# Subspaces of Vector Spaces

One of the most interesting topics from Linear Algebra is that of $\textbf{Subspaces}$ of $\textit{Vector Spaces}$. The introduction of these two ideas into our study allows for a more thorough, and dynamic view of many of previous covered ideas. Not only that, the insight of $\textbf{Subspaces}$ and $\textit{Vector Spaces}$ in general allows us to take prior concepts and generalize them in a more broader sense. To begin, we look at the definition of a $\textbf{Vector Subspace}$.

Let $V$ be a Vector Space, and $H$ be a non-empty set. We call $H$ a $\textbf{Subspace}$ of $V$ provided that $H$ meets the following conditions:

* $H$ is a $\textit{subset}$ of $V$.
* $H$ is $\textit{closed}$ under Vector Addition.
* $H$ is $\textit{closed}$ under Scalar Mulitplication.

It is important to note here that the assertion that $H$ be non-empty is not trivial; that is it is important that $H$ contains at least one element. The conditions listed above would hold $\textit{vacously}$ for the empty set if we did not include the aformentioned caveat.

From just these three conditions, we can prove an interesting result of subspaces in general.

$\textbf{Theorem 1}:$ If $H$ is a subspace of some vector space $V$, then the the zero vector ($\vec 0$) of $V$ is in $H$.

$\textit{Proof}:$

Suppose $H$ is a subspace of $V$. This would mean that $H$ is a non-empty set of $H$. By the fact that $H$ is non-empty, we see that there exists some vector $\vec x$ that is in $H$. Furthermore, since $H$ is a subspace of $V$, we see that $H$ is closed under scalar multiplication. It follows then that

$$ 0 \cdot \vec x = \vec 0.$$

Thus, we see that $\vec 0 \in H$ by the closure of $H$.

The result above gives us a relatively easy way to check if a subset $T$ of a Vector space is a subspace: determine whether or not the zero vector is in $T$. If the zero vector is NOT in $T$, then $T$ is NOT a vector space.Nonetheless, if the zero vector is in $T$, it is still nessary to check to see if $H$ satisfies the closure properties for all elements in $H$.

## The Column Space and Null Space as Subspaces

Generally when the term subspace is used, it is typically associated with the concepts of the $\textbf{Null Space}$ of a matrix and the $\textbf{Column Space}$ of a matrix. This is quite natural since both of these sets are actually $\textit{subspaces of different vector spaces}$. This is elaborated below.

Let $A$ be $m \times n$ matrix. Then, the $\textit{Column Space}$ of $A$ is the set of all vectors $\vec b \in \mathbb{R^m}$ such that

$$A \vec x = \vec y.$$

The Column Space of $A$ is denoted by $Col (A)$.

We can say that the $\textit{Column Space}$ of a matrix $A$ is the set of all vectors $\vec b$ that can be generated as a $\textit{linear combination}$ of the columns of $A$. Next, we look at the definition of the $\textbf{Null Space}$ of $A$.

The $\textbf{Null Space}$ of $A$ is given by the set

$$\{\vec x \in \mathbb{R^{n}} : A
\vec x = \vec 0\}.$$

The Null Space of $A$ is denoted by $Nul (A)$.
From this defintion, we can see that the $\textbf{Null Space}$ of $A$ is the $\textit{solutions to the homongenous system}$ $A \vec x = \vec 0$.

But why is it that these particular sets are considered vector subspaces? Firstly, it is clear that both the $Col (A)$ and $Nul (A)$ are $\textit{subsets}$ of particular vector vspaces (those being $\mathbb{R^{m}}$ and $\mathbb{R^{n}}$ respectively). However, the properties of these sets being non-empty and closed under vector addition and scalar multiplication are not quite as obvious. The former property is considered next.

# The Existence of Elements in $Col (A)$ and $Nul (A)$

For the $Col (A)$ to be non-empty would mean that at least one vector in $\mathbb{R^{m}}$ must be a linear combination of the columns of $A$. Note however, that the zero vector of $\mathbb{R^{m}}$ is $\textit{always}$ a linear combination of the columns of an arbitary matrix $A$ (we need only to set all the weights $c_i$ equal to 0). That is if $\vec a_1$, $\vec a_2$, ... $\vec a_i$ are the columns of $A$, then

$$0 \vec a_1 + 0 \vec a_2 + \cdots + 0 \vec a_i = \vec 0.$$

Thus, $\vec 0 \in \mathbb{R^{m}}$ is a linear combination of the columns of $A$. Consequently, the column space of $A$ is non-empty.

