Hand an engineer a few building blocks and the first question is always the same: what can I build with these? In linear algebra the building blocks are vectors, and "building" means stretching them and adding the results. Every vector you can make that way is a linear combination, and the set of everything you can make is the span. Both ideas run the whole rest of the course, so here's the cleanest picture of them there is: a superhero with two ways to fly.
A superhero stands at the origin with two gadgets. A hoverboard can ride any distance along the arrow $h = (2, 1)$, forward or backward, and a magic carpet can ride any distance along $m = (1, 3)$. The villain is hiding at $(-1, 7)$. Can the hero get there? And is there anywhere in the plane the villain could hide where the hero can't follow?
A hoverboard rides along $h = (2, 1)$ and a magic carpet along $m = (1, 3)$, any distance, forward or back. Two rides backward on the hoverboard and three forward on the carpet reach the villain: $-2h + 3m = (-1, 7)$. Every trip the hero can take is some $ah + bm$, and the question "where can the hero go?" is the question this page answers.
Every trip the hero can take is "ride the hoverboard $a$ times, then the carpet $b$ times," which lands at $ah + bm$. The villain was at $-2h + 3m$. That expression, numbers times arrows added up, is what this whole page is about.
On the Vector Spaces page you were handed the numbers and the vectors and asked for something like $3u - 2v$. This page mostly runs that question backward: here's the answer, now find the numbers.
A sum of stretched vectors is the most basic thing you can build. Here's its official name.
Let $\V$ be a vector space and $u_1, u_2, \ldots, u_k$ vectors in $\V$. A vector $w$ is a linear combination of $u_1, \ldots, u_k$ if $w = c_1u_1 + c_2u_2 + \cdots + c_ku_k$ for some scalars $c_1, \ldots, c_k$, called the coefficients of the combination. For a set $S$ of vectors in $\V$, $w$ is a linear combination of vectors of $S$ if it is a linear combination of finitely many vectors from $S$.
A linear combination of arrows $u_1, u_2, \ldots, u_k$ in $\R^2$ or $\R^3$ is any arrow $c_1u_1 + c_2u_2 + \cdots + c_ku_k$ you can build by stretching each $u_i$ by a number $c_i$ (shrinking, flipping, or zeroing it out is fine) and adding the results tip to tail. The numbers $c_1, \ldots, c_k$ are its coefficients.
Notice the word finitely in the last sentence. $S$ may hold infinitely many vectors, and you may use as many of them as you like, but only finitely many in any one combination. One freebie: since $0u = 0$, the zero vector is a linear combination of any nonempty set. Just use the coefficient $0$.
Building one is a recipe: stretch, then lay the pieces tip to tail, and wherever the chain ends is the combination. One freebie: set every coefficient to $0$ and the chain never leaves the origin, so the zero vector is a combination of any arrows at all.
Linear combinations aren't only about arrows. Record the vitamins in 100 grams of a food as a vector in $\R^5$, one entry per vitamin, and a meal becomes a linear combination of foods.
200 grams of soy sauce. For your health.
Here are the vitamin vectors of five foods (100 grams each). Is the vitamin vector of raw wild rice a linear combination of the other four?
| Food | A | B$_1$ | B$_2$ | Niacin | C |
|---|---|---|---|---|---|
| Cupcake from mix | 0 | 0.05 | 0.06 | 0.3 | 0 |
| Coconut custard pie | 0 | 0.02 | 0.02 | 0.4 | 0 |
| Raw brown rice | 0 | 0.34 | 0.05 | 4.7 | 0 |
| Soy sauce | 0 | 0.02 | 0.25 | 0.4 | 0 |
| Raw wild rice | 0 | 0.45 | 0.63 | 6.2 | 0 |
Units: A in units, the rest in mg. Source: Watt and Merrill, Composition of Foods, USDA, 1963.
