# Basis of a Vector Space Examples 2

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 more problems regarding bases of vector spaces.

## Example 1

**Consider the vector space $\wp_2 (\mathbb{R})$ of polynomials of degree less than or equal to $2$. Determine whether or not there exists a basis of $\wp_2 (\mathbb{R})$ of three polynomials $\{ p_0(x), p_1(x), p_2(x) \}$ for which none of these polynomials is of degree $1$.**

Consider the set of polynomials $\{ 1, x + x^2, x – x^2 \}$. Let $p(x) = a_0 + a_1x + a_2x^2 \in \wp_2 (\mathbb{R})$. Then for scalars $b_0, b_1, b_2 \in \mathbb{F}$ we will consider the following equation:

(1)

Thus we let:

(2)

The third equation can be rewritten as $b_2 = a_2 + b_1$, and plugging this into the second equation gives us that $2b_1 + a_2 = a_1$ so $b_1 = \frac{a_1 – a_2}{2}$. Plugging this back into the third equation gives us $b_2 = a_2 + \frac{a_1 – a_2}{2} = \frac{2a_2 + a_1 – a_2}{2} = \frac{a_1 + a_2}{2}$, and so we have that:

(3)

Therefore $\{ 1, x – x^2, x + x^2 \}$ spans $\wp_2 (\mathbb{R})$.

Now for $c_0, c_1, c_2 \in \mathbb{F}$, consider the following vector equation:

(4)

We thus get that:

(5)

The third equation implies that $c_1 = c_2$ and plugging this into the second equation gives us $2c_1 = 0$ so $c_1 = 0$ and thus $c_2 = 0$. The first equation says that $c_0 = 0$, so indeed, $\{ 1, x – x^2, x + x^2 \}$ is a linearly independent set of vectors.

Therefore $\{ 1, x – x^2, x + x^2 \}$ is a basis of $\wp_2 (\mathbb{R})$ and none of the polynomials in this basis have degree $1$.

## Example 2

**Let $U$ and $W$ be subspaces of the finite-dimensional vector space $V$ such that $V = U \oplus W$. Let $\{ u_1, u_2, …, u_n \}$ and $\{ w_1, w_2, …, w_m \}$ be bases of $U$ and $W$ respectively. Show that then $\{ u_1, u_2, …, u_n, w_1, w_2, …, w_m \}$ is a basis of $V$.**

If $V = U \oplus W$ then for every vector $v \in V$ we have that $v$ can be written uniquely as $v = u + w$ where $u \in U$ and $w \in W$. Since $\{ u_1, u_2, …, u_n \}$ is a basis of $U$ we have that for $u \in U$ there exists scalars $a_1, a_2, …, a_n \in \mathbb{F}$ such that $u = a_1u_1 + a_2u_2 + … + a_nu_n$. Furthermore, since $\{ w_1, w_2, …, w_m \}$ is a basis of $W$ we have that for $w \in W$ there exists scalars $b_1, b_2, …, b_m \in \mathbb{F}$ such that $w = b_1w_1 + b_2w_2 + … + b_nw_n$. Therefore, for any vector $v \in V$ we have that:

(6)

Therefore $\{ u_1, u_2, …, u_n, w_1, w_2, …, w_m \}$ spans $V$. Now consider the following vector equation:

(7)

We thus have that $u = -w$ which implies that $u, w \in U \cap W$. However, $U \cap W = \{ 0 \}$ since $V = U \oplus W$, and so $u = 0$ and $w = 0$, that is, $a_1u_1 + a_2u_2 + … + a_nu_n = 0$ and $b_1w_1 + b_2w_2 + … + b_mw_m = 0$. But $\{ u_1, u_2, …, u_n \}$ and $\{ w_1, w_2, …, w_m \}$ are linearly independent sets which imply that $a_1 = a_2 = … = a_n = 0$ and $b_1 = b_2 = … = b_n = 0$.

Therefore $\{ u_1, u_2, …, u_n, w_1, w_2, …, w_m \}$ is a linearly independent set.

Since $\{ u_1, u_2, …, u_n, w_1, w_2, …, w_m \}$ spans $V$ and is linearly independent in $V$, we have by definition that $\{ u_1, u_2, …, u_n, w_1, w_2, …, w_m \}$ is a basis of $V$.