Spanning sets

A linear combination is an expression of one vector in terms of other vectors. In general, linear combinations take the form

$\overrightarrow{v}=a_1{\overrightarrow{u}}_1+a_2{\overrightarrow{u}}_2+\cdots +a_n{\overrightarrow{u}}_n\text{.}$

In ${ℝ}^2\text{,}$ we know that all vectors can be expressed in terms of the standard basis vectors $\hat{i}$ and $\hat{j}\text{.}$ There are a few ways we can describe these two vectors: we can say $\hat{i}$ and $\hat{j}$ are a basis for ${ℝ}^2$; we can say $\{\hat{i},\hat{j}\}$ form a spanning set for ${ℝ}^2$; and we can say that $\{\hat{i},\hat{j}\}$ spans ${ℝ}^2\text{.}$ They all mean the same thing—namely, that you can represent all possible vectors in ${ℝ}^2$ with linear combinations of $\hat{i}$ and $\hat{j}\text{.}$

In this case, we have an orthogonal basis because $\hat{i}$ and $\hat{j}$ are perpendicular (orthogonal is another way of saying perpendicular, but it is a bit more general). However, this is not a requirement for all spanning sets. There is only one requirement for a pair of two vectors to span ${ℝ}^2$: they must be nonzero and non-collinear. (Two vectors are collinear if they are parallel or antiparallel.) The zero restriction is fairly obvious. But why can’t they be collinear? Just try imaging it. If two vectors have the same direction or opposite directions, you are locked onto a line. There is no way you can move outside it. But if the directions differ even by just a billionth of a degree, you have a spanning set and you are free to roam anywhere you please on the Cartesian plane.

There are also spanning sets in ${ℝ}^3\text{.}$ A triplet of three-component vectors forms a basis for ${ℝ}^3$ if they are nonzero and non-coplanar. What does coplanar mean? Head on over to the next section.

Example

Express the vector $[\mathrm{6,7}]$ in terms of the spanning set $\{[\mathrm{10,5}],[-\mathrm{2,1}]\}\text{.}$

Referring to the definition of a linear combination, we can write an equation with unknown coefficients $a$ and $b$:

$[\mathrm{6,7}]=a[\mathrm{10,5}]+b[-\mathrm{2,1}]\text{.}$

We can unpack this into two equations—one for each component:

$6=10a-2b$ and $7=5a+b\text{.}$

Solving by elimination tells us that $a=1$ and $b=2\text{,}$ therefore

$[\mathrm{6,7}]=[\mathrm{10,5}]+2[-\mathrm{2,1}]\text{.}$

This works the same way in ${ℝ}^3$ (three variables, three equations).