Yes:
$$\begin{aligned} &\text{cupcake} + \text{pie} + \text{brown rice} \\ &\quad + 2 \cdot \text{soy sauce} = \text{wild rice}. \end{aligned}$$Check it one vitamin at a time. B$_1$: $0.05 + 0.02 + 0.34 + 2(0.02) = 0.45$. B$_2$: $0.06 + 0.02 + 0.05 + 2(0.25) = 0.63$. Niacin: $0.3 + 0.4 + 4.7 + 2(0.4) = 6.2$. A and C are $0$ all the way down.
So the coefficients are $1, 1, 1, 2$: 100 grams each of cupcake, pie, and brown rice, plus 200 grams of soy sauce, give exactly the same vitamins as 100 grams of wild rice.
Arrows don't have to be directions in space. Record what's in a handful of food as (protein, carbs, fat) in grams, and every snack becomes an arrow in $\R^3$.
A handful of peanuts has about $P = (7, 4, 14)$, a handful of raisins $R = (1, 22, 0)$, and a handful of chocolate chips $K = (2, 18, 8)$. What's in a bag of trail mix made from 2 handfuls of peanuts, 3 of raisins, and 1 of chips?
The bag is the linear combination $2P + 3R + K$, with coefficients $2$, $3$, $1$: the handful counts.
$$\begin{aligned} 2P + 3R + K &= (14, 8, 28) + (3, 66, 0) + (2, 18, 8) \\ &= (19, 92, 36). \end{aligned}$$So the bag has 19 g of protein, 92 g of carbs, and 36 g of fat. Here's the same sum as a chain of arrows, one handful at a time (drawn at a tenth of the size):
Three handfuls of raisins and one of chips. Someone is going to pick the chips out, and the vector is going to be wrong.
Back to arrows. With two arrows $u$ and $v$ that point different ways, every $su + tv$ sits at the far corner of a parallelogram with sides $su$ and $tv$. Dial $s$ and $t$ and watch where it goes.
The far corner wanders all over one flat plane through the origin, and never leaves it. In $\R^2$ the same game paints the whole plane, unless the two arrows line up. On the left of this widget is every pair of coefficients $(a, b)$; on the right is where $au + bv$ lands.
When $u$ and $v$ point different ways, the paint covers everything. Line them up and the paint never leaves one line, no matter how wild the coefficients get. Keep that picture in mind: which vectors you can build depends on the vectors you start with, not just on how many there are.
The question you'll ask over and over is the backward one: "Is $w$ a linear combination of these vectors, and if so, with what coefficients?" Write the coefficients as unknowns, then match coordinates. Each coordinate gives one equation, so the question turns into a system of linear equations.
Every system below gets solved with the row operations from Systems & Row Reduction: swap two equations, scale one by a nonzero number, or add a multiple of one to another. None of them changes the solutions, and a row that turns into $0 = c$ with $c \neq 0$ means there are none.
Let $u_1 = (1, -1, 2)$ and $u_2 = (2, 1, 1)$.
(a) Is $w = (4, 5, -1)$ a linear combination of $u_1$ and $u_2$?
(b) Is $w' = (4, 5, 3)$?
(a) We want numbers $a$ and $b$ with $au_1 + bu_2 = w$, that is,
$$(a + 2b,\ -a + b,\ 2a + b) = (4, 5, -1).$$Matching the three coordinates gives the system
$$\begin{aligned} a + 2b &= 4 \\ -a + b &= 5 \\ 2a + b &= -1. \end{aligned}$$Clear $a$ out of the last two equations with $R_2 \to R_2 + R_1$ and $R_3 \to R_3 - 2R_1$, then clear $b$ out of the last one with $R_3 \to R_3 + R_2$:
$$\begin{aligned} a + 2b &= 4 \\ 3b &= 9 \\ 0 &= 0. \end{aligned}$$So $b = 3$ and $a = 4 - 2(3) = -2$, and $w = -2u_1 + 3u_2$. Check: $-2(1, -1, 2) + 3(2, 1, 1) = (-2 + 6,\ 2 + 3,\ -4 + 3) = (4, 5, -1)$. $\checkmark$
(b) Same left sides, new right sides. The same three row operations give
$$\begin{aligned} a + 2b &= 4 \\ 3b &= 9 \\ 0 &= 4. \end{aligned}$$The last line says $0 = 4$. No choice of $a$ and $b$ makes that true, so the system has no solutions, and $w'$ is not a linear combination of $u_1$ and $u_2$.
