Linear algebra is the math of lists of numbers that you can add and stretch. A vector in $\R^n$ is just $n$ numbers in order, like $(2, -1, 3)$ in $\R^3$ or $(1, 0, 4, -2)$ in $\R^4$. That's the whole object. What makes it worth a course is the picture: a vector is a move, an arrow you can slide anywhere, and the arithmetic of lists turns out to be the geometry of arrows. This page builds that picture, so every later page has something to see.
Linear algebra is the math of lists of numbers that you can add and stretch. A vector is just numbers in order: two for the plane, like $(2, -1)$, or three for space, like $(2, -1, 3)$. (The set of all two-entry vectors is called $\R^2$, and the set of all three-entry vectors is $\R^3$.) That's the whole object. What makes it worth a course is the picture: a vector is a move, an arrow you can slide anywhere, and the arithmetic of lists turns out to be the geometry of arrows. This page builds that picture, so every later page has something to see.
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. Flip the switch on a question to type your answer instead: vectors and points as entries like (2, -1, 4), and fractions like 2/3.
On the precalculus Vectors page, a vector was an arrow in the plane, written with a little arrow on top and its components in parentheses. Linear algebra keeps the idea, drops the arrow, and lets the list be as long as it likes: plain letters, any number of entries.
On the precalculus Vectors page, a vector was an arrow in the plane, written with a little arrow on top and its components in parentheses. Linear algebra keeps the idea and drops the arrow: plain letters, and a third entry when the arrow lives in space.
Read $(2, 1, 1)$ as instructions: go $2$ in the $x$ direction, $1$ in $y$, $1$ in $z$. Those instructions don't care where you start. Follow them from anywhere and you've made the same move:
Every arrow in that loop is the same vector. The one that starts at the origin is special only because its tip lands on the point $(2, 1, 1)$. That's why linear algebra writes points and vectors the same way: the point $(2, 1, 1)$ is where the vector $(2, 1, 1)$ takes you from $0$.
A vector is an arrow I'm allowed to pick up and carry anywhere, as long as I don't turn it or stretch it. The world's strictest moving company.
Run it backward. If you know where an arrow starts and where it ends, which move is it? Whatever gets you from the start to the end: the end minus the start.
The displacement vector from the point $A$ to the point $B$ in $\R^n$ is $B - A$.
The displacement vector from the point $A$ to the point $B$ (in the plane or in space) is $B - A$.
Tip minus tail, every time. It's the right move because $A + (B - A) = B$: start at $A$, make the move, and you're at $B$.
(a) Find the vector from $A = (1, 4, -2)$ to $B = (3, -1, 5)$. (b) The vector $(4, 0, -3)$ starts at $(-1, 2, 2)$. Where does it end? (c) The vector $(1, 2, 3)$ ends at $(5, 5, 5)$. Where does it start?
(a) $B - A = (3 - 1, \ -1 - 4, \ 5 - (-2)) = (2, -5, 7)$.
(b) Start plus move: $(-1, 2, 2) + (4, 0, -3) = (3, 2, -1)$.
(c) End minus move: $(5, 5, 5) - (1, 2, 3) = (4, 3, 2)$. Check: $(4, 3, 2) + (1, 2, 3) = (5, 5, 5)$.
All three answers are one relation, $A + v = B$, each time missing a different piece. Watch each part happen in space:
Drag the two points and watch the move between them. The missions are about which part of the picture the vector actually cares about:
The second mission is the point of this whole section: you moved both points and the vector didn't change, because the vector only knows the move.
Tail minus tip, $A - B$, gives an arrow of the right length pointing exactly backward, and it's the most common slip on this page. Say "tip minus tail" out loud. (Other books write the vector from A to B as AB with an arrow over it. Same thing as B − A.)
Your turn. Some ask for the move, some for an endpoint.
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Vectors add entry by entry, $(1, 2, 0) + (3, -1, 4) = (4, 1, 4)$, and a number scales every entry, $3(1, 2, 0) = (3, 6, 0)$. That's all the arithmetic there is. (The official definitions live on Vector Spaces.) The pictures are what's new. Adding two moves means making one and then the other, head to tail:
Either order lands on the same corner, which is why $u + v = v + u$, and the two routes outline a parallelogram with $u + v$ as its diagonal. In $\R^5$ there's no picture to draw, but the rule is the same: walk one move, then the other, entry by entry.
Let $u = (2, 0, 1)$ and $v = (0, 2, 1)$. Find $u + v$, and check that walking $v$ first gets you to the same place.
