A line through the origin in $\R^2$ is a pretty small thing compared to the whole plane. But add two arrows that lie along the line and you stay on the line. Stretch one and you stay on the line. So the line is a little vector space living inside a bigger one, borrowing its addition and scaling. This page is about spotting those little vector spaces, about a three-item test that replaces the eight rules, and about what happens when two of them overlap.
Draw a line through the origin. Add two arrows that lie along it, and the sum lies along it too. Stretch one, or flip it around: still on the line. You can't fall off. This page is about the sets in the plane and in space that you can't fall off of, and it turns out there are far fewer kinds of them than you'd guess.
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.
So a subspace moves into someone else's vector space and uses their addition and their scalar multiplication. Rent-free.
Here's the official name for a vector space hiding inside another one.
A subset $W$ of a vector space $\V$ is called a subspace of $\V$ if $W$ is a vector space with the operations of addition and scalar multiplication defined on $\V$.
A set $W$ of vectors in $\R^2$ (or in $\R^3$) is called a subspace if $W$ is a vector space with the usual adding and scaling of $\R^2$ (or of $\R^3$).
The important words are "the operations defined on $\V$." You don't get to invent new operations for $W$. You add and scale exactly the way $\V$ does, and the question is whether $W$ is still a vector space when you do.
The important word is "usual." You don't get to invent a new way of adding for $W$. You add arrows tip to tail and stretch them like always, and the question is whether you ever leave $W$ by doing it.
That word matters more than it looks. Here's a set that was a vector space on the last page and still isn't a subspace.
On the Vector Spaces page, the positive numbers $\R_{>0}$ were a vector space. Are they a subspace of $\R$?
No. $\R_{>0}$ was a vector space only because its "addition" was multiplication, $x + y = xy$, and its "scaling" was a power, $r \cdot x = x^r$. A subspace of $\R$ has to use $\R$'s own adding and scaling, and with those, $(-1) \cdot 2 = -2$ is not positive. It falls out.
Same set, different operations, different answer:
Two subspaces come free with every vector space. $\V$ is a subspace of itself, which is a little silly but true.
Two subspaces come free. $\R^3$ is a subspace of itself, which is a little silly but true.
And the set holding nothing but the zero vector is a subspace too: $0 + 0 = 0$ and $c0 = 0$, so you can never get out of it.
The subspace $\{0\}$, holding only the zero vector, is called the zero subspace.
Now, checking that $W$ is a vector space sounds like checking all eight rules again. It isn't.
On Vector Spaces, a set borrowing the usual operations could only fail in four ways: a sum leaves the set, a multiple leaves the set, $0$ is missing, or some $-u$ is missing. The other six rules are equations that already hold in the bigger space.
The first two of those four have names, because you'll say them constantly.
A set $W$ of vectors is closed under addition if $u + v$ is in $W$ whenever $u$ and $v$ are in $W$.
"Closed" means the door is shut: whatever you build out of vectors in $W$ stays in $W$. Here's a set with a strange shape that passes anyway.
Is the V-shaped region $W = \{(x, y) : y \geq |x|\}$ closed under addition?
Yes. Take $(x_1, y_1)$ and $(x_2, y_2)$ in $W$, so $y_1 \geq |x_1|$ and $y_2 \geq |x_2|$. Adding,
$$y_1 + y_2 \geq |x_1| + |x_2| \geq |x_1 + x_2|,$$since adding two numbers never gives something bigger in size than their sizes added. So the sum $(x_1 + x_2,\ y_1 + y_2)$ is in $W$ too.
The same word works for scaling.
A set $W$ of vectors is closed under scalar multiplication if $cu$ is in $W$ whenever $c$ is a real number and $u$ is in $W$.
The V fails this one: $(0, 1)$ is in it, but $(-1)(0, 1) = (0, -1)$ sits below the point of the V. Being closed under one operation says nothing about the other.
Four conditions is already better than eight, but it gets better. The zero vector of $W$ has no choice but to be the zero vector of $\V$, and additive inverses come for free from scaling by $-1$. That leaves three checks.
Four checks is already better than eight, but it gets better: scaling by $-1$ hands you every "undo" arrow for free. That leaves three checks.