Here are both answers in the picture. The arrows $u_1$ and $u_2$ span a plane, the one with equation $-x + y + z = 0$, and $w$ sits on it while $w'$ floats above it:
The $4$ in $0 = 4$ isn't random. The three row operations added up to $-R_1 + R_2 + R_3$, and $-1, 1, 1$ are the coefficients of the plane's equation. So the leftover row is that equation evaluated at $w'$: $-4 + 5 + 3 = 4$. It measures how far off the plane $w'$ is.
The system just looked me in the eye and said $0 = 4$. Bold.
That example is the whole method, and it's worth stating once and for all. Put the vectors side by side as the columns of a matrix, and the question "is $b$ a combination of them?" becomes "is this system consistent?"
Let $u_1, \ldots, u_k$ and $b$ be vectors in $\R^n$$\R^2$ or $\R^3$. Then $b$ is a linear combination of $u_1, \ldots, u_k$ exactly when the system with augmented matrix $(u_1 \ \cdots \ u_k \mid b)$, the $u_i$ written as columns, is consistent. Its solutions $(c_1, \ldots, c_k)$ are exactly the coefficients with $c_1u_1 + \cdots + c_ku_k = b$.
Two vectors in $\R^n$ are equal exactly when their entries match, one coordinate at a time. The $i$th entry of $c_1u_1 + \cdots + c_ku_k$ is $c_1 (u_1)_i + \cdots + c_k (u_k)_i$, where $(u_j)_i$ is the $i$th entry of $u_j$. So $c_1u_1 + \cdots + c_ku_k = b$ says
$$c_1 (u_1)_i + c_2 (u_2)_i + \cdots + c_k (u_k)_i = b_i \quad \text{for } i = 1, \ldots, n,$$and that list of $n$ equations is the system whose $i$th row is the $i$th entries of $u_1, \ldots, u_k$ and $b$, which is the augmented matrix $(u_1 \ \cdots \ u_k \mid b)$. So the solutions of the system are exactly the lists of coefficients that work, and $b$ is a linear combination exactly when there is at least one. $\blacksquare$
This is a visual argument, not a formal proof. Build the chain $c_1u_1 + c_2u_2$ and read off where it ends: its $x$, $y$, and $z$ coordinates are exactly the left sides of the system's three equations. Turn the dials until the end lands on the target, and you've made all three equations true at once.
Landing on $b$ and solving the system are the same job, one done with arrows and one with equations. The picture shows two arrows in $\R^3$; with more arrows the chain just gets longer and the system gets more unknowns. The full proof is in the n version.
That's the column picture from Matrices read backward: $Ax$ is a combination of the columns of $A$, so asking whether $b$ is one is asking whether $Ax = b$ has a solution. And Rank & Inverses already told you how to decide that without finishing the solve.
That's the column picture from Matrices read backward: $Ax$ is a chain of the columns of $A$, so asking whether $b$ is one is asking whether $Ax = b$ has a solution. And Rank & Inverses already told you how to decide that without finishing the solve.
Don't panic at a row of zeros. $0 = 0$ is harmless: it's always true, so it puts no limit on the coefficients, and you just cross it out. Only $0 = c$ with $c \neq 0$ means "not a combination." And when you do find coefficients, plug them back in. One sign slip during elimination gives coefficients that look perfectly reasonable and are completely wrong.
With more vectors there are more unknowns than the coordinates can pin down, and then the coefficients stop being unique.
Write $w = (3, 1, 5)$ as a linear combination of $u_1 = (1, 0, 2)$, $u_2 = (0, 1, -1)$, and $u_3 = (1, 2, 0)$.
