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)

\begin{align} \quad a_0 + a_1x + a_2x^2 = b_0 + b_1(x - x^2) + b_2(x + x^2) \\ \quad a_0 + a_1x + a_2x^2 = b_0 + (b_1 + b_2)x + (b_2 - b_1)x^2 \end{align}

Thus we let:

(2)

\begin{align} \quad b_0 = a_0 \\ \quad b_1 + b_2 = a_1 \\ \quad b_2 - b_1 = a_2 \end{align}

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)

\begin{align} \quad b_0 = a_0 \\ \quad b_1 = \frac{a_1 - a_2}{2} \\ \quad b_2 = \frac{a_1 + a_2}{2} \end{align}

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)

\begin{align} \quad c_0(1) + c_1(x - x^2) + c_2(x + x^2) = 0 \\ \quad c_0 + (c_1 + c_2)x + (c_2 - c_1)x^2 = 0 \end{align}

We thus get that:

(5)

\begin{align} \quad c_0 = 0 \\ \quad c_1 + c_2 = 0 \\ \quad c_2 - c_1 = 0 \end{align}

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)

\begin{align} \quad v = u + w \\ \quad v = a_1u_1 + a_2u_2 + ... + a_nu_n + b_1w_1 + b_2w_2 + ... + b_mw_m \end{align}

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

(7)

\begin{align} \quad a_1u_1 + a_2u_2 + ... + a_nu_n + b_1w_1 + b_2w_2 + ... + b_mw_m = 0 \\ \quad \underbrace{a_1u_1 + a_2u_2 + ... + a_nu_n}_{= u \in U} = \underbrace{-b_1w_1 - b_2w_2 - ... - b_mw_m}_{=w \in W} \end{align}

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 $V$.