By now you've added and scaled three different kinds of things: vectors in $\R^n$, matrices, and the solutions of a homogeneous system $Ax = 0$. Each time, the same handful of rules made the algebra work. This page writes those rules down and gives anything that obeys them a name, a vector space. Then it asks a sneaky question: what else obeys them? A few of the answers look nothing like arrows. Everything later in the course, from span to eigenvectors, is built on these rules, so this page is about learning them cold.
You can add arrows tip to tail and stretch them, in the plane and in space. You've also added matrices and added solutions of systems, and the same moves worked every time. This page collects the rules those moves follow, and then goes hunting with them. Which collections of arrows can you add and stretch without ever falling out? And what happens if you change what "add" means?
Here is the same move, adding two things, in three places you've already been. Watch how little changes:
The practice is mixed into the lesson: each question sits right after the idea it needs, and the Mixed Practice at the bottom takes that hint away.
The Vectors page used $\R^n$ without fuss. Before generalizing it, here are the official definitions.
An n-tuple is an ordered list $(a_1, a_2, \ldots, a_n)$ of $n$ real numbers. The numbers $a_1, a_2, \ldots, a_n$ are its entries (or components). Two $n$-tuples $(a_1, \ldots, a_n)$ and $(b_1, \ldots, b_n)$ are equal if $a_i = b_i$ for every $i = 1, 2, \ldots, n$.
"Ordered" is doing real work there: $(1, 2)$ and $(2, 1)$ are different 2-tuples, because their first entries don't match. The set of all $n$-tuples of real numbers is $\R^n$. So $\R^2$ is the plane, $\R^3$ is space, and $\R^5$ is five numbers in a row: you can't picture it, but you can compute with it just fine. Both operations work one entry at a time.
The coordinatewise operations on $\R^n$ are defined as follows. If $u = (a_1, a_2, \ldots, a_n)$ and $v = (b_1, b_2, \ldots, b_n)$ are in $\R^n$ and $c$ is a real number, then
$u + v = (a_1 + b_1,\ a_2 + b_2,\ \ldots,\ a_n + b_n)$ and $cu = (ca_1,\ ca_2,\ \ldots,\ ca_n)$.
A vector in $\R^2$ is an ordered pair and a vector in $\R^3$ is an ordered triple. "Ordered" matters: $(1, 2)$ and $(2, 1)$ are different points, so they're different vectors. Adding and stretching work one entry at a time, which is exactly what you did with arrows on the Vectors page.
The coordinatewise operations on $\R^3$: if $u = (a_1, a_2, a_3)$ and $v = (b_1, b_2, b_3)$ and $c$ is a real number, then
$u + v = (a_1 + b_1,\ a_2 + b_2,\ a_3 + b_3)$ and $cu = (ca_1,\ ca_2,\ ca_3)$.
In $\R^2$ it's the same with the third entry left off.
Long lists get cramped written sideways, so you'll often see them stood up on end.
A column vector is an $n$-tuple written vertically:
$\begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix}$ instead of the row vector $(a_1, a_2, \ldots, a_n)$.
Same object, different handwriting. $(2, -1, 4)$ and $\begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}$ are the same vector, and you add and scale column vectors the same way, top to bottom. Here's the arithmetic once, slowly.
Here's the arithmetic once, slowly.
In $\R^3$, compute $(3, -2, 0) + (-1, 1, 4)$ and $-5(1, -2, 0)$.
Add matching entries: $(3 + (-1),\ -2 + 1,\ 0 + 4) = (2, -1, 4)$.
Multiply every entry by $-5$: $(-5 \cdot 1,\ -5 \cdot (-2),\ -5 \cdot 0) = (-5, 10, 0)$. Watch the middle entry: negative times negative is positive.
Longer expressions are where the slips happen.
In an expression like $3u - 2v$, scale first, then add. Find $3u$ and $-2v$ as their own vectors, then add them entry by entry. Doing it all in one go is where the sign slips sneak in, especially on entries like $-2 \cdot (-5) = 10$.
Your turn: scale, then add.
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Here's the big move. Instead of saying what a vector is, linear algebra says what you can do with vectors: add two of them and scale one, following eight rules. Any set where that works gets to be called a vector space.