We want $a_1u_1 + a_2u_2 + a_3u_3 = w$. Matching coordinates:
$$\begin{aligned} a_1 + a_3 &= 3 \\ a_2 + 2a_3 &= 1 \\ 2a_1 - a_2 &= 5. \end{aligned}$$$R_3 \to R_3 - 2R_1$ turns the last equation into $-a_2 - 2a_3 = -1$, and then $R_3 \to R_3 + R_2$ turns it into $0 = 0$. Two equations are left for three unknowns, so $a_3$ is free:
$$a_1 = 3 - a_3, \qquad a_2 = 1 - 2a_3.$$Every choice of $a_3$ gives a recipe. With $a_3 = 0$: $w = 3u_1 + u_2$. With $a_3 = 1$: $w = 2u_1 - u_2 + u_3$. Check the second one: $(2, 0, 4) + (0, -1, 1) + (1, 2, 0) = (3, 1, 5)$. $\checkmark$
Free unknowns? Set them to zero. Lazy, and completely legal.
To show $w$ is a combination you only need one recipe, so the lazy choice is fine.
Why so many recipes? All three arrows lie on one plane ($u_3 = u_1 + 2u_2$), so there's more than one way to walk to the same spot:
One more way to see all of this, with pictures. Each mission gives you $u$, $v$, and a target; slide $a$ and $b$ until $au + bv$ hits it. One target can't be hit. When you find it, say so.
The impossible mission is the $0 = 4$ example in picture form. When $u$ and $v$ lie on one line, every combination stays on that line, and anything off it gives a system with no solution.
Your turn. In free response, type the coefficients in order, separated by commas, like -2, 3, or a single number when the question asks for one.
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So far the question has been "can I reach this vector?" Zoom out: what's the set of all the vectors you can reach?
The span of vectors $u_1, \ldots, u_k$ in a vector space $\V$, written $\operatorname{span}\{u_1, \ldots, u_k\}$, is the set of all their linear combinations. More generally, for any set $S$ of vectors in $\V$, $\operatorname{span}(S)$ is the set of all linear combinations of finitely many vectors of $S$, and by convention $\operatorname{span}(\emptyset) = \{0\}$.
The span of arrows $u_1, \ldots, u_k$ in $\R^2$ or $\R^3$, written $\operatorname{span}\{u_1, \ldots, u_k\}$, is the set of every arrow you can build from them as a linear combination: every place a chain of them can end. The span of no arrows at all is $\{0\}$, since you never leave the origin.
The span of nothing still has something in it. The zero vector shows up to every party, even the empty ones.
In the $au + bv$ widget, the paint was the span: all of $\R^2$, or just a line once $u$ and $v$ lined up. Here are a few you can write down by hand.
Describe $\operatorname{span}\{(1, 2, 2)\}$ and $\operatorname{span}\{(1, 0, 0), (0, 1, 0)\}$ in $\R^3$.
The combinations of one vector are its multiples $c(1, 2, 2)$, so the first span is the line through the origin in the direction $(1, 2, 2)$.
Every combination of the other two looks like
$$a(1, 0, 0) + b(0, 1, 0) = (a, b, 0),$$so the second span is every vector whose third entry is $0$: the $xy$-plane.
To decide what a span is, take a general target and ask which ones you can reach. Hefferon's first example of this lives in the plane.
Show that $\operatorname{span}\{(1, 1), (1, -1)\} = \R^2$.
Take any target $(x, y)$ and look for $c_1$, $c_2$ with $c_1(1, 1) + c_2(1, -1) = (x, y)$:
$$\begin{aligned} c_1 + c_2 &= x \\ c_1 - c_2 &= y. \end{aligned}$$$R_2 \to R_2 - R_1$ gives $-2c_2 = y - x$, so $c_2 = \frac{x - y}{2}$, and then $c_1 = x - c_2 = \frac{x + y}{2}$. Those work for every $x$ and $y$, so every vector of $\R^2$ is in the span. For instance, $(1, 2) = \frac{3}{2}(1, 1) - \frac{1}{2}(1, -1)$.