The Subspace Test. Let $\V$ be a vector space and $W$ a subset of $\V$. Then $W$ is a subspace of $\V$ if and only if the following three conditions hold for the operations defined in $\V$.
(a) $0 \in W$.
(b) $u + v \in W$ whenever $u \in W$ and $v \in W$.
(c) $cu \in W$ whenever $c$ is a real number and $u \in W$.
If $W$ is a subspace of $\V$, then $W$ is a vector space with the operations of $\V$. Hence (b) and (c) hold, and there is a vector $0' \in W$ with $u + 0' = u$ for each $u \in W$. But also $u + 0 = u$, so $u + 0' = u + 0$, and $0' = 0$ by the cancellation law. So (a) holds.
Conversely, suppose (a), (b), and (c) hold. Then adding and scaling never leave $W$, $W$ contains $0$, and the six rules that are equations, (VS 1), (VS 2), and (VS 5) through (VS 8), hold in $W$ because they hold for every vector of $\V$. That leaves (VS 4). If $u \in W$, then $(-1)u \in W$ by (c), and $(-1)u = -u$ by the scalar rules. Hence $W$ is a subspace of $\V$. $\blacksquare$
The Subspace Test. A set $W$ of vectors in $\R^2$ or $\R^3$ is a subspace exactly when all three of these hold:
(a) the origin $0$ is in $W$;
(b) $W$ is closed under addition;
(c) $W$ is closed under scalar multiplication.
This is a visual argument, not a formal proof. Take any $u$ in $W$ and slide the stretch $c$ from $2$ down through $0$ to $-1$. If $W$ is closed under scaling, every stop along the way is in $W$:
At $c = -1$ you're holding $-u$, the arrow that undoes $u$. So (c) hands you every undo arrow, and the six rules that are equations hold because they hold everywhere in $\R^3$. Why keep (a), then? Because a set with nothing in it at all passes (b) and (c) without ever being tested. Check (a), and $W$ has at least one vector to slide. The full proof is in the n version.
Eight axioms, down to three checks. Whoever noticed this deserves a raise.
This is the tool for the rest of the course. Whenever someone asks "is this a subspace?", you run (a), (b), (c), in that order.
Always check (a) first. It's one plug-in, and it kills a ton of sets on the spot: if $0$ isn't in $W$, you're done, no closure work needed. Then, for (b) and (c), remember the rule from last time. Showing a condition fails takes one example with actual numbers. Showing it holds takes letters, because it has to work for every $u$, $v$, and $c$.
Here's what that looks like when a set passes.
Show that $W = \{(x, y, z) : 3x - y + 2z = 0\}$ is a subspace of $\R^3$.
(a) Plug in $0 = (0, 0, 0)$: $3 \cdot 0 - 0 + 2 \cdot 0 = 0$. So $0 \in W$.
(b) Let $u = (x_1, y_1, z_1)$ and $v = (x_2, y_2, z_2)$ be in $W$, so $3x_1 - y_1 + 2z_1 = 0$ and $3x_2 - y_2 + 2z_2 = 0$. Then $u + v = (x_1 + x_2,\ y_1 + y_2,\ z_1 + z_2)$, and
$$\begin{aligned} &3(x_1 + x_2) - (y_1 + y_2) + 2(z_1 + z_2) \\ &\quad = (3x_1 - y_1 + 2z_1) + (3x_2 - y_2 + 2z_2) \\ &\quad = 0 + 0 = 0. \end{aligned}$$So $u + v \in W$.
(c) Let $u = (x_1, y_1, z_1) \in W$ and let $c$ be a real number. Then $cu = (cx_1, cy_1, cz_1)$, and
$$\begin{aligned} 3(cx_1) - cy_1 + 2(cz_1) &= c(3x_1 - y_1 + 2z_1) \\ &= c \cdot 0 = 0. \end{aligned}$$So $cu \in W$. All three hold, so $W$ is a subspace of $\R^3$ by the subspace test. $\blacksquare$
Read the equation as a dot product. With $n = (3, -1, 2)$, it says $n \cdot (x, y, z) = 0$: the vectors in $W$ are exactly the ones square to $n$. That's a plane through the origin, and the three checks are three facts about dot products.
(a) $n \cdot 0 = 0$, so $0$ is in $W$.