A vector space $\V$ (over $\R$) is a set with two operations, called addition and scalar multiplication, so that for each pair of elements $x, y$ in $\V$ there is a unique element $x + y$ in $\V$, and for each real number $a$ and each $x$ in $\V$ there is a unique element $ax$ in $\V$ (adding and scaling never leave $\V$), and the following conditions hold.
Take a set $\V$ of points in $\R^2$ or $\R^3$, together with a way to add two of its points and a way to scale one by a real number. (Usually these are the coordinatewise operations, but not always.) $\V$ is a vector space when adding and scaling never leave $\V$, and these eight rules hold for all $x, y, z$ in $\V$ and all real numbers $a, b$:
Nothing in the eight rules needs the scalars to be real numbers, only that they can be added, multiplied, and divided like real numbers: any field $\mathbb{F}$ works. Over $\mathbb{F}_2 = \{0, 1\}$, the triples $(\mathbb{F}_2)^3$ are three light switches, and adding a vector to itself gives $0$: flip the same switches twice and you're back where you started. The same definition with a ring in place of the field gives a module.
If you've seen groups, look at (VS 1) through (VS 4) again: they say exactly that $\V$ with $+$ is an abelian group. A vector space is an abelian group that also knows how to be scaled.
Read them in two batches. (VS 1) through (VS 4) are only about adding: order doesn't matter, grouping doesn't matter, there's a "do nothing" vector $0$, and you can always undo an addition. (VS 5) through (VS 8) are about how scaling gets along with everything else: scaling by $1$ changes nothing, scaling twice is scaling by the product, and scaling distributes in both directions.
Every one of them is a picture you can check with arrows. Here are all eight, one after another:
This is a visual argument, not a formal proof. With the usual adding (tip to tail) and stretching, each rule is a fact about parallelograms and stretched arrows:
Each rule is drawn for one choice of arrows, and one drawing never covers every case. What covers every case is that the operations work one entry at a time, so every rule comes down to a fact about plain numbers in each entry: $a + b = b + a$, $c(a + b) = ca + cb$, and so on. The n version writes that out for (VS 7).
On Matrices you checked these same eight rules for matrices, entry by entry. So the set $M_{m \times n}(\R)$ of all $m \times n$ real matrices is a vector space, and a matrix is a vector that happens to be shaped like a rectangle.
On Matrices, matrices added and scaled entry by entry and followed these same eight rules. That's no accident: a matrix is a team of arrows, its columns, and everything happens one column at a time.
The two kinds of things in a vector space get their own names.
In a vector space, the real numbers we multiply by are called scalars.
A scalar "scales": it stretches, shrinks, or flips. And the things being scaled get the name you'd expect.
The elements of a vector space $\V$ are called vectors.
Careful: "vector" now means any element of a vector space, not "an arrow with a length and a direction." Arrows are one example. Matrices are another, and so are the solutions of a homogeneous system. Later on this page, even positive numbers get a turn.
Our main example, $\R^n$ with the coordinatewise operations, is a vector space. Checking all eight rules sounds like a lot, but every one comes down to a fact about plain real numbers, one entry at a time. Here's one, so you can see the pattern.
One example of a rule, checked with numbers and then with a picture.
Show that $\R^n$ with the coordinatewise operations satisfies (VS 7).
Let $x = (a_1, \ldots, a_n)$ and $y = (b_1, \ldots, b_n)$, and let $c$ be a real number. Work out both sides.
$c(x + y) = c(a_1 + b_1, \ldots, a_n + b_n) = \big(c(a_1 + b_1), \ldots, c(a_n + b_n)\big)$
$cx + cy = (ca_1, \ldots, ca_n) + (cb_1, \ldots, cb_n) = (ca_1 + cb_1, \ldots, ca_n + cb_n)$
Entry $i$ is $c(a_i + b_i)$ on the left and $ca_i + cb_i$ on the right. Those are equal because real numbers distribute. Every entry matches, so $c(x + y) = cx + cy$. $\blacksquare$
Check (VS 7) for $x = (1, 2)$, $y = (3, -1)$, and $c = 2$.