Notice the letters. To show a span is everything, one target isn't enough; you need a recipe that works for all of them at once. Now try putting a vector into a span. In free response, type the number.
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Every span so far has been a point, a line, a plane, or everything, and each of those is a subspace. That's no accident.
The Subspaces page ended on unions: two lines together aren't a subspace, because they're missing the sums. A span throws in every sum and every stretch from the start.
For any set $S$ of vectors in a vector space $\V$$\R^2$ or $\R^3$, $\operatorname{span}(S)$ is a subspace. Moreover, every subspace that contains $S$ also contains $\operatorname{span}(S)$, so the span is the smallest subspace containing $S$.
The span is a subspace. If $S = \emptyset$, then $\operatorname{span}(S) = \{0\}$, which is a subspace. Otherwise $S$ contains some vector $z$, and $0 = 0z$ is in $\operatorname{span}(S)$. Let $x, y \in \operatorname{span}(S)$, say
$$x = a_1u_1 + \cdots + a_mu_m, \qquad y = b_1v_1 + \cdots + b_nv_n$$with every $u_i$ and $v_j$ in $S$. Then $x + y = a_1u_1 + \cdots + a_mu_m + b_1v_1 + \cdots + b_nv_n$ is a linear combination of finitely many vectors of $S$ (if some $u_i$ equals some $v_j$, combine their terms), and for any scalar $c$, $cx = (ca_1)u_1 + \cdots + (ca_m)u_m$ is one too. So $\operatorname{span}(S)$ contains $0$ and is closed under addition and scalar multiplication, and it's a subspace by the subspace test.
It is the smallest. Let $W$ be a subspace of $\V$ containing $S$, and let $w = c_1w_1 + \cdots + c_kw_k$ with each $w_i \in S$. Each $w_i$ is in $W$, so each $c_iw_i$ is in $W$ by closure under scalar multiplication. Now add the terms one at a time: if $c_1w_1 + \cdots + c_jw_j$ is in $W$, then so is $(c_1w_1 + \cdots + c_jw_j) + c_{j+1}w_{j+1}$, by closure under addition. After $k - 1$ additions, $w \in W$. (If $S = \emptyset$, then $\operatorname{span}(S) = \{0\} \subseteq W$ since every subspace contains $0$.) $\blacksquare$
This is a visual argument, not a formal proof. Take two points of $\operatorname{span}\{u, v\}$, each the end of a chain of $u$'s and $v$'s. Lay one chain at the end of the other and you reach their sum with a longer chain of $u$'s and $v$'s, which regroups into a single combination. Double a point and every arrow in its chain doubles. Then bring in any subspace $W$ that holds $u$ and $v$:
Adding and stretching never leave the span, so it passes the subspace test. And $W$, being closed under stretching and adding, holds every $su$, $tv$, and $su + tv$, so the whole span sits inside $W$. The picture uses two arrows; with more, the chains are longer and nothing else changes. The full proof is in the n version.
The first sentence is the one you'll use constantly: a span is never just "some set," it's always a subspace, so everything from the Subspaces page applies to it. The second says a span takes nothing it doesn't need. And it runs the other way too: every subspace is the span of something (of itself, if nothing smaller), so "subspace" and "span" describe the same objects.
That gives every kind of subspace a recipe. The four kinds are exactly the spans of zero arrows ($\{0\}$), one arrow (a line), two arrows pointing different ways (a plane), and three arrows that don't all lie in one plane (all of $\R^3$).
On Solution Sets, the homogeneous solutions came out as every $s(1, 1, 0) + t(-1, 0, 1)$. Now that set has a name: $\operatorname{span}\{(1, 1, 0), (-1, 0, 1)\}$. The solutions of a homogeneous system are always the span of the vectors you read off its free variables.
If you've seen subgroups, this is the vector space version of the subgroup generated by a subset: build everything you can out of $S$, and you get a subgroup that sits inside every subgroup containing $S$.
The best case is when the span is everything.
