore \vec{0} \in S.

Showing these three properties hold proves that S is a subspace in \mathbb{R}^3

The concept of a subspace is important because it allows us to conclude information about subsets of the vector space. A vector space has many properties that we will discuss throughout this series of articles. If we can show that a set is a subspace of the vector space, then we can make conclusions about properties that apply without needing the reprove them for the subspace.

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