Span of a Set of Vectors

Be sure to review what a linear combination of a vector is before continuing on this page.

Definition: Suppose that $\{ v_1, v_2, ..., v_n \}$ is a set of vectors of the vector space $V$. Then the Span of the Set $\{ v_1, v_2, ..., v_n \}$ denoted $\mathrm{span} (v_1, v_2, ..., v_n)$ and $\sum_{i=1}^{n} \mathbb{F}v_i$ is the set of all linear combinations of the vectors in $\{ v_1, v_2, ..., v_n \}$, that is, for any scalars $k_1, k_2, ..., k_n$, $(k_1v_1 + k_2v_2 + ... + k_nv_n ) \in \mathrm{span}(V)$.

Let's first look at an example. Suppose that we have a set of scalars $V = \{ v_1, v_2 \}$ where $v_1 = (1, 2)$ and $v_2 = (3, 5)$. We thus note that $\mathrm{span}(V) = c_1(1, 2) + c_2(3, 5)$. For example, suppose we choose $k_1 = 1$ and $k_2 = 2$, and thus, $1(1, 2) + 2(3, 5) = (1, 2) + (6, 10) = (7, 12)$. Therefore, $(7, 12) \in \mathrm{span}(V)$.

Note: Consider the zero vector space $\{ 0 \}$, i.e., the vector space that contains only the zero vector. We have show that this set is in fact a vector space, and by convention we say that $\mathrm{span} \{ 0 \} = \emptyset$, that is, the the set of all linear combinations of the zero vector is the empty set.

Example 1

Let $V = \{ (2, 3), (1, 2) \}$. Show whether or not the vector $(19, 3) \in \mathrm{span}(V)$.

By the definition of a vector existing within the span of $V$, we must find scalars $c_1$ and $c_2$ such that:

(1)

\begin{equation} c_1(2, 3) + c_2(1, 2) = (19, 3) \end{equation}

If we cannot such scalars, then $(19, 3) \not\in \mathrm{span}(V)$ and similarly, if we can find such scalars, then $(19, 3) \in \mathrm{span}(V)$. We thus obtain the following system of linear equations:

(2)

\begin{align} 2c_1 + c_2 =19 \\ 3c_1 + 2c_2 = 3 \end{align}

When we reduce the augment matrix of this system to reduced row echelon form, we get that:

(3)

\begin{align} c_1 + 0c_2 = 35 \\ 0c_1 + c_2 = -51 \end{align}

Therefore, we have found a set of scalars $c_1, c_2$ which satisfy our condition, and therefore, $(19, 3) \in \mathrm{span} (V)$ since $35(2, 3) + -51(1, 2) = (19, 3)$.

Example 2

Let $V = \{ (-1, 4), (3, -5) \}$. Show whether or not the vector $(10, 1) \in \mathrm{span}(V)$.

For the vector $S (10, 1)$ to be in the span if $V$, we must show that $(10, 1)$ is a linear combination of the vectors in $V$ so that there exists scalars $c_1, c_2$ such that $(10, 1) = c_1(-1, 4) + c_2(3, -5)$. We thus get the following system of equations:

(4)

\begin{align} -c_1 + 3c_2 = 10 \\ 4c_1 - 5c_2 = 1 \\ \end{align}

When we reduce this system to RREF, we obtain that:

(5)

\begin{align} c_1 + 0c_2 = \frac{53}{7} \\ 0c_1 + c_2 = \frac{41}{7} \end{align}

Therefore there exists scalars $c_1$ and $c_2$ that make $(10, 1)$ a linear combination of the vectors in $V$ so $(10, 1) \in \mathrm{span}(V)$.

Example 3

Let $V = \{ (0, a), (0, b) \}$ where $a \neq b$ and $a, b \in \mathbb{R}$. Show that any vector in the form $(c, 0) \not \in \mathrm{span}(V)$ where $c \neq 0$.

For $(c, 0) \in \mathrm{span}(V)$, $(c, 0)$ must be a linear combination of the vectors in $V$, that is there must exist some scalars $c_1, c_2$ such that $(c, 0) = c_1(0, a) + c_2(0, b)$. However, note that then:

(6)

\begin{align} (c, 0) = (0c_1 + 0c_2, ac_1 + bc_2) \\ (c, 0) = (0, ac_1 + bc_2) \end{align}

From this we see that $c = 0$. But in our problem, we said that $c \neq 0$ so $(c, 0) \not \in \mathrm{span}(V)$.

Example 4

Show that if $A$ is an $n \times n$ so that $Ax = b$ is consistent for every $n \times 1$ matrix $b$, then the column vectors of $A$ span $\mathbb{R}^n$.

Let $C = \{ C_1, C_2, ..., C_n \}$ be the set of column vectors of $A$ and let $b$ be any $n \times 1$ vector in $\mathbb{R}^n$. We note that $Ax = b$ is consistent, so there exists a vector $x = \begin{bmatrix} x_1\\ x_2\\ \vdots\\ x_n \end{bmatrix}$ such that $x_1C_1 + x_2C_2 + ... + x_nC_n = b$ So $b$ is a linear combination of the column vectors of $A$ so $C$ spans $\mathbb{R}^n$.