A set $S$ of vectors in a vector space $\V$ spans $\V$ if $\operatorname{span}(S) = \V$. Then $S$ is a spanning set for $\V$. (Some books say $S$ generates $\V$.)
A set $S$ of arrows in $\R^3$ spans $\R^3$ if $\operatorname{span}(S)$ is all of $\R^3$: every arrow in space can be built from them. Then $S$ is a spanning set for $\R^3$. The same goes for $\R^2$. (Some books say $S$ generates the space.)
To show a set spans $\R^3$, you have to reach every vector, so you work with letters: take a general $(x, y, z)$ and find coefficients that work for it.
Show that $u = (1, 0, 0)$, $v = (2, 1, 0)$, and $w = (3, 2, 1)$ span $\R^3$.
Take any $(x, y, z)$ in $\R^3$. We want $r$, $s$, $t$ with
$$\begin{aligned} &r(1, 0, 0) + s(2, 1, 0) + t(3, 2, 1) \\ &\quad = (r + 2s + 3t,\ s + 2t,\ t) \\ &\quad = (x, y, z). \end{aligned}$$Work from the bottom up. The third coordinate says $t = z$. The second says $s + 2t = y$, so $s = y - 2z$. The first says $r + 2s + 3t = x$, so
$$r = x - 2(y - 2z) - 3z = x - 2y + z.$$So every vector in $\R^3$ is a linear combination of the three:
$$\begin{aligned} (x, y, z) = {} &(x - 2y + z)(1, 0, 0) \\ &+ (y - 2z)(2, 1, 0) \\ &+ z(3, 2, 1). \end{aligned}$$For instance, $(4, 3, 1) = -(1, 0, 0) + (2, 1, 0) + (3, 2, 1)$. $\checkmark$
Bottom up has a picture: only $w$ climbs, so it goes first and gets the height right; $v$ fixes $y$ without spoiling the height; $u$ finishes off $x$.
Three vectors in $\R^3$ don't always do it, though. It depends on which three.
Do $(1, 2, 1)$, $(3, -1, 3)$, and $(0, 1, 0)$ span $\R^3$?
Look at the entries: in every one of these vectors, the first entry equals the third. Scaling keeps that true, and so does adding, so every linear combination of them has first entry equal to third entry too.
But $(1, 0, 0)$ doesn't. So $(1, 0, 0)$ is not in the span, and these three vectors do not span $\R^3$. Their span is only the plane $x = z$.
Those two examples are the two halves of every "does it span?" question, and they don't take the same kind of work.
Same rule as on the Subspaces page. Showing a set spans takes letters, because it has to work for every vector. Showing it doesn't takes one vector it can't reach. Find a pattern every vector in the set shares (like "first entry = third entry"), or run elimination on a target and get a $0 = c$.
Before the widget draws each span, call it yourself: a line, a plane, or all of $\R^3$? Then drag the picture around to check from a few angles.
The counting is the trap. Two vectors can give only a line, and three vectors can give only a plane. What matters is how many genuinely new directions the set has, not how many vectors are in it. Here's the precise version: a new vector grows the span exactly when it wasn't already in it.
Let $S$ be a set of vectors in a vector space $\V$$\R^2$ or $\R^3$ and $v$ another vector. Then $\operatorname{span}(S \cup \{v\}) = \operatorname{span}(S)$ exactly when $v \in \operatorname{span}(S)$.
Every linear combination of vectors of $S$ is also one of vectors of $S \cup \{v\}$, so $\operatorname{span}(S) \subseteq \operatorname{span}(S \cup \{v\})$ always.
If $\operatorname{span}(S \cup \{v\}) = \operatorname{span}(S)$, then $v = 1v$ is in the left side, so $v \in \operatorname{span}(S)$.
