Linear combinations and span

You may travel along two arrows, any amount, forwards or back. Watch where you can get to: the dots fill the plane.

Travelling with two arrows

Here is a game. You stand at the origin and hold two arrows, $\mathbf{v}=(2,1)$ and $\mathbf{w}=(1,2)$. You may walk along $\mathbf{v}$ as far as you like, forwards or backwards, and then along $\mathbf{w}$ as far as you like. Where can you end up?

Walk 2 lengths of $\mathbf{v}$ and 1 length of $\mathbf{w}$, and you arrive at

Walk half of $\mathbf{v}$ backwards and 3 of $\mathbf{w}$: $-\tfrac12\mathbf{v}+3\,\mathbf{w}=(-1,-0.5)+(3,6)=(2,5.5)$. Each choice of two amounts lands you on one point, and each dot in the picture at the top of the page is one such choice.

Every trip has the same shape: scale each arrow by a number, then add. The result

is a linear combination of $\mathbf{v}$ and $\mathbf{w}$. The two numbers $c$ and $d$ are its weights. “Linear” means only these two things happen: scaling and adding. Nothing is squared and nothing is multiplied by another arrow.

The span: every place you can reach

Watch the clip again. The dots spread out in every direction and keep filling the gaps. Given enough trips, you can reach any point of the plane with these two arrows.

The set of all the places you can reach is the span of the arrows, written

where $c$ and $d$ run over all numbers.

For $\mathbf{v}=(2,1)$ and $\mathbf{w}=(1,2)$ the span is the whole plane, $\mathbb{R}^2$.

Fewer arrows reach less. With one arrow $\mathbf{v}$ you can only walk along it, so its span is the line through the origin in the direction of $\mathbf{v}$. The zero vector is a stranger case: every multiple of $\mathbf{0}$ is $\mathbf{0}$, so its span is a single point, the origin. Every span contains the origin, because you can always choose all the weights to be 0.

When an arrow adds nothing

Now drag the tip of $\mathbf{w}$ onto the line of $\mathbf{v}$ and watch the cloud. The dots stop filling the plane and fall onto one line, and the readout changes to “a line”.

Here is why. Put $\mathbf{w}$ at $(4,2)$, which is $2\mathbf{v}$. Then every trip is

a multiple of $(2,1)$. Walking along $\mathbf{w}$ only moves you along the line you could already reach with $\mathbf{v}$. The second arrow adds nothing new, and two arrows span only a line.

So the count of arrows is not what matters. What matters is whether each new arrow points somewhere the others can’t reach. Two arrows that are not on one line span the plane; two arrows on one line span only that line.

Three arrows in space

In space, points have three components. One arrow spans a line, and two arrows not on one line span a flat plane through the origin. A third arrow fills all of space, but only if it points out of that plane.

Watch the dots before and after the swap. The arrow $(1,1,0)$ is $(1,0,0)+(0,1,0)$, so it lies in the floor already, and three arrows paint only the floor. The arrow $(1,1,1)$ points up out of the floor, and with it the dots reach every height.

Reaching a given point

Often the question comes the other way round: here is a target, can I reach it, and with which weights? Take the target $\mathbf{b}=(5,4)$ and the arrows $(2,1)$ and $(1,2)$. I need weights $c$ and $d$ with

Comparing the two components gives two equations, one per component:

The worked example below solves them. If the equations have a solution, the target is in the span; if they have none, it is not. In A matrix moves the plane the two arrows become the columns of a table of numbers, and this same question gets its own short notation.

Worked example

1. Reaching $(5,4)$ with $(2,1)$ and $(1,2)$. From $2c+d=5$, write $d=5-2c$. Put that into $c+2d=4$:

Then $d=5-2\cdot2=1$. Check: $2(2,1)+1(1,2)=(4,2)+(1,2)=(5,4)$. The target is in the span.

2. The same target with $(2,1)$ and $(4,2)$. Every combination is a multiple of $(2,1)$, so its second component is half its first. For $(5,4)$, half of 5 is 2.5, not 4. No weights work: $(5,4)$ is not in the span of $(2,1)$ and $(4,2)$.

3. Three arrows in space. With three arrows I need three weights; I’ll call them $a$, $b$ and $c$. Reach $(2,3,5)$ with $(1,0,0)$, $(0,1,0)$ and $(1,1,1)$:

The last component gives $c=5$ at once. Then $a+5=2$ gives $a=-3$, and $b+5=3$ gives $b=-2$. Check: $(-3,0,0)+(0,-2,0)+(5,5,5)=(2,3,5)$.

Predict before you drag: if $\mathbf{w}=3\mathbf{v}$, what does the cloud look like? A line, the line of $\mathbf{v}$.

Where you’ll meet this

Colour on a screen. Every pixel mixes red, green and blue light. The colour you see is a linear combination of three fixed colours, with the three brightnesses as weights. The span of those three is every colour the screen can show.

Chords and superposition. A chord is a mix of notes sounding together. Its waveform is the sum of the notes’ waveforms, each scaled by how loudly it is played: a linear combination of signals.

Equalisers. An audio equaliser splits sound into bands, scales each band by a gain and adds them back. The output is a linear combination of the band signals, with the slider settings as weights, as in Audio equalisers and biquads.

Reference card