(b) If $n \cdot u = 0$ and $n \cdot v = 0$, then $n \cdot (u + v) = n \cdot u + n \cdot v = 0 + 0 = 0$.
(c) If $n \cdot u = 0$, then $n \cdot (cu) = c(n \cdot u) = c \cdot 0 = 0$.
Here it is with real numbers, $u = (1, 3, 0)$ and $v = (1, -1, -2)$:
The whole thing hinged on that $= 0$ on the right side. Regroup, and the two zeros add up to zero; pull out the $c$, and $c$ times zero is zero. Change the right side and everything falls apart.
You can fold (b) and (c) into one check: $W$ is closed under both exactly when $cu + dv$ is in $W$ for every $u$ and $v$ in $W$ and every pair of real numbers $c$, $d$. Taking $c = d = 1$ gives (b), and taking $d = 0$ gives (c). Plenty of books test subspaces this way.
The line below is just $y = x$ shifted up ONE unit. One! Can't we just let it in?
Is $W = \{(x, y) : y = x + 1\}$ a subspace of $\R^2$?
Check (a): plug in $(0, 0)$ and you get $0 = 0 + 1$, which is false. So $0$ is not in $W$, and $W$ is not a subspace. Done in one line.
Geometrically, $W$ is a line that misses the origin. Every subspace has to pass through $0$, so a shifted line never makes it. The next two do contain $0$, and they still fail. (The quadrant failed this way on the Vector Spaces page.)
Is the half-plane $W = \{(x, y) : y \geq 2x\}$ a subspace of $\R^2$?
(a) passes: $0 \geq 2 \cdot 0$. (b) passes too: if $y_1 \geq 2x_1$ and $y_2 \geq 2x_2$, adding gives $y_1 + y_2 \geq 2(x_1 + x_2)$.
But (c) fails. Take $u = (0, 1)$, which is in $W$, and $c = -1$. Then $cu = (0, -1)$, and $-1 \geq 0$ is false. So $W$ is not a subspace.
That's the classic trap: a set can survive addition and still die to one negative scalar. And it can go the other way around.
Is $W = \{(x, y) : xy = 0\}$ a subspace of $\R^2$?
$xy = 0$ means one of the entries is $0$, so $W$ is the two axes together. (a) passes: $0 \cdot 0 = 0$. (c) passes: if $xy = 0$ then $(cx)(cy) = c^2 xy = 0$.
But (b) fails. Take $u = (1, 0)$ and $v = (0, 1)$, both in $W$. Then $u + v = (1, 1)$, and $1 \cdot 1 = 1 \neq 0$. So $W$ is not a subspace.
Here are four near misses, each with the one move that sinks it.
Now sort some on your own. For each set, decide before you drag: is it a subspace? If you call one "not," be ready to name the vector that escapes.
Every failure so far came from one number or one sign. Change it, and some of these sets turn into subspaces. Some can't be saved.
Your turn with the test. The answers here are sets and sentences, so this one stays multiple choice.
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Every subspace contains $0$, so in pictures, every subspace of $\R^2$ or $\R^3$ goes through the origin. The friendliest ones are lines.
Show that the line $W = \{t(2, -1, 4) : t \in \R\}$, all multiples of $(2, -1, 4)$, is a subspace of $\R^3$.
(a) $t = 0$ gives $0(2, -1, 4) = (0, 0, 0)$, so $0$ is in $W$.
(b) Two vectors of $W$ look like $s(2, -1, 4)$ and $t(2, -1, 4)$. Their sum is $(s + t)(2, -1, 4)$, another multiple, so it's in $W$.
(c) $c \cdot t(2, -1, 4) = (ct)(2, -1, 4)$, another multiple. So $W$ is a subspace of $\R^3$. $\blacksquare$
Every vector on the line is some number of copies of $(2, -1, 4)$, so adding two of them just adds their counts:
A plane through the origin can be written the same way, as everything you can build out of a couple of arrows. Solve its equation for one letter and the arrows fall out.
Write the plane $W = \{(x, y, z) : x - 2y + z = 0\}$ as all the combinations of two vectors.