Now consider the $Nul (A)$. Since there is always the trivial solution $\vec 0 \in \mathbb{R^{n}}$ to the homogenous equation $A \vec x = \vec 0$, we see that the $\textbf{Null Space}$ of $A$ must be non-empty. That is since

$$A (\vec 0) = \vec 0 $$

there exists at least one element ($\vec 0 \in \mathbb{R^{n}}$) in the $Nul (A)$. Thus, the $Nul A$ is non-empty.

We need only to show that the $\textit{Col (A)}$ and the $\textit{Nul (A)}$ are closed under vector addition and scalar multiplication. We will prove this below.

# Proofs of Closure Properties of $Col (A)$ and $Nul (A)$

$\textit{Proof}$: (Closure of the $Col(A)$)

Suppose that $A$ is an $m \times n$ matrix. Let $\vec b \in Col (A)$, and $c$ be a scalar. Since $\vec b$ is in $Col (A)$, there exists an $\vec x$ that is an element of $\mathbb{R^{n}}$ such that

$$A \vec x = \vec b$$.

We see then that

$$c \cdot A \vec x = c \cdot \vec b .$$

By the properties of matrices, we see that we could write

$$A (c \vec x) = c \vec b.$$

The eqation above indictates that vector $c \vec b$ is an element of $Col (A)$. This proves that the $Col (A)$ is closed under scalar multiplication.

Now, suppose that vector $\vec c$ is an element of the $Col (A)$. This would mean that there exists a vector $\vec y \in \mathbb{R^{n}}$ such that

$$A \vec y = \vec c$$.

By the property of matices, we see then that

$$ A \vec x + A \vec y = \vec b + \vec c. \\ A(\vec x + \vec y) = \vec b + \vec c. $$

The equation above implies that the vector $\vec b + \vec c$ is in $Col (A)$. This proves that $Col (A)$ is closed under vector addition.

$\textit{Proof}:$ (Closure of $Nul(A)$)

Let $A$ be an $m \times n$, and $c$ is scalar. Suppose that $x \in Nul(A)$. This would mean that

$$A \vec x = \vec 0.$$

By the properties of matrices, we see that

$$c \cdot A = c \cdot \vec 0. \\ A(c \vec x) = \vec 0 .$$

The equation agove indicates that vector $c \vec x$ is in the $Nul (A)$. This proves that $Nul (A)$ is closed under scalar multiplication.

Now, suppose that $\vec y \in Nul (A)$. This would mean that

$$A \vec y = \vec 0.$$

We see then by the properties of matrices that

$$A \vec x + A \vec y = \vec 0 + \vec 0. \\ A \vec x + A \vec y = \vec 0. \\ A( \vec x + \vec y) = \vec 0.$$

The equation above indicates that vector $\vec x + \vec y$ is in the $Nul(A)$. This proves that the $Nul(A)$ is closed under vector addition.

$\textbf{Exam Question}$

Let $A$ be an $m \times n$. Explain why the $\textbf{Row Space}$ of $A$ is actually a subspace $\mathbb{R^{n}}$.

$$\text{Citations}$$

Cheung, K. (n.d.). Two additional vector spaces associated with a matrix. Retrieved from https://people.math.carleton.ca/~kcheung/math/notes/MATH1107/wk09/09_column_space_row_space.html.

Ikenaga, B. (n.d.). Row Space, Column Space, and Null Space. Retrieved from http://sites.millersville.edu/bikenaga/linear-algebra/matrix-subspaces/matrix-subspaces.html.

Lay, D. C. (2006). Linear Algebra and its Applications. Reading: Pearson AddisonWesley Publishing Company

Row Space, Column Space, and Rank-Nullity Theorem. (n.d.). Retrieved from https://www.math.upenn.edu/~moose/240S2013/slides7-22.pdf.