Vector Subspaces Examples 1

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)
\begin{align} a\mathbf{x} + b\mathbf{y} \in U \end{align}

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

(2)
\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:

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

(4)
\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 $3(ax_1 + by_1) - 2(ax_2 + by_2) + 4(ax_4 + by_4) = 0$.

(5)
\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

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)
\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 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.