Solve for $x$: $x = 2y - z$. Call the two free letters $y = s$ and $z = t$. Then every point of $W$ is
$$(2s - t,\ s,\ t) = s(2, 1, 0) + t(-1, 0, 1).$$So $W = \{s(2, 1, 0) + t(-1, 0, 1) : s, t \in \R\}$. Check one: $s = 1$, $t = 1$ gives $(1, 1, 1)$, and $1 - 2 + 1 = 0$. $\checkmark$
Turning the dials $s$ and $t$ walks a point all over the plane, and it never leaves:
Put it all together and the subspaces of $\R^2$ are: the zero subspace, lines through the origin, and $\R^2$ itself. In $\R^3$: the zero subspace, lines through the origin, planes through the origin, and $\R^3$.
Is that list really complete? Could there be some weird subspace of $\R^3$ that isn't a point, line, or plane? No, and the proof comes on the Bases & Dimension page, once we can count how many directions a subspace has.
So far every subspace on this page has been a point, a line, a plane, or everything. That's not luck.
The Four Kinds. Every subspace of $\R^3$ is one of these: the zero subspace $\{0\}$, a line through $0$, a plane through $0$, or all of $\R^3$. (In $\R^2$ there are three kinds: $\{0\}$, a line through $0$, or all of $\R^2$.)
This is a visual argument, not a formal proof. Start with a subspace $W$. If $0$ is all it holds, it's $\{0\}$. Otherwise grab a vector $u \neq 0$ in $W$: scaling puts the whole line through $u$ into $W$. If that's everything, $W$ is a line. If not, grab a vector $v$ off that line: adding and scaling now reach every point of the plane through $u$ and $v$. If $W$ holds a vector off that plane too, the three directions reach every point in space.
Each grab either finds nothing new or jumps up a whole kind, so there's no room for a fifth kind. The picture leans on two facts that get checked properly later: two directions really do fill a plane, and three directions off each other really do fill space. The full proof is on Bases & Dimension.
Here's the whole family, smallest at the bottom, with a line from each subspace up to the bigger ones that contain it. There are infinitely many lines and planes, so the picture shows a few of each.
On Lines & Planes in Space, a plane was $ax + by + cz = d$. The planes that are subspaces are exactly the ones with $d = 0$.
So what happens when two subspaces overlap? Take two planes through the origin in $\R^3$. The vectors they share form a line through the origin, which is another subspace. That's not a coincidence.
Any intersection of subspaces of a vector space $\V$ is a subspace of $\V$.
Let $\mathcal{C}$ be a collection of subspaces of $\V$, and let $W$ denote the intersection of the subspaces in $\mathcal{C}$. Since every subspace contains the zero vector, $0 \in W$. Let $c$ be a real number and $u, v \in W$. Then $u$ and $v$ are contained in each subspace in $\mathcal{C}$. Because each subspace in $\mathcal{C}$ is closed under addition and scalar multiplication, $u + v$ and $cu$ are contained in each subspace in $\mathcal{C}$. Hence $u + v$ and $cu$ are also contained in $W$, so $W$ is a subspace of $\V$ by the subspace test. $\blacksquare$
If $W_1$ and $W_2$ are subspaces of $\R^3$, then so is their intersection $W_1 \cap W_2$, the vectors that are in both.
This is a visual argument, not a formal proof. Take two vectors that are in both planes. Their sum is in the first plane, because the first plane is closed under adding, and it's in the second plane for the same reason. So it's in both:
Stretches work the same way, and $0$ is in both. Nothing in that argument used planes: it works for any two subspaces, and for three or more. The full proof is in the n version.
If you've seen subgroups, this one should feel familiar: an intersection of subgroups is a subgroup, and the proof is nearly word for word the same.
The reason is really one sentence: anything you build from vectors in all the subspaces lands in each one of them, so it lands in all of them. Finding the intersection in practice means solving the equations together.
Let $W_1 = \{(x, y, z) : x - y + 2z = 0\}$ and $W_2 = \{(x, y, z) : x + y = 0\}$. Describe $W_1 \cap W_2$.
A vector is in $W_1 \cap W_2$ when it satisfies both equations. The second says $y = -x$. Put that into the first: $x - (-x) + 2z = 0$, so $2x + 2z = 0$ and $z = -x$.