Left side: $x + y = (4, 1)$, so $2(x + y) = (8, 2)$.
Right side: $2x = (2, 4)$ and $2y = (6, -2)$, so $2x + 2y = (8, 2)$. They match.
One example isn't a reason, but the picture is. Doubling a parallelogram gives a parallelogram twice as big with the same shape, so the doubled diagonal is the diagonal of the doubled sides:
The other seven go the same way: write both sides out entry by entry and point at the real-number rule that makes them match. (VS 1), for instance, is just $a_i + b_i = b_i + a_i$ in every entry, and the zero vector in (VS 3) is $(0, 0, \ldots, 0)$. That gives the first entry in a list of vector spaces you already know.
So $\R^2$ and $\R^3$ are vector spaces. A few smaller sets are too, and the smallest of all gets a name.
The trivial space $\{0\}$ is the vector space with only one vector, its zero vector, where $0 + 0 = 0$ and $a0 = 0$ for every scalar $a$.
A vector space can't be any smaller: (VS 3) demands at least one vector, the zero vector. Here's the list.
Each of these is a vector space:
(a) $\R^n$, with the coordinatewise operations.
(b) $M_{m \times n}(\R)$, all $m \times n$ real matrices, with entrywise addition and scaling.
(c) The solution set of a homogeneous system $Ax = 0$ in $n$ unknowns, with the operations of $\R^n$.
(d) The trivial space $\{0\}$.
(a) Each rule comes down to the matching rule for real numbers in every entry, as in the (VS 7) example; the zero vector is $(0, \ldots, 0)$ and the partner in (VS 4) is $(-a_1, \ldots, -a_n)$.
(b) An $m \times n$ matrix is a list of $mn$ real numbers, added and scaled one entry at a time, so the argument in (a) applies word for word.
(c) Call the solution set $S$. If $x$ and $y$ are in $S$ and $c$ is real, then $A(x + y) = Ax + Ay = 0 + 0 = 0$ and $A(cx) = c(Ax) = c0 = 0$, so adding and scaling never leave $S$. The zero vector of $\R^n$ is in $S$, since $A0 = 0$, and for $x$ in $S$ its partner $-x = (-1)x$ is in $S$ by what we just showed. The six rules that are equations, (VS 1), (VS 2), and (VS 5) through (VS 8), hold for all vectors of $\R^n$, so in particular for those in $S$.
(d) Every rule reads $0 = 0$. $\blacksquare$
Each of these is a vector space:
(a) $\R^2$ and $\R^3$, with the coordinatewise operations.
(b) The $2 \times 2$ matrices, and the $3 \times 3$ matrices, with entrywise adding and scaling.
(c) A line or a plane through $0$, cut out by equations with $0$ on the right side (like $x - 2y + z = 0$), with the usual operations.
(d) The trivial space $\{0\}$.
This is a visual argument, not a formal proof. Parts (a) and (b) are the eight-rules picture above, one arrow (or one column) at a time. Part (c) is the interesting one. Take the plane $x - 2y + z = 0$ and two points on it, $p = (1, 1, 1)$ and $q = (2, 1, 0)$. Their sum and their stretches stay on the plane. Then slide the plane off the origin, to $x - 2y + z = 1$, and watch the sum fall off:
Why the difference? With $0$ on the right, plugging in a sum gives $0 + 0 = 0$, and plugging in $c$ times a point gives $c \cdot 0 = 0$: still on the plane. With $1$ on the right, a sum gives $1 + 1 = 2$, too much. The rules that are equations (like $x + y = y + x$) hold for all of $\R^3$, so they hold on the plane too. Part (d) is one point where every rule reads $0 = 0$. The full proof is in the n version.
On Solution Sets, sums and multiples of solutions of a homogeneous system were solutions again. That's exactly what "adding and scaling never leave the set" asks for.
Part (c) of the theorem is a pattern worth stealing. A set that sits inside a bigger vector space can borrow its adding and scaling, and then most of the eight rules come for free, because they're equations that already hold everywhere. Here's a line, checked by hand.
Is the line $y = 3x$ in $\R^2$ a vector space, with the usual operations?
