4 Linear algebra 4.6 Linear independence 4.8 Bases

4.7 Spanning sequences

4.7.1 Definition of span

Definition 4.7.1.

Let $V$ be an $\mathbb{F}$ -vector space and $\mathbf{v}_{1},\ldots,\mathbf{v}_{n}\in V$ . The span of $\mathbf{v}_{1},\ldots,\mathbf{v}_{n}$ , written $\operatorname{span}(\mathbf{v}_{1},\ldots,\mathbf{v}_{n})$ is the set of all linear combinations of $\mathbf{v}_{1},\ldots,\mathbf{v}_{n}$ , so

\operatorname{span}(\mathbf{v}_{1},\ldots,\mathbf{v}_{n})=\{\lambda_{1}\mathbf% {v}_{1}+\cdots+\lambda_{n}\mathbf{v}_{n}:\lambda_{1},\ldots,\lambda_{n}\in% \mathbb{F}\}.

For technical reasons we define the span of the empty sequence of vectors to be $\{\textbf{0}_{V}\}$ .

To understand the definition a bit better, let’s look at two simple special cases. The span of a single element $\mathbf{s}$ of an $\mathbb{F}$ -vector space $V$ is

\{\lambda\mathbf{s}:\lambda\in\mathbb{F}\},

since any linear combination of $\mathbf{s}$ is just a scalar multiple of $\mathbf{s}$ . The span of two elements $\mathbf{u},\mathbf{v}$ of $V$ is

\{a\mathbf{u}+b\mathbf{v}:a,b,\in\mathbb{F}\}.

4.7.2 Spans are subspaces

Proposition 4.7.1.

If $\mathbf{s}_{1},\ldots,\mathbf{s}_{n}$ are elements of a vector space $V$ then $\operatorname{span}(\mathbf{s}_{1},\ldots,\mathbf{s}_{n})$ is a subspace of $V$ .

Proof.

Write $S$ for $\operatorname{span}\{\mathbf{s}_{1},\ldots,\mathbf{s}_{n}\}$ . Recall that $S$ consists of every linear combination $\sum_{i=1}^{n}\lambda_{i}\mathbf{s}_{i}$ , where the $\lambda_{i}$ are scalars.

1.

$S$ contains the zero vector because it contains $\sum_{i=1}^{n}0\mathbf{s}_{i}$ , and each $0\mathbf{s}_{i}$ is the zero vector.
2.

$S$ is closed under addition because if $\sum_{i=1}^{n}\lambda_{i}\mathbf{s}_{i}$ and $\sum_{i=1}^{n}\mu_{i}\mathbf{s}_{i}$ are any two elements of $S$ then

$\sum_{i=1}^{n}\lambda_{i}\mathbf{s}_{i}+\sum_{i=1}^{n}\mu_{i}\mathbf{s}_{i}=% \sum_{i=1}^{n}(\lambda_{i}+\mu_{i})\mathbf{s}_{i}$

is in $S$ .
3.

$S$ is closed under scalar multiplication because if $\sum_{i=1}^{n}\lambda_{i}\mathbf{s}_{i}$ is in $S$ and $\lambda$ is a scalar then

$\lambda\sum_{i=1}^{n}\lambda_{i}\mathbf{s}_{i}=\sum_{i=1}^{n}(\lambda\lambda_{% i})\mathbf{s}_{i}$

is also in $S$ .

$S$ fulfils all three conditions in the Definition 4.4.1 of a subspace, so $S\leqslant V$ . ∎

4.7.3 Spanning sequences

Definition 4.7.2.

Elements $\mathbf{v}_{1},\ldots,\mathbf{v}_{n}$ of a vector space $V$ are a spanning sequence for $V$ if and only if $\operatorname{span}(\mathbf{v}_{1},\ldots,\mathbf{v}_{n})=V$ .

The term spanning set is also used.

We also say $\mathbf{v}_{1},\ldots,\mathbf{v}_{n}$ spans $V$ to mean that it is a spanning sequence.

Often deciding whether or not a sequence of vectors is a spanning sequence is equivalent to solving some linear equations.

Example 4.7.1.

If you want to check whether $\begin{pmatrix}1\\ 1\end{pmatrix}$ and $\begin{pmatrix}-1\\ 1\end{pmatrix}$ are a spanning sequence for $\mathbb{R}^{2}$ , what you need to do is to verify that for every $\begin{pmatrix}x\\ y\end{pmatrix}\in\mathbb{R}^{2}$ there are real numbers $\alpha$ and $\beta$ such that

\alpha\begin{pmatrix}1\\ 1\end{pmatrix}+\beta\begin{pmatrix}-1\\ 1\end{pmatrix}=\begin{pmatrix}x\\ y\end{pmatrix}.

In other words, you have to prove that for every $x,y\in\mathbb{R}$ the system of linear equations

	$\displaystyle\alpha+\beta$	$\displaystyle=x$
	$\displaystyle\alpha-\beta$	$\displaystyle=y$

has a solution. That’s easy in this case, because you can just notice that $\alpha=(x+y)/2,\beta=(x-y)/2$ is a solution, but for bigger and more complicated systems you can use the method of RREF.

Example 4.7.2.

$\begin{pmatrix}1\\ 1\end{pmatrix}$ and $\begin{pmatrix}-1\\ 1\end{pmatrix}$ are a spanning sequence for $\mathbb{R}^{2}$ , as we have just seen.

Example 4.7.3.

Let’s try to determine whether $\mathbf{v}_{1}=\begin{pmatrix}1\\ -1\\ 0\end{pmatrix}$ , $\mathbf{v}_{2}=\begin{pmatrix}0\\ 1\\ -1\end{pmatrix}$ , $\mathbf{v}_{3}=\begin{pmatrix}1\\ 0\\ -1\end{pmatrix}$ are a spanning sequence for $\mathbb{R}^{3}$ . We need to find out whether it’s true that for all $\begin{pmatrix}x\\ y\\ z\end{pmatrix}\in\mathbb{R}^{3}$ there exist $\alpha,\beta,\gamma\in\mathbb{R}^{3}$ such that

\alpha\begin{pmatrix}1\\ -1\\ 0\end{pmatrix}+\beta\begin{pmatrix}0\\ 1\\ -1\end{pmatrix}+\gamma\begin{pmatrix}1\\ 0\\ -1\end{pmatrix}=\begin{pmatrix}x\\ y\\ z\end{pmatrix}.

This is equivalent to asking whether for every $x, y, z$ the simultaneous equations

	$\displaystyle\alpha+\gamma$	$\displaystyle=x$
	$\displaystyle-\alpha+\beta$	$\displaystyle=y$
	$\displaystyle-\beta-\gamma$	$\displaystyle=z$

have a solution. Again, in this special case you might just notice that (adding the three equations) there is no solution unless $x+y+z=0$ , so this collection of vectors is not a spanning sequence. In general, to find out if a system of linear equations has a solution you can put the augmented matrix into row reduced echelon form. In this case the augmented matrix is

\begin{pmatrix}1&0&1&x\\ -1&1&0&y\\ 0&-1&-1&z\end{pmatrix}

Doing the row operations $r_{2}\mapsto r_{2}+r_{1}$ followed by $r_{3}\mapsto r_{3}+r_{2}$ leads to

\begin{pmatrix}1&0&1&x\\ 0&1&1&y+x\\ 0&0&0&z+y+x\end{pmatrix}

These equations have no solutions if $x+y+z\neq 0$ , so for example $\begin{pmatrix}1\\ 0\\ 0\end{pmatrix}$ is not in the span of $\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3}$ because $1+0+0\neq 0$ .