$u + v = (2 + 0, \ 0 + 2, \ 1 + 1) = (2, 2, 2)$. The other way, $v + u = (0 + 2, \ 2 + 0, \ 1 + 1) = (2, 2, 2)$. Same corner of the parallelogram.
These are exactly the two arrows in the animation above: watch the parallelogram close on $(2, 2, 2)$ whichever way you walk it.
Scaling is just as visual. Multiplying by $c$ stretches the arrow by $|c|$, and a negative $c$ turns it around. Watch the tips of $cv$ as $c$ slides:
The tips never leave one straight line through the origin. Every multiple of $v$ lives on that line, and every point of the line is a multiple of $v$. Hold on to that; it comes back twice.
That line, all the multiples $cv$, is where Lines & Planes in Space begins: slide it off the origin by a point $p$ and you get every line in space, $p + tv$. On Span it gets a name, $\operatorname{span}\{v\}$.
Subtraction is the last picture, and it's tip minus tail in disguise. $u - v$ is the move that takes you from the tip of $v$ to the tip of $u$, since $v + (u - v) = u$:
So $u - v$ runs tip to tip, from $v$ toward $u$, and $v - u$ is the same arrow flipped. Both pictures in one example:
Let $u = (3, 1)$ and $v = (1, 2)$. Find $u - v$ and $v - u$, and say where each one runs when $u$ and $v$ are drawn from the origin.
$u - v = (3 - 1, \ 1 - 2) = (2, -1)$: the arrow from the tip of $v$, $(1, 2)$, to the tip of $u$, $(3, 1)$. Check: $(1, 2) + (2, -1) = (3, 1)$.
$v - u = (-2, 1)$: the same arrow pointing back, from the tip of $u$ to the tip of $v$.
Now match expressions to arrows. Each arrow is drawn wherever it's most natural, not always from the origin, so you'll need the head-to-tail and tip-to-tip pictures:
$\R^4$ has a fourth axis and I have no idea which way to point it. The arithmetic doesn't care.
Walking $-v$ is just walking $v$ backward. I've decided to call it moonwalking and nobody can stop me.
Your turn, from a picture. Read $u$ and $v$ off the grid, then find the arrow.
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Adding and scaling arrows obey eight rules you've been using without thinking: $u + v = v + u$, $c(u + v) = cu + cv$, and six more. On Vector Spaces, those eight rules become the definition, and suddenly polynomials and matrices count as vectors too.
Adding and scaling arrows obey eight rules you've been using without thinking: $u + v = v + u$, $c(u + v) = cu + cv$, and six more. On Vector Spaces, those eight rules take center stage: each one is a picture about parallelograms and stretching, and that page checks them all.
Two arrows on that line through the origin point along the same line, possibly in opposite directions. That's the idea of parallel, and the scaling picture makes it exact.
Two nonzero vectors $u$ and $v$ in $\R^n$ are parallel if $u = cv$ for some real number $c$.
Two nonzero vectors $u$ and $v$ (in the plane or in space) are parallel if $u = cv$ for some real number $c$.
A positive $c$ means they point the same way, a negative $c$ the opposite way. The test is one division per entry, and every entry has to give the same $c$:The test in symbols is one division per entry, every entry giving the same $c$ — and in pictures, it's whether one dial setting can land $cv$ exactly on $u$:
Are these pairs parallel? (a) $(-6, 3, 9)$ and $(2, -1, -3)$ (b) $(2, 4, 6)$ and $(1, 2, 4)$ (c) For which $k$ is $(3, k, -6)$ parallel to $(-1, 4, 2)$?
(a) $\frac{-6}{2} = -3$, $\frac{3}{-1} = -3$, $\frac{9}{-3} = -3$. One scalar for every entry: $(-6, 3, 9) = -3(2, -1, -3)$. Parallel, pointing opposite ways.
(b) $\frac{2}{1} = 2$ and $\frac{4}{2} = 2$, but $\frac{6}{4} = \frac{3}{2}$. The last entry breaks it: not parallel.
(c) The entries we know fix the scalar: $\frac{3}{-1} = -3$ (and $\frac{-6}{2} = -3$ agrees). So $k = -3 \cdot 4 = -12$.
Parts (a) and (b), drawn. One dial $c$ stretches $v$; parallel means some single setting lands every entry at once:
Slide $c$ and try to land $cv$ on each star. One of the stars is a trap:
The trap star is off the line, so no scalar reaches it: it isn't parallel to $v$, no matter how close it looks.
