# Vector Subspaces Examples 1

We will now look at some more vector subspaces and verify that they are in fact subspaces of another vector space. Recall the following theorem and lemma regarding vector subspaces:

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 $( a\mathbf{x} + b\mathbf{y} ) \in U$. |

It is important to keep these in mind when verifying that a subset $U$ is a vector subspace.

## 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 $V$?**

We should note that $U$ 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.

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 $\mathbf{y} = (y_1, y_2, y_3, y_4)$. Applying the lemma above, we want to show that:

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

(2)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:

(3)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 $(ax_1 + by_1) + (ax_2 + by_2) - (ax_3 + by_3) + (ax_4 + by_4) = 0$.

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

(5)Therefore $U$ is a vector subspace of $V$.

## Example 3

**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:

(6)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 4

**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.