Worked examples — Span — definition

Before we start, three words we will keep using, each anchored to a picture in your head:

The span is the collection of all the landing spots you can produce this way.

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The scenario matrix

Every span question falls into one of these cells. We will hit each one with a worked example.

Cell	Situation	What can go wrong / what to watch	Example
A	1 vector, nonzero	Always a line through origin	Ex 1
B	1 vector, the zero vector (degenerate)	Span collapses to a single point	Ex 2
C	2 vectors, independent, in ${\mathbb{R}}^2$	Fill the whole plane — test with Determinant	Ex 3
D	2 vectors, dependent (one is a multiple)	Collapses to a line, not a plane	Ex 4
E	2 vectors in ${\mathbb{R}}^3$	A tilted plane through origin, not all of space	Ex 5
F	3 vectors in ${\mathbb{R}}^3$, independent	Fill all of ${\mathbb{R}}^3$	Ex 6
G	3 vectors in ${\mathbb{R}}^3$, one redundant	Still only a plane — count real directions	Ex 7
H	Word problem (real-world reachability)	Translate "amounts" into scalars	Ex 8
I	Exam twist: negative-scalar / "positive only" trap	Scalars range over all of $\mathbb{R}$	Ex 9

The signs of the scalars ($+$, $-$, $0$) are covered inside the examples because a span always allows all three — that is exactly what makes it symmetric through the origin.

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Cell A — one nonzero vector: a line

What the figure shows: a single cyan straight line passing through the origin along the direction $(3,1)$, extending to the upper-right and lower-left. The white arrow marks $v=(3,1)$; the amber dot at $(6,2)$ sits exactly on the line, while the red dot at $(3,-1)$ sits clearly off it.

1. Write the span. With one vector there is one scalar $c$: $\mathrm{span}\{v\}=\left\{c\left[\begin{array}{c}3\\ 1\end{array}\right]:c\in \mathbb{R}\right\}.$ Why this step? The definition says span = all linear combinations; with one generator a linear combination is just $cv$.
2. See the shape. As $c$ runs through every real number, $cv$ slides the tip along the straight direction $(3,1)$. Positive $c$ goes one way, negative $c$ flips it to the other side, $c=0$ sits at the origin. Look at the cyan line in the figure (described above) — it passes through the origin in both directions. So the span is a line through the origin.
3. Test $(6,2)$. Need $c(3,1)=(6,2)$: from the first slot $3c=6\Rightarrow c=2$; check second slot $1\cdot 2=2$ ✔. So $(6,2)$ is in the span (amber tip in figure).
4. Test $(3,-1)$. Need $3c=3\Rightarrow c=1$, but then second slot gives $1\cdot 1=1\ne -1$. No consistent $c$. So $(3,-1)$ is not in the span.

Verify: $2\cdot (3,1)=(6,2)$ ✔. For $(3,-1)$: the two slots demand $c=1$ and $c=-1$ simultaneously — impossible, so correctly rejected.

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Cell B — the zero vector: a single point

1. Apply the definition. $\mathrm{span}\{0\}=\{c0:c\in \mathbb{R}\}$. Why this step? Same rule as always — all scalar multiples.
2. Compute the multiples. For every scalar $c$, $c\cdot (0,0)=(0,0)$. Stretching nothing gives nothing; flipping nothing gives nothing. Why this step? $c\cdot 0=0$ for any real $c$, so the "set of all landing spots" has exactly one member.
3. Conclusion. $\mathrm{span}\{0\}=\{0\}$ — a single point, the origin.

Verify: the span is still a valid Subspace (it contains $0$, is closed under $+$ and scaling). It has Dimension $0$. This is the smallest possible span.

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Cell C — two independent vectors fill ${\mathbb{R}}^2$

What the figure shows: a lattice of cyan dots — the points $i{v}_1+j{v}_2$ for integer $i,j$ — spreading evenly across the whole plane (a tilted grid). The amber arrow is $v_1=(1,2)$, the white arrow is $v_2=(3,1)$, and a red dot marks the target $(0,1)$ that the grid reaches.

1. Set up the reachability question. "Span ${\mathbb{R}}^2$" means every target $(x,y)$ is hit: $c_1\left[\begin{array}{c}1\\ 2\end{array}\right]+c_2\left[\begin{array}{c}3\\ 1\end{array}\right]=\left[\begin{array}{c}x\\ y\end{array}\right].$ Why this step? Membership in a span is always a system of linear equations in the scalars $c_1,c_2$.
2. Read it as a matrix. Put the vectors as columns: $M=\left[\begin{array}{cc}1& 3\\ 2& 1\end{array}\right]$. The system is $M\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right]=\left[\begin{array}{c}x\\ y\end{array}\right]$. Why this step? Stacking the vectors as columns is precisely how a linear combination $c_1{v}_1+c_2{v}_2$ becomes the single product $M\mathbf{c}$ — so "is $(x,y)$ reachable?" turns into the standard solvability question "does $M\mathbf{c}=(x,y)$ have a solution?", which the determinant can then answer in one number.
3. Ask the decisive question with the Determinant. Why the determinant and not something else? Because for a square matrix, $\det \ne 0$ is exactly the condition that "$M$ can be undone" (invertible) — meaning every $(x,y)$ has a solution. $\det M=1\cdot 1-3\cdot 2=1-6=-5\ne 0.$
4. Conclude. Nonzero determinant ⇒ invertible ⇒ every $(x,y)$ reachable ⇒ they span ${\mathbb{R}}^2$ (full span). In the figure, the two cyan arrows form a genuine "grid" of tip-to-tail parallelograms that tile the whole plane.