Conversely, suppose $v \in \operatorname{span}(S)$. Then $\operatorname{span}(S)$ is a subspace (by the previous theorem) that contains every vector of $S \cup \{v\}$, so it contains $\operatorname{span}(S \cup \{v\})$ by the same theorem. Together with the first paragraph, the two spans are equal. $\blacksquare$
This is a visual argument, not a formal proof. Two arrows $u$ and $v$ span a plane. Add a third arrow $x$ that already lies on that plane, and every chain that uses it still ends on the plane. Tip $x$ off the plane, and the chains climb out:
An arrow already in the span is a chain of the old arrows, so anything built with it can be rebuilt without it. An arrow outside the span is a place the old arrows couldn't reach, so the span has to grow. The picture starts from a plane; starting from a line or from $\{0\}$ works the same way. The full proof is in the n version.
Build some spans yourself. Each tap puts an arrow into $S$ or takes it out, and the picture redraws $\operatorname{span}(S)$ along with a count of arrows against directions.
Different sets can span the same thing. Find one that spans exactly the same plane as the one you're given. The answers here are sets, so this one stays multiple choice.
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No notation for the span is completely standard. Besides span(S), some books write [S] (Hefferon's choice), sp(S), or ⟨S⟩. This site always writes span(S), and span{u, v} for a listed set.
Lots of different sets span the same subspace, and the theorem above says which vectors are dead weight: the ones already in the span of the others. Throw those out and you're left with the smallest set that still does the job. Telling when that's happened is the next page.
Back to the superhero. In the plane, the villain never stood a chance.
(a) The hoverboard rides along $h = (2, 1)$ and the magic carpet along $m = (1, 3)$. Is there anywhere in $\R^2$ the villain can hide?
(b) In space, the hero has a hoverboard $h = (1, 2, 0)$, a magic carpet $m = (0, 1, 1)$, and a scooter $s = (2, 5, 1)$. Is there anywhere in $\R^3$ the villain can hide?
(a) Take any spot $(x, y)$ and solve $a(2, 1) + b(1, 3) = (x, y)$:
$$\begin{aligned} 2a + b &= x \\ a + 3b &= y. \end{aligned}$$The second equation gives $a = y - 3b$; putting that into the first gives $2y - 5b = x$, so $b = \frac{2y - x}{5}$ and $a = \frac{3x - y}{5}$. Those work for every $x$ and $y$, so $h$ and $m$ span $\R^2$: no hiding place. The villain at $(-1, 7)$ gets $a = -2$, $b = 3$, just like in the opening picture.
(b) Three gadgets in three dimensions sounds like plenty. But look: $2h + m = (2, 4, 0) + (0, 1, 1) = (2, 5, 1) = s$. The scooter is two hoverboard rides and a carpet ride, so by the theorem above it adds nothing: $\operatorname{span}\{h, m, s\} = \operatorname{span}\{h, m\}$, a plane. Its equation is $2x - y + z = 0$ (check: $h$ and $m$ both satisfy it).
So the villain hides anywhere off that plane, like $(0, 0, 1)$, where $2x - y + z = 1$. Row reduction agrees. Solving $ah + bm + cs = (0, 0, 1)$:
$$\begin{aligned} a + 2c &= 0 \\ 2a + b + 5c &= 0 \\ b + c &= 1. \end{aligned}$$$R_2 \to R_2 - 2R_1$ gives $b + c = 0$, and then $R_3 \to R_3 - R_2$ gives $0 = 1$. No trip reaches $(0, 0, 1)$.
Swap the superhero for a robot arm whose joints each push the hand along one direction, and (b) is a real design problem: a set of moves that spans only a plane leaves a whole direction the hand can never correct. Engineers check that the moves span the space before they build anything.
Every question so far came right after its idea, which is a hint a quiz never gives you. Down here the hint is gone. First, true or false: a claim from anywhere on this page, and you decide whether it holds.
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Next, a mixed set: every question type on this page, shuffled. Type coefficients on one line, with commas between them, like -2, 3, and a number as itself. The same-span questions stay multiple choice.
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Made it this far? Time for something with a little heat. Two planes through the origin, each handed to you as the span of two vectors, meet in a line. Find it. In free response, type any vector along the line, like 2, -1, 3; any nonzero multiple counts.
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