Its points are $(s, 3s)$ for real numbers $s$. Add two: $(s, 3s) + (t, 3t) = (s + t,\ 3(s + t))$, which is on the line. Scale one: $c(s, 3s) = (cs,\ 3(cs))$, also on the line. The zero vector $(0, 0)$ is on the line ($s = 0$), and the partner of $(s, 3s)$ is $(-s, 3(-s))$, on the line too.
The other six rules are equations that hold for every vector in $\R^2$, so they hold on the line. So yes: the line is a vector space.
Now nudge the line up by $1$, to $y = 3x + 1$, and the same check fails on the very first step:
That leaves only a short list of things that can go wrong.
When a set borrows the usual operations, only four things can fail: a sum leaves the set, a multiple leaves the set, $0$ is missing, or some $-x$ is missing. Check those four. The other rules are equations that hold in the bigger space already, so there's nothing to check.
Here are four sets that each fail one of the four, for different reasons.
With the usual operations, decide which of these is a vector space: (a) the points of $\R^2$ with whole-number coordinates; (b) the plane $x + y + z = 1$ in $\R^3$; (c) the first quadrant, $x \ge 0$ and $y \ge 0$; (d) the unit circle $x^2 + y^2 = 1$.
(a) No. $(4, 3)$ is in it, but $\frac{1}{2}(4, 3) = (2, 1.5)$ isn't: scaling leaves the set.
(b) No. $(1, 0, 0)$ and $(0, 0, 1)$ are on the plane, but their sum $(1, 0, 1)$ gives $1 + 0 + 1 = 2$, not $1$. (It doesn't even contain $0$.)
(c) No. Adding never leaves it, but $(-1)(2, 1) = (-2, -1)$ does, and $(2, 1)$ has no partner in the quadrant to cancel it.
(d) No. $(1, 0) + (0, 1) = (1, 1)$, which is $\sqrt{2}$ from the origin, off the circle.
None of them is a vector space. One escape with actual numbers settles each one.
Here are the escapes from (a), (c), and (d):
Now you try to escape. Each mission draws a set; drag $u$ and $v$ and slide $c$ until a sum or a multiple leaves it.
Every set on this page that passed was sitting inside a bigger vector space and borrowing its operations. On Subspaces those get a name, and the check shrinks to three items.
Lines and planes through $0$ kept passing, and the ones that miss $0$ kept failing. On Subspaces it turns out that's the whole story in $\R^3$: the only sets that pass are $\{0\}$, lines through $0$, planes through $0$, and all of $\R^3$.
So far the operations were always the usual ones. But a vector space isn't just a set; it's a set plus its two operations. Change the operations and the same set $\R^2$ can stop being a vector space. Here are two rule-breakers.
On $\R^2$, define $(a_1, a_2) + (b_1, b_2) = (a_1 + b_1,\ a_2 - b_2)$ and $c(a_1, a_2) = (ca_1,\ ca_2)$. Show that (VS 1) fails.
Try $x = (1, 2)$ and $y = (1, 5)$. Then
$x + y = (1 + 1,\ 2 - 5) = (2, -3)$ but $y + x = (1 + 1,\ 5 - 2) = (2, 3)$.
Those aren't equal, so (VS 1) fails, and $\R^2$ is not a vector space with these operations. (VS 2) and (VS 8) fail too, but one broken rule is already enough.
The two orders land on mirror images of each other, reflected across the $x$-axis:
Notice what the counterexample needed: the second entries of $x$ and $y$ had to be different. If you'd tried $x = (1, 2)$ and $y = (7, 2)$, both orders give $(8, 0)$, and you'd have learned nothing. Here's the other rule-breaker.
On $\R^2$, define $(a_1, a_2) + (b_1, b_2) = (a_1 + b_1,\ 0)$ and $c(a_1, a_2) = (ca_1,\ 0)$. Show that (VS 5) fails.
Take $x = (3, 4)$. Then $1x = (1 \cdot 3,\ 0) = (3, 0)$, which is not $(3, 4)$. So $1x \neq x$, and (VS 5) fails.
(VS 3) fails here too: $x + z$ always has second entry $0$, so no $z$ can give back $(3, 4)$. With no zero vector, (VS 4) doesn't even make sense.