Verify: solve for a specific hard target, say $(0,1)$. $M^{-1}=\frac{1}{-5}\left[\begin{array}{cc}1& -3\\ -2& 1\end{array}\right]$, so $\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right]=\frac{1}{-5}\left[\begin{array}{cc}1& -3\\ -2& 1\end{array}\right]\left[\begin{array}{c}0\\ 1\end{array}\right]=\frac{1}{-5}\left[\begin{array}{c}-3\\ 1\end{array}\right]=\left[\begin{array}{c}3/5\\ -1/5\end{array}\right]$. Check: $\frac{3}{5}(1,2)-\frac{1}{5}(3,1)=(\frac{3}{5}-\frac{3}{5},\frac{6}{5}-\frac{1}{5})=(0,1)$ ✔.

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Cell D — two dependent vectors: collapse to a line

What the figure shows: one single cyan line through the origin. Both the amber arrow $v_1=(2,-1)$ and the white arrow $v_2=(-4,2)$ lie along that same line — $v_2$ just points the opposite way and is twice as long. A red dot at $(1,1)$ sits off the line, showing it is unreachable.

1. Check for a hidden multiple. Is $v_2=\lambda v_1$? $(-4,2)=\lambda (2,-1)$ needs $\lambda =-2$ from the first slot; check second: $-2\cdot (-1)=2$ ✔. So $v_2=-2v_1$ — they are linearly dependent (see Linear independence).
2. Simplify the linear combination. Every combination collapses: $c_1{v}_1+c_2{v}_2=c_1{v}_1+c_2(-2v_1)=(c_1-2c_2)v_1.$ Why this step? Substituting $v_2=-2v_1$ turns two free scalars into a single effective scalar $(c_1-2c_2)$, which can still be any real number.
3. Conclude. The span is $\{t{v}_1:t\in \mathbb{R}\}$ — just the line through $(2,-1)$, not the plane. In the figure both arrows lie on the same cyan line (one just points the opposite way).
4. Confirm with the determinant. $\det \left[\begin{array}{cc}2& -4\\ -1& 2\end{array}\right]=2\cdot 2-(-4)(-1)=4-4=0$. Zero determinant ⇒ the matrix is not invertible ⇒ they do not achieve full span (they cover only a line, not all of ${\mathbb{R}}^2$), matching our finding.

Verify: is $(1,1)$ in the span? It would need $t(2,-1)=(1,1)$: $2t=1\Rightarrow t=1/2$, but then $-1\cdot \frac{1}{2}=-\frac{1}{2}\ne 1$. No — correctly, the plane is not filled. Determinant $=0$ ✔ agrees.

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Cell E — two vectors in ${\mathbb{R}}^3$: a tilted plane

What the figure shows: a flat, tilted cyan sheet passing through the origin in 3D — this is the plane $x+y-z=0$. The amber arrow $v_1=(1,0,1)$ and the white arrow $v_2=(0,1,1)$ both lie in the sheet, and the amber dot $(2,3,5)$ sits right on it, confirming reachability.

1. Write a general point. $a(1,0,1)+b(0,1,1)=(a,b,a+b)$. Why this step? Add coordinate-by-coordinate to see the pattern.
2. Spot the constraint. The third coordinate is forced: $z=a+b=x+y$. So every point in the span obeys $x+y-z=0$. Two free numbers $a,b$ sweep out a 2-dimensional flat — a plane through the origin, tilted (not one of the coordinate planes). See the cyan sheet described above. Why can't it be all of ${\mathbb{R}}^3$? Two arrows give at most two independent directions; filling 3D needs three. See Dimension.
3. Membership becomes a plane test. A point is in the span iff $x+y-z=0$.

Verify: is $(2,3,5)$ in it? Check $2+3-5=0$ ✔ — yes, with $a=2,b=3$ giving $(2,3,5)$. Is $(1,1,1)$ in it? $1+1-1=1\ne 0$ — no. Both agree with the coordinate formula.

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Cell F — three independent vectors fill ${\mathbb{R}}^3$

1. Stack as columns and test with the determinant. Why the determinant? For a square $3\times 3$ matrix, $\det \ne 0$ ⇔ invertible ⇔ every target reachable, exactly as in Cell C but one dimension up. $M=\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 1\\ 0& 0& 1\end{array}\right].$
2. Compute the determinant. $M$ is upper-triangular, so $\det M=$ product of the diagonal $=1\cdot 1\cdot 1=1\ne 0$. Why this shortcut? For a triangular matrix the determinant is just the diagonal product — no row reduction needed.
3. Conclude. $\det \ne 0$ ⇒ the three columns are independent ⇒ they span all of ${\mathbb{R}}^3$ (full span). They also form a Basis (spanning and independent).

