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)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)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)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)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)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)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 $U$.