Every sum and every multiple gets squashed onto the $x$-axis, so $(3, 4)$ can never be reached again:
(VS 5) looks too obvious to bother writing down. The second example is why it's there: without it, "scalar multiplication" could secretly squash every vector flat and still pass most of the other rules.
Each mission below starts with the rule working, because the two sides happen to agree for the starting vectors. Drag until they don't.
Proving a rule fails takes one counterexample with actual numbers. Proving a rule holds takes letters: you have to show it works for every $x$, $y$, $a$, $b$ at once. Testing three examples that happen to work proves nothing. The first rule-breaker above passed (VS 1) for $(1, 2)$ and $(7, 2)$!
Your turn to judge. Each rule below changes the operations on $\R^2$. Decide whether you still get a vector space, then watch all eight rules get checked. At least one of them will surprise you.
Your turn: find the rule that breaks.
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Rule D in that widget is the surprise: strange operations, and still all eight rules, with a zero vector at $(1, 0)$. Here's a stranger one, where "adding" doesn't add at all.
Let $\R_{>0}$ be the positive real numbers, with new operations: "adding" $x$ and $y$ means multiplying, $x + y = xy$, and "scaling" $x$ by a real number $r$ means taking a power, $rx = x^r$. So in this space $2 + 3 = 6$ and $2 \cdot 3 = 3^2 = 9$. Find the zero vector and the partner of $x$, and check (VS 8).
The zero vector $z$ has to do nothing: $x + z = x$, which here says $xz = x$. So $z = 1$. The number $1$ is this space's zero vector.
The partner $y$ of $x$ has to bring it back to the zero vector: $x + y = 1$, which says $xy = 1$. So the partner of $x$ is $\frac{1}{x}$. For example, the partner of $4$ is $\frac{1}{4}$.
(VS 8) asks whether $(a + b)x = ax + bx$. The left side is $x^{a + b}$, and the right side is $x^a$ "plus" $x^b$, which is $x^a x^b$. Those are equal by the exponent rules. Every other rule is an exponent rule too, and products and powers of positive numbers stay positive, so $\R_{>0}$ with these operations is a vector space.
Why does something this odd work? A log ruler explains it: on a ruler marked by logarithms, multiplying is laying lengths end to end, and taking a power is stretching a length. So this space is just the number line in disguise:
In a strange space, find the zero vector by solving $x + z = x$ with that space's own adding, and never assume it's $0$ or $(0, 0)$. Here it was the number $1$, and in Rule D it was the point $(1, 0)$. The partner of $x$ works the same way: solve $x + y = $ (the zero vector).
What if "adding" meant averaging? $x + y = \frac{x + y}{2}$. It's fair, it commutes... but which $z$ does nothing to $x$? The average of $x$ and $z$ is $x$ only when $z = x$. So every vector would need its own personal zero, and (VS 3) says no.
Your turn: compute in a space with strange operations.
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Once a set passes all eight rules, a bunch of other facts come along at no extra charge. You never check them separately; they're consequences of the eight, so they hold in every vector space, strange ones included. The first one is the vector version of "subtract the same thing from both sides."
Cancellation Law for Vector Addition. If $x$, $y$, and $z$ are vectors in a vector space $\V$ such that $x + z = y + z$, then $x = y$.
By (VS 4) there is a vector $v$ in $\V$ with $z + v = 0$. Then
$x = x + 0 = x + (z + v) = (x + z) + v = (y + z) + v = y + (z + v) = y + 0 = y$
by (VS 2) and (VS 3). $\blacksquare$
This is a visual argument, not a formal proof. $x + z$ and $y + z$ are the same point. Now walk back from that point along $-z$. There's only one walk and only one place to land, and it's both $x$ and $y$:
The picture uses ordinary arrows, but the walk back only uses the rules: add the partner of $z$ to both sides, regroup, and let the zero vector clean up. So it works in strange spaces too. In $\R_{>0}$ it says: if $xz = yz$, then $x = y$. The full proof is in the n version.
