Spanning Set of a Vector Space Examples 1

# Spanning Set of a Vector Space Examples 1

Recall from the Spanning Set of a Vector Space page that if for every vector $v \in V$ we have that $v$ can be written as a linear combination of the set of vectors in the set $\{ v_1, v_2, ..., v_n \}$ then we say that the set of vectors $\{ v_1, v_2, ..., v_n \}$ spans $V$ and write $V = \mathrm{span} (v_1, v_2, ..., v_n)$.

We will now look at some examples of spanning sets of vectors.

## Example 1

Prove that if $\{ v_1, v_2, ..., v_n \}$ is a set of vectors from the vector space $V$ then $\mathrm{span} (v_1, v_2, ..., v_n)$ is a subspace of $V$.

Let $\{ v_1, v_2, ..., v_n \}$ be a set of vectors in $V$. To show that $\mathrm{span} (v_1, v_2, ..., v_n)$ is a subspace of $V$ we must show that $0 \in \mathrm{span} (v_1, v_2, ..., v_n)$, and that $\mathrm{span} (v_1, v_2, ..., v_n)$ is closed under addition and scalar multiplication.

Clearly $0 \in \mathrm{span} (v_1, v_2, ..., v_n)$ since we can write:

(1)
\begin{align} \quad 0 = 0v_1 + 0v_2 + ... + 0v_n \end{align}

Now let $u, v \in \mathrm{span} (v_1, v_2, ..., v_n)$. Then for some set of scalars $a_1, a_2, ..., a_n, b_1, b_2, ..., b_n \in \mathbb{F}$ we have that $u = a_1v_1 + a_2u_2 + ... + a_nu_n$ and $w = b_1v_1 + b_2v_2 + ... + b_nv_n$ and so for $c,d \in \mathbb{F}$ we have that:

(2)
\begin{align} \quad cu + dv = c(a_1v_1 + a_2v_2 + ... + a_nv_n) + d(b_1v_1 + b_2v_2 + ... + b_nv_n) \\ \quad cu + dv = (ca_1 + db_1)v_1 + (ca_2 + db_2)v_2 + ... + (ca_n + db_n)v_n \end{align}

Therefore $(cu + dv) \in \mathrm{span} (v_1, v_2, ..., v_n)$ so $\mathrm{span} (v_1, v_2, ..., v_n)$ is closed under addition and scalar multiplication.

## Example 2

Show that if the set of vectors $\{ v_1, v_2, v_3, v_4 \}$ spans $V$ then $\{ v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4 \}$ spans $V$.

Let $v \in \mathrm{span} (v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4 )$. Then for constants $a_1, a_2, a_3, a_4 \in \mathbb{F}$ we have that:

(3)
\begin{align} \quad v = a_1(v_1 - v_2) + a_2(v_2 - v_3) + a_3(v_3 - v_4) + a_4v_4 \\ \quad v = a_1v_1 - a_1v_2 + a_2v_2 - a_2v_3 + a_3v_3 - a_3v_4 + a_4v_4 \\ \quad v = a_1v_1 + (a_2 - a_1)v_2 + (a_3 - a_2)v_3 + (a_4 - a_3)v_4 \end{align}

Therefore $v \in \mathrm{span} (v_1, v_2, v_3, v_4) = V$, so $\mathrm{span} (v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4) \subseteq V$.

Now let $v \in V = \mathrm{span} (v_1, v_2, v_3, v_4)$. We want to show that for some set of scalars $a_1, a_2, a_3, a_4 \in \mathbb{F}$ that we have that:

(4)
\begin{align} \quad v = a_1(v_1 - v_2) + a_2(v_2 - v_3) + a_3(v_3 - v_4) + a_4v_4 \\ \quad v = a_1v_1 + (a_2 - a_1)v_2 + (a_3 - a_2)v_3 + (a_4 - a_3)v_4 \end{align}

Since $v \in \mathrm{span} (v_1, v_2, v_3, v_4)$ then for $b_1, b_2, b_3, b_4 \in \mathbb{F}$ we have that:

(5)
\begin{align} \quad v = b_1v_1 + b_2v_2 + b_3v_3 + b_4v_4 \end{align}

Thus if we let $b_1 = a_1$, $b_2 = a_2 - a_1$, $b_3 = a_3 - a_2$ and $b_4 = a_4 - a_3$ then:

(6)
\begin{align} \quad v = a_1v_1 + (a_2 - a_1)v_2 + (a_3 - a_2)v_3 + (a_4 - a_3)v_4 \end{align}

So therefore $a_1 = b_1$, $a_2 = b_2 + a_1$, $a_3 = b_3 + a_2$, and $a_4 = b_4 + a_3$, so $v \in \mathrm{span} (v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4)$, so $V \subseteq \mathrm{span} (v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4)$.