So every vector in the intersection looks like $(x, -x, -x) = x(1, -1, -1)$. That means $W_1 \cap W_2 = \{t(1, -1, -1) : t \in \R\}$, a line through the origin.
Check with $t = 1$: $1 - (-1) + 2(-1) = 0$ and $1 + (-1) = 0$. $\checkmark$
The same steps as a picture: two planes, and the one line they share.
Play with two planes yourself. Each slider tilts one of them; drag the picture to turn it around and see the purple line from different sides.
Two different planes through the origin always meet in exactly a line. The only way to get more is to make them the same plane. Lines get in on it too: a line and a plane share either the whole line or just $0$, and so do two different lines. Pick a pair, call what they share, then look.
Now find some intersections by hand. In free response, type a vector's entries separated by commas, like 1, 3, -2.
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So intersections are always subspaces. What about unions, where you throw the two sets together? A union of subspaces always contains $0$, and it's always closed under scalar multiplication: scaling a vector keeps it in whichever subspace it came from. But it usually fails closure under addition. You've already seen it happen: the two axes are the union of two lines through the origin, and $(1, 0) + (0, 1)$ escaped.
Let $W_1$ and $W_2$ be subspaces of a vector space $\V$$\R^3$. Then $W_1 \cup W_2$ is a subspace exactly when one of them contains the other.
If $W_1 \subseteq W_2$, then $W_1 \cup W_2 = W_2$, a subspace; the case $W_2 \subseteq W_1$ is the same. Conversely, suppose neither contains the other. Pick $u \in W_1$ with $u \notin W_2$ and $v \in W_2$ with $v \notin W_1$. If $u + v$ were in $W_1$, then $v = (u + v) + (-1)u$ would be in $W_1$, since $W_1$ is closed; it isn't. If $u + v$ were in $W_2$, then $u = (u + v) + (-1)v$ would be in $W_2$; it isn't. So $u + v \notin W_1 \cup W_2$, and the union is not closed under addition. $\blacksquare$
This is a visual argument, not a formal proof. Put $u$ on one line and $v$ on another. If $u + v$ landed back on $u$'s line, then walking back along $-u$ would put $v$ on that line too, and it isn't. The same goes for $v$'s line. So $u + v$ lands on neither:
When one sits inside the other, like a line lying in a plane, the union is just the bigger one, and that's a subspace. The full proof is in the n version.
So for two different lines, or two different planes, the union is never a subspace. Watch the symbols:
Don't mix up $\cap$ and $\cup$. The intersection $W_1 \cap W_2$ (vectors in both) is always a subspace. The union $W_1 \cup W_2$ (vectors in either) usually isn't, because adding one vector from each side can land you in neither.
Unions fail because they're missing sums. Next page: start from a few vectors and throw in every sum and multiple you can build, the way the plane above was built from $(2, 1, 0)$ and $(-1, 0, 1)$. What comes out is called the span, and it's always a subspace.
Three friends, Ana, Ben, and Cy, settle up after a trip. Write each payment as a vector of money changes, in the order (Ana, Ben, Cy). Ana paying Ben $\$5$ is $(-5, 5, 0)$: Ana is down five, Ben is up five. Money only moves between them, so the three entries of every payment add up to $0$. Every payment lives on the plane $x + y + z = 0$.
Is the set of payments $\{(x, y, z) : x + y + z = 0\}$ a subspace of $\R^3$? What about the set $\{(x, y, z) : x + y + z = 15\}$, the ways to split a $\$15$ gift card among them?
Payments: yes. It's a plane through the origin, and every check has a meaning. Doing nothing is $(0, 0, 0)$, which is a payment. Two payments back to back add up to one combined payment. Doing a payment twice, or undoing it, is scaling by $2$ or by $-1$.
Gift-card splits: no. Doing nothing isn't a split, since $0 + 0 + 0 \neq 15$. And two splits add up to $\$30$ handed out, not $\$15$.
Bookkeepers lean on exactly this. A ledger whose every row of changes adds up to zero balances, and sums of balanced ledgers still balance. The balanced ones form a subspace, which is why checking them is so easy.
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 subspace-test questions stay multiple choice.
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Made it this far? Time for something with a little heat. Two planes through the origin meet in a line, unless they're secretly the same plane wearing a different equation. Find the $k$ that unmasks them.
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