Verify: hit the target $(0,0,1)$. Solve $c_1(1,0,0)+c_2(1,1,0)+c_3(1,1,1)=(0,0,1)$. Third slot: $c_3=1$. Second: $c_2+c_3=0\Rightarrow c_2=-1$. First: $c_1+c_2+c_3=0\Rightarrow c_1=0$. Check: $0(1,0,0)-1(1,1,0)+1(1,1,1)=(0,0,1)$ ✔.

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Cell G — three vectors, one redundant: still a plane

1. Hunt for redundancy. Notice $(1,1,0)=(1,0,0)+(0,1,0)$. The third vector is already a combination of the first two — it adds no new direction. Why this step? Span grows only when a new vector points somewhere unreachable; here it points into the plane we already had.
2. Reduce to the essentials. So $\mathrm{span}$ of all three $=\mathrm{span}\{(1,0,0),(0,1,0)\}$, which is the set $\{(a,b,0)\}$ — the $xy$-plane, a 2D flat.
3. Confirm with the determinant. $\det \left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 1\\ 0& 0& 0\end{array}\right]=0$ (a whole row of the third-coordinate is zero). Zero ⇒ not full span (only a plane, not all of ${\mathbb{R}}^3$), matching. The number of genuinely independent columns is $2$, so Dimension $=2$.

Verify: is $(0,0,7)$ in the span? Any combination keeps the third coordinate $0\cdot (\cdots )=0$, so third coordinate is always $0\ne 7$. No. Three vectors did not guarantee ${\mathbb{R}}^3$.

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Cell H — a real-world word problem

1. Translate to span. "$c_1$ of packet $P$ plus $c_2$ of packet $Q$" is exactly $c_1(1,2)+c_2(3,1)$. The reachable blends are $\mathrm{span}\{(1,2),(3,1)\}$. Why this step? "Use some amount of each ingredient and combine" is scaling-and-adding — the definition of a linear combination.
2. Test full span. These are the same vectors as Ex 3, $\det =-5\ne 0$, so every $(s,c)$ is reachable — the packets span ${\mathbb{R}}^2$.
3. Concrete order. Want $\left[\begin{array}{c}s\\ c\end{array}\right]=\left[\begin{array}{c}7\\ 4\end{array}\right]$. Solve $c_1(1,2)+c_2(3,1)=(7,4)$: from Ex 3's inverse, $\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right]=\frac{1}{-5}\left[\begin{array}{cc}1& -3\\ -2& 1\end{array}\right]\left[\begin{array}{c}7\\ 4\end{array}\right]=\frac{1}{-5}\left[\begin{array}{c}7-12\\ -14+4\end{array}\right]=\frac{1}{-5}\left[\begin{array}{c}-5\\ -10\end{array}\right]=\left[\begin{array}{c}1\\ 2\end{array}\right]$.

Verify: $1\cdot (1,2)+2\cdot (3,1)=(1+6,2+2)=(7,4)$ ✔. Units: $1$ packet $P$ + $2$ packets $Q$ → $7$ kg sand, $4$ kg cement. (Physically you'd restrict to $c_i\ge 0$; mathematically the span allows all reals — see the next trap.)

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Cell I — the exam twist (negative-scalar trap)

1. Recall the exact definition. In $\mathrm{span}$, the scalars range over all of $\mathbb{R}$ — positive, negative, and zero. This is not optional; it is built into the definition (see Linear combination). Why this matters: allowing $c<0$ is precisely what makes the span symmetric through the origin.
2. Show a negative combination. Take $c=-4$: $-4(1,0)=(-4,0)$, which has $x=-4<0$. That point is in the span, but it is not in the proposed ray. Why this step? One concrete counterexample is enough to break a "for all" claim: to disprove "the span is only the ray $x\ge 0$", we just exhibit a single legitimate span element that lies outside that ray. $(-4,0)$ does exactly that.
3. Name the correct object for the restricted version. If you did forbid negatives ($c\ge 0$), you'd get a ray (in general a cone), not a span. The statement confuses a cone with a span.
4. Deliver the verdict. The statement is FALSE. The true span is $\mathrm{span}\{(1,0)\}=\{(x,0):x\in \mathbb{R}\}$ — the entire horizontal axis, extending in both directions through the origin, not a one-sided ray.

Verify: $\mathrm{span}\{(1,0)\}=\{(x,0):x\in \mathbb{R}\}$ — the whole horizontal axis. The point $(-4,0)$ is reachable with $c=-4$ ✔, yet $-4<0$ places it outside the claimed ray $\{x\ge 0\}$ — so the claim is correctly rejected.

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Active recall

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Connections

• Linear combination — every worked example is a membership question phrased as a linear combination.
• Linear independence — the hidden pivot deciding line vs plane (Cells D, G).
• Determinant — the yes/no full-span test for square cases (Cells C, F).
• Subspace — even $\{0\}$ (Cell B) is a valid one.
• Basis — Cell F's three independent vectors form one.
• Dimension — counts real directions across all cells.
• Column space — the span of a matrix's columns, exactly what the determinant tests probe.