Basis of a Vector Space Examples 1

# Basis of a Vector Space Examples 1

Recall from the Basis of a Vector Space that if $V$ is a finite-dimensional vector space, then a set of vectors $\{ v_1, v_2, ..., v_n \}$ is said to be a basis of $V$ if $\{ v_1, v_2, ..., v_n \}$ spans $V$ and $\{ v_1, v_2, ..., v_n \}$ is a linearly independent set of vectors in $V$.

We will now look at some problems regarding bases of vector spaces.

## Example 1

Let $U = \{ (x_1, x_2, x_3) \in \mathbb{R}^3 : x_1 = 2x_2 \}$ be a subspace of $\mathbb{R}^3$. Find a basis of $U$.

If $U = \{ (x_1, x_2, x_3) \in \mathbb{R}^3 : x_1 = 2x_2 \}$ then $U = \{ (2x_2, x_2, x_3) \in \mathbb{R}^3 : x_2, x_3 \in \mathbb{R} \}$, and so one such basis of $U$ is:

(1)
\begin{align} \quad \{ (2, 1, 0), (0, 0, 1) \} \end{align}

To verify this set of vectors is a basis of $U$ we must show that $U = \mathrm{span} ((2, 1, 0), (0, 0, 1))$ and that $\{ (2, 1, 0), (0, 0, 1) \}$ is a linearly independent set of vectors in $V$. Let $x = (x_1, x_2, x_3) \in U$. Then we have that:

(2)
\begin{align} \quad x = (x_1, x_2, x_3) = (2x_2, x_2, x_3) = x_2(2, 1, 0) + x_3(0, 0, 1) \end{align}

So $U = \mathrm{span} ((2, 1, 0), (0, 0, 1))$. Now consider the following vector equation for $a_1, a_2 \in \mathbb{F}$:

(3)
\begin{align} \quad a_1(2, 1, 0) + a_2(0, 0, 1) = 0 \\ \quad (2a_1, a_1, 0) + (0, 0, a_2) = 0 \\ \quad (2a_1, a_1, a_2) = (0, 0, 0) \end{align}

The equation above implies that $2a_1 = 0$, $a_1 = 0$, and $a_2 = 0$, so $a_1 = a_2 = 0$ and $\{ (2, 1, 0), (0, 0, 1) \}$ is a linearly independent set of vectors in $\mathbb{R}^3$.

Thus $\{ (2, 1, 0), (0, 0, 1) \}$ is a basis of $U$.

## Example 2

Let $U = \{ (x_1, x_2, x_3, x_4, x_5) \in \mathbb{R}^5 : x_1 = 2x_2, x_3 = 4x_4, x_5 = 2x_4 \}$ be a subspace of $\mathbb{R}^5$. Find a basis for $U$.

We can rewrite the subspace $U$ as:

(4)
\begin{align} \quad U = \{ (2x_2, x_2, 4x_4, x_4, 2x_4) : x_2, x_4 \in \mathbb{R} \} \end{align}

Therefore we have that $\{ (2, 1, 0, 0, 0), (0, 0, 4, 1, 2) \}$ is a basis of $U$. To verify this, let $x = (x_1, x_2, x_3, x_4, x_5) \in \mathbb{R}^5$. Then we have that:

(5)
\begin{align} \quad x = (x_1, x_2, x_3, x_4, x_5) = (2x_2, x_2, 4x_4, x_4, 2x_4) = x_2(2, 1, 0, 0, 0) + x_4(0, 0, 4, 1, 2) \end{align}

So $U = \mathrm{span} ((2, 1, 0, 0, 0), (0, 0, 4, 1, 2))$. Now consider the following vector equation for $a_1, a_2 \in \mathbb{R}$:

(6)
\begin{align} \quad a_1(2, 1, 0, 0, 0) + a_2(0, 0, 4, 1, 2) = 0 \\ \quad (2a_1, a_1, 0, 0, 0) + (0, 0, 4a_2, a_2, 2a_2) = 0 \\ \quad (2a_1, a_1, 4a_2, a_2, 2a_2) = (0, 0, 0, 0, 0) \end{align}

The equation above implies that:

(7)
Thus $a_1 = a_2 = 0$ and so $\{ (2, 1, 0, 0, 0), (0, 0, 4, 1, 2) \}$ is a linearly independent set of vectors in $U$. Thus $\{ (2, 1, 0, 0, 0), (0, 0, 4, 1, 2) \}$ is a basis of $U$.