Vector Subspaces of Homogenous Systems of Rn

# Vector Subspaces of Homogenous Systems of Rn

We will now look at some more vector subspaces and verify that they are in fact subspaces of another vector space. Instead of verify axioms 9 and 10 however, we will instead utilize the following lemma (proven on the Vector Subspaces page) to show that these sets are subspaces of a larger vector space:

 Lemma: A nonempty subset $U$ of an $\mathbb{F}$-vector space $V$ is a subspace of $V$ if and only if for all $a, b \in \mathbb{F}$ (where $\mathbb{F} = \mathbb{R}$ or $\mathbb{F} = \mathbb{C}$) and for all vectors $\mathbf{x}, \mathbf{y} \in U$, then $( a\mathbf{x} + b\mathbf{y} ) \in U$.

## Example 1

Let $V = \mathbb{R}^4$ be a vector space. Is $U = \{ (x_1, x_2, x_3, x_4 ) : x_1 - 2x_2 + x_3 + 4x_4 = 0 \}$ a subspace of $V$?

$U$ is a vector subspace of $V$ if it satisfies all of vector space axioms. We will need to check that this space is closed under addition and scalar multiplication which we will show using the lemma from above.

First let $a, b \in \mathbb{F}$ and let $\mathbf{x}, \mathbf{y} \in U$ such that $\mathbf{x} = (x_1, x_2, x_3, x_4)$ and $\mathbf{y} = (y_1, y_2, y_3, y_4)$. From the lemma above, we want to show that $(a \mathbf{x} + b \mathbf{y} ) \in U$

Expanding $a\mathbf{x} + b\mathbf{y}$ we get:

(1)
\begin{align} a_1\mathbf{x} + a_2\mathbf{y} = a(x_1, x_2, x_3, x_4) + b(y_1, y_2, y_3, y_4) \\ = (ax_1, ax_2, ax_3, ax_4) + (by_1, by_2, by_3, by_4) \\ = (ax_1 + by_1, ax_2 + by_2, ax_3 + by_3, ax_4 + by_4) \end{align}

Now we want to check to see if if this vector satisfies the condition of our set, that is if $(ax_1 + by_1) - 2(ax_2 + by_2) + (ax_3 + by_3) + 4(ax_4 + by_4) = 0$. We note that:

(2)
\begin{align} (ax_1 + by_1) - 2(ax_2 + by_2) + (ax_3 + by_3) + 4(ax_4 + by_4) \\ = ax_1 - 2ax_2 + ax_3 + 4x_4 + by_1 -2by_2 + by_3 + 4by_4 \\ = a\underbrace{(x_1 - 2x_2 + x_3 + 4x_4)}_{\in U} + b\underbrace{(y_1 - 2y_2 + y_3 + 4y_4)}_{\in U} \\ = a (0) + b(0) \\ = 0 \end{align}

Therefore we conclude that $U$ is a vector subspace of $V$, that is $U \subset V$.

## Example 2

Let $V = \mathbb{R}^4$ and let $U = \{ (x_1, x_2, x_3, x_4) : x_1 + x_2 - x_3 + x_4 = 0 \: \mathrm{and} \: 3x_1 -2x_2 + 4x_4 = 0 \}$. If $U$ a subspace of $V$?

We first let $\mathbf{x}, \mathbf{y} \in U$ such that $x = (x_1, x_2, x_3, x_4)$ and $y = (y_1, y_2, y_3, y_4)$. Now let $a, b \in \mathbb{R}$. We want to show that $a\mathbf{x} + b\mathbf{y} \in U$, that is we want to show that $(ax_1 + by_1, ax_2 + by_2, ax_3 + by_3, ax_4 + by_4) \in U$. We will have to show this vector satisfies the two conditions defining the set $U$.

For the first condition we need to show that $(ax_1 + by_1) + (ax_2 + by_2) - (ax_3 + by_3) + (ax_4 + by_4) = 0$.

(3)
\begin{align} (ax_1 + by_1) + (ax_2 + by_2) - (ax_3 + by_3) + (ax_4 + by_4) \\ = a\underbrace{(x_1 + x_2 - x_3 + x_4)}_{=0} + b\underbrace{(y_1 + y_2 - y_3 + y_4)}_{=0} \\ = a \cdot 0 + b \cdot 0 \\ = 0 \end{align}

Now for the second condition we need to show that $3(ax_1 + by_1) - 2(ax_2 + by_2) + 4(ax_4 + by_4) = 0$.

(4)
\begin{align} 3(ax_1 + by_1) - 2(ax_2 + by_2) + 4(ax_4 + by_4) \\ = a\underbrace{(3x_1 -2x_2 + 4x_4)}{=0} +b\underbrace{(3y_1 - 2y_2 + 4y_4)}_{=0} \\ = a \cdot 0 + b \cdot 0 \\ = 0 \end{align}

Therefore $U$ is a vector subspace of $V$.

## Example 3

We will now look at a subset of $V = \mathbb{R}^3$ that is not a vector subspace and is also not a homogenous linear equation. For example, consider the set $U = \{ (x_1, x_2, x_3) : x_1 + x_2 + x_3 = 4 \}$. We can think of $U$ as the set of all coordinates $(x_1, x_2, x_3)$ that satisfy the linear equation $x_1 + x_2 + x_3 = 4$. This collection of points form a plane. But notice that the the zero list $(0, 0, 0) \not \in U$ since $0 + 0 + 0 \neq 4$, and so this plane does not pass through the origin. Recall that if $0 \not \in U$ then $U$ cannot be a subspace of $V$.

There's also another reason as to why $U$ is not a subspace of $V$}]. We note that the pair of coordinates [[$(1, 1, 2) \in U$. And so any scalar multiple $k(1, 1, 2)$ should be contained within $U$ since $U$ needs to be closed under scalar multiplication to be a vector subspace of $V$. If $k = 2$, then the point $2(1,1,2) = (2,2,4) \not \in U$ since $2 + 2 + 4 \neq 4$. Therefore $U$ is not a vector subspace of $\mathbb{R}^3$.

## Example 4

At minimum, how many vector subspaces must a vector space $V$ have?

We note that the set containing only the zero element, that is $U_{0} = \{ 0 \}$ and the entire vector space set $U_{V} = V$ are vector subspaces of $V$ that satisfy all the axioms. We will show this now.

Clearly $U_{V} = V \subseteq V$ is a vector space since $V$ is a vector space. We now need to show $U_{0} = \{ 0 \}$ is a vector subspace.

Let $a, b \in \mathbb{F}$ and let $\mathbf{x}, \mathbf{y}, \in U_{0}$. Clearly $x = 0$ and $y = 0$ since $U_{0}$ contains only one element, namely $0$. Therefore:

(5)
\begin{align} a\mathbf{x} + b\mathbf{y} = a \cdot 0 + b \cdot 0 = 0 \in U \end{align}

And so $U_{0}$ is a vector subspace of $V$.

We conclude that thus every vector space $V$ has at minimum $2$ vector subspaces.

## Example 5

List all of the vector subspaces for $\mathbb{R}^3$.

We note that all vector subspaces must contain the zero element $(0, 0, 0) \in \mathbb{R}^3$, namely, we must have each vector subspace set contain the zero element. We verify that all of the vector subspaces for $\mathbb{R}^3$ is the set $\mathbb{R}^3$ itself, $\{ 0 \}$, any line $L$ that passes through the origin, and any plane $\Pi$ that also passes through the origin.