The trick in that proof is worth stealing: add the "undo" vector $v$ to both sides, regroup with (VS 2), and let (VS 3) clean up. Two quick consequences:
Walking back along $-z$ is a trick worth stealing. Two quick consequences:
(a) A vector space has only one zero vector: the vector $0$ described in (VS 3) is unique. (b) Each vector $x$ has only one partner $y$ with $x + y = 0$.
(a) Suppose $0$ and $0'$ both satisfy (VS 3). Then $0 + 0' = 0$ (because $0'$ does nothing) and $0' + 0 = 0'$ (because $0$ does nothing). By (VS 1) the left sides are equal, so $0 = 0'$.
(b) Suppose $x + y = 0$ and $x + y' = 0$. By (VS 1), $y + x = y' + x$, and the cancellation law gives $y = y'$. $\blacksquare$
This is a visual argument, not a formal proof. (a) Suppose two different points both "do nothing." Add them to each other. The first does nothing to the second, so the sum is the second; the second does nothing to the first, so the sum is the first. One sum, so they were the same point all along. (b) If $y$ and $y'$ both bring $x$ home to $0$, then $x + y$ and $x + y'$ are the same point, and walking back along $-x$, as in the picture above, lands $y$ and $y'$ on the same spot. The full proof is in the n version.
So there's exactly one "do nothing" vector, and every vector has exactly one "undo" partner. Since they're unique, they get names.
The vector $0$ in (VS 3) is called the zero vector of $\V$.
In $\R^n$ it's $(0, 0, \ldots, 0)$In $\R^2$ and $\R^3$ it's the origin, but in $\R_{>0}$ it was the number $1$. The "undo" partner gets a name too.
The unique vector $y$ in (VS 4) with $x + y = 0$ is called the additive inverse of $x$ and is denoted by $-x$.
Last free gift: the scalar rules you'd hope for are all true.
In any vector space $\V$, the following statements are true:
(a) $0x = 0$ for each $x \in \V$.
(b) $(-a)x = -(ax) = a(-x)$ for each real number $a$ and each $x \in \V$.
(c) $a0 = 0$ for each real number $a$.
(a) By (VS 8), (VS 3), and (VS 1),
$0x + 0x = (0 + 0)x = 0x = 0x + 0 = 0 + 0x.$
Hence $0x = 0$ by the cancellation law.
(b) The vector $-(ax)$ is the unique vector with $ax + [-(ax)] = 0$. So if $ax + (-a)x = 0$, uniqueness gives $(-a)x = -(ax)$. And by (VS 8) and part (a),
$ax + (-a)x = [a + (-a)]x = 0x = 0.$
So $(-a)x = -(ax)$. In particular, $(-1)x = -x$. Then by (VS 6),
$a(-x) = a[(-1)x] = [a(-1)]x = (-a)x.$
(c) By (VS 7), (VS 3), and (VS 1), $a0 + a0 = a(0 + 0) = a0 = a0 + 0 = 0 + a0$, so $a0 = 0$ by the cancellation law. $\blacksquare$
This is a visual argument, not a formal proof. Turn a dial $c$ from $1$ down to $-1$ and watch $cx$. At $c = 0$ the arrow shrinks to a single point, the zero vector, and at $c = -1$ it has flipped around into $-x$. Then the same dial in Rule D's strange space, whose zero vector is $(1, 0)$: at $c = 0$ the point lands on $(1, 0)$, that space's zero, not on the origin.
So "$0x = 0$" means "scaling by the number zero lands on this space's zero vector," wherever that is. The full proof, from the eight rules alone, is in the n version.
Read (a) slowly: the $0$ on the left is the number zero, and the $0$ on the right is the zero vector. And (b) with $a = 1$ says $(-1)x = -x$: the additive inverse is just "scale by $-1$."
These rules are exactly what you need to solve equations where the unknown is a vector. It's the same algebra you've always done, except now every step has a rule behind it.
In $\R^2$, find the vector $x$ with $3x + (1, -2) = (7, 4)$.
Add the additive inverse $-(1, -2) = (-1, 2)$ to both sides (VS 4 and VS 3): $3x = (7, 4) + (-1, 2) = (6, 6)$.
