Vector Subspaces Examples 1
 Table of Contents

# Vector Subspaces Examples 1

Recall from the Vector Subspaces page that a subset $U$ of the subspace $V$ is said to be a vector subspace of $V$ if $U$ contains the zero vector of $V$ and is closed under both addition and scalar multiplication defined on $V$.

We also saw an important lemma which says that if $V$ is a vector space, then a subset $U$ of $V$ is a subspace of $V$ if and only if $(ax + by) \in U$ for all $x, y \in U$ and for all $a, b \in \mathbb{F}$.

We will now look at some examples and non-examples of vector subspaces.

## Example 1

Explain why $U = \{ (x_1, x_2, x_3, x_4) : x_1 = 2x_2 + 2 \}$ is not a subspace of $\mathbb{F}^4$.

Suppose that $x, y \in U$ where $x = (x_1, x_2, x_3, x_4)$ and $y = (y_1, y_2, y_3, y_4)$. Then we have that $x_1 = 2x_2 + 2$ and $y_1 = 2y_2 + 2$.

Now consider the vector $x + y = (x_1 + y_1, x_2 + y_2, x_3 + y_3, x_4 + y_4)$. If $(x + y) \in U$ then $x_1 + y_1 = 2x_1 + 2y_2 + 2$. However, we instead have that $x_1 + y_1 = 2x_2 + 2y_2 + 4$. Therefore $(x + y) \not \in U$ and so $U$ is not closed under addition and is hence not a subspace of $\mathbb{F}^4$.

## Example 2

Explain why the set $U$ of differentiable real-valued functions on the interval $[0, 1]$ where $f'(1) = 1$ is not a subspace of the set of continuous real-valued functions on $[0, 1]$.

We note that the the zero function $f(x) = 0$ is a differentiable real-valued function on the closed interval $[0, 1]$. We also note that $f'(x) = 0$. So $f'(1) = 0$. Therefore the zero function $f(x) = 0 \not \in U$. Since $U$ does not contain the zero function we have that $U$ is not a subspace of the set of continuous real-valued functions on $[0, 1]$.

## Example 3

Determine whether the subspace $U = \{ (x_1, x_2, x_3) : x_1 + x_2 + x_3 = 0$ of $\mathbb{F}^3$ is a subspace of $\mathbb{F}^3$.

Clearly the zero vector $0 = (0, 0, 0) \in U$ since $0 + 0 + 0 = 0$.

Let $x, y \in U$ and let $a, b \in \mathbb{F}$. Then we have that:

(1)
\begin{align} \quad ax + by = a(x_1, x_2, x_3) + b(y_1, y_2, y_3) = (ax_1 + by_1, ax_2+by_2, ax_3+by_3) \end{align}

Now we have that:

(2)
\begin{align} \quad (ax_1 + by_1) + (ax_2 + by_2) + (ax_3 + by_3) = a\underbrace{(x_1 + x_2 + x_3)}_{=0} + b\underbrace{(y_1 + y_2 + y_3)}_{=0} = 0 \end{align}

Therefore $(ax + by) \in U$, so indeed, $U$ is a subspace of $\mathbb{F}^3$.

## Example 4

Determine whether the subspace $U = \{ (x_1, x_2, x_3) : x_1x_2x_3 = 0$ of $\mathbb{F}^3$ is a subspace of $\mathbb{F}^3$.

Consider the vectors $x = (1, 0, 0)$ and $y = (0, 1, 1)$. Clearly $x, y \in U$ since $1 * 0 * 0 = 0$ and $0 * 1 * 1 = 0$.

Now consider the sum:

(3)
\begin{align} \quad x + y = (1, 0, 0) + (0, 1, 1) = (1, 1, 1) \end{align}

Note that $(x + y) \not \in U$ since $(x_1 + y_1)(x_2 + y_2)(x_3 + y_3) = 1 * 1 * 1 = 1 \neq 0$. Therefore $U$ is not closed under addition and is hence not a subspace of $\mathbb{F}^3$.

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