Checking one entry proves nothing: $(2, 4, 6)$ and $(1, 2, 4)$ agree on the first two ratios and still aren't parallel. Check every entry, and watch for zeros: $(0, 3)$ and $(0, 5)$ are parallel ($c = \frac{3}{5}$), but $(0, 3)$ and $(1, 5)$ aren't, because no scalar turns $1$ into $0$. (Some books also call the zero vector parallel to everything. We leave it out, since it has no direction.)
Your turn: find the scalar, or the missing entry.
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Here's the payoff of thinking in moves. To find the point a third of the way from $A$ to $B$, take a third of the move $B - A$ and make it from $A$:
As $t$ runs from $0$ to $1$, the point $A + t(B - A)$ slides from $A$ all the way to $B$. Here it is as a theorem.
Let $A$ and $B$ be points in $\R^n$ and $0 \le t \le 1$. The point $$P = A + t(B - A) = (1 - t)A + tB$$ is $t$ of the way from $A$ to $B$: the move from $A$ to $P$ is $t(B - A)$, and the move from $P$ to $B$ is $(1 - t)(B - A)$. In particular, the midpoint of the segment is $\frac{1}{2}(A + B)$.
Expanding, $A + t(B - A) = A + tB - tA = (1 - t)A + tB$. The move from $A$ to $P$ is $P - A = t(B - A)$, and the move from $P$ to $B$ is $B - P = (B - A) - t(B - A) = (1 - t)(B - A)$. Both are nonnegative multiples of the same move $B - A$, so $P$ sits on the segment, splitting the move into a $t$ part and a $1 - t$ part. At $t = \frac{1}{2}$ the two parts are equal, and $P = \frac{1}{2}A + \frac{1}{2}B = \frac{1}{2}(A + B)$. $\blacksquare$
Let $A$ and $B$ be points in the plane or in space, and let $0 \le t \le 1$. The point $$P = A + t(B - A) = (1 - t)A + tB$$ is $t$ of the way from $A$ to $B$: the move from $A$ to $P$ is $t(B - A)$, and the move from $P$ to $B$ is $(1 - t)(B - A)$. In particular, the midpoint of the segment is $\frac{1}{2}(A + B)$.
This is a visual argument, not a formal proof, but it carries the whole idea. $P = A + t(B - A)$ says: start at $A$ and make $t$ of the move to $B$. Watch the move split in two:
The green piece from $A$ to $P$ is $t(B - A)$, and what's left, from $P$ to $B$, is the other $(1 - t)(B - A)$. The two pieces point the same way, and laid end to end they make the whole move, so $P$ really does sit on the segment, $t$ of the way along. The second form of $P$ is one distribute: $A + t(B - A) = A + tB - tA = (1 - t)A + tB$. And at $t = \frac{1}{2}$ the two pieces match, so the midpoint is $\frac{1}{2}A + \frac{1}{2}B = \frac{1}{2}(A + B)$. The full proof is in the n version.
The midpoint is the friendliest case: average the coordinates. The general case is the same move-from-$A$ idea.
Let $A = (1, 2, -1)$ and $B = (7, -4, 5)$. Find (a) the midpoint (b) the point $\frac{1}{3}$ of the way from $A$ to $B$ (c) the point $B'$ that makes $A$ the midpoint of $BB'$.
(a) $\frac{1}{2}(A + B) = \frac{1}{2}(8, -2, 4) = (4, -1, 2)$.
(b) The move is $B - A = (6, -6, 6)$, a third of it is $(2, -2, 2)$, so the point is $(1, 2, -1) + (2, -2, 2) = (3, 0, 1)$.
(c) $A$ is halfway from $B$ to $B'$, so $B' = A + (A - B) = 2A - B = (2, 4, -2) - (7, -4, 5) = (-5, 8, -7)$.
All three parts in motion. For (c), watch the move from $B$ to $A$ get made twice: that's what "$A$ is the midpoint" means.
Let $t$ run past $0$ and $1$ to every real number, and $A + t(B - A)$ traces the whole line through $A$ and $B$. That's exactly how Lines & Planes in Space writes a line.
Drag the endpoints until the points that split the segment into thirds land on the stars:
The last mission runs the theorem backward: you know where the thirds have to be and you're solving for the ends. Your turn, with numbers.
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That's the picture: vectors are moves, points are where moves take you from $0$, adding is walking head to tail, scaling stretches along a line, subtracting runs tip to tip, and a fraction of a move lands you a fraction of the way. Next up, Dot & Cross Products gives these arrows lengths and angles.
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.
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Made it this far? Time for something with a little heat: a triangle, the midpoint of one side, and a point two thirds of the way to it. Simplify at the end and something pretty falls out.
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