Multiply both sides by $\frac{1}{3}$. By (VS 6) and (VS 5), $\frac{1}{3}(3x) = \left(\frac{1}{3} \cdot 3\right)x = 1x = x$, so $x = \frac{1}{3}(6, 6) = (2, 2)$.
Check: $3(2, 2) + (1, -2) = (6, 6) + (1, -2) = (7, 4)$. $\checkmark$
The same three steps as moves: slide back by $(1, -2)$, shrink to a third, then rebuild to check.
Your turn: undo the operations on $x$.
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Engineers and physicists keep track of units. A speed is a length per time, written $\mathrm{L}\mathrm{T}^{-1}$; a force is a mass times an acceleration, $\mathrm{M}\mathrm{L}\mathrm{T}^{-2}$. Every unit built from length, mass, and time looks like $\mathrm{L}^a \mathrm{M}^b \mathrm{T}^c$, so you can record it as the vector $(a, b, c)$ of its exponents.
| quantity | units | vector $(a, b, c)$ |
|---|---|---|
| length $\ell$ | $\mathrm{L}$ | $(1, 0, 0)$ |
| mass $m$ | $\mathrm{M}$ | $(0, 1, 0)$ |
| time $t$ | $\mathrm{T}$ | $(0, 0, 1)$ |
| speed | $\mathrm{L}\mathrm{T}^{-1}$ | $(1, 0, -1)$ |
| gravity $g$ | $\mathrm{L}\mathrm{T}^{-2}$ | $(1, 0, -2)$ |
| force | $\mathrm{M}\mathrm{L}\mathrm{T}^{-2}$ | $(1, 1, -2)$ |
Multiplying two quantities adds their exponents, so it adds their vectors: mass times gravity is $\mathrm{M} \cdot \mathrm{L}\mathrm{T}^{-2}$, and $(0, 1, 0) + (1, 0, -2) = (1, 1, -2)$, a force. Raising a quantity to a power $r$ multiplies every exponent by $r$, so it scales the vector. That's the $\R_{>0}$ trick again: "adding" is multiplying and "scaling" is a power. Units form a vector space, and it's a copy of $\R^3$ with $\mathrm{L}$, $\mathrm{M}$, and $\mathrm{T}$ as its three axes. Here's what that buys you.
A pendulum's period $p$ (a time) might depend on the string's length $\ell$, the bob's mass $m$, and gravity $g$. If $p = k\,\ell^a m^b g^c$ for some number $k$ with no units, what must $a$, $b$, and $c$ be?
The units of $\ell^a m^b g^c$ are the vector
$a(1, 0, 0) + b(0, 1, 0) + c(1, 0, -2) = (a + c,\ b,\ -2c),$
and they have to match the units of a time, $(0, 0, 1)$. So $a + c = 0$, $b = 0$, and $-2c = 1$. That gives $c = -\frac{1}{2}$, $a = \frac{1}{2}$, and $b = 0$:
$p = k\,\ell^{1/2} g^{-1/2} = k\sqrt{\dfrac{\ell}{g}}.$
Two things fall out for free. The mass doesn't matter at all, since nothing else in the list has an $\mathrm{M}$ to cancel it. And a string four times as long swings with twice the period. (Physics finds $k \approx 2\pi$ for small swings; units alone can't.)
In the space of units, that's a walk along $\ell$ and $g$ that has to land exactly on the time axis:
Matching the units was a linear system in $a$, $b$, $c$, one equation per unit. With more quantities than units it has free variables, and row reduction finds every combination that works at once.
The toolkit for this page: a vector space is a set with an adding and a scaling that never leave it and follow the eight rules. With the usual operations, check that sums, multiples, $0$, and every $-x$ stay inside. With strange operations, test each rule with actual numbers, and find the zero vector by solving $x + z = x$. And once the eight rules hold, cancellation, $0x = 0$, and $(-1)x = -x$ come free.
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 a vector on one line, with commas between entries, like 2, -1, 4, and a number as itself. The questions about the rules stay multiple choice.
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Made it this far? Time for something with a little heat. The pendulum had exactly as many unknown exponents as units, so the units pinned them down. Here are more quantities from physics, each with its own unknown exponents.
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