Geometric interpretation — signed volume: Visual walkthrough

Before Step 1, two words we will lean on constantly:

Step 1 — The two basic arrows and the unit square

WHAT. Draw the two simplest arrows: $e_{1} = (1, 0)$ (one step right) and $e_{2} = (0, 1)$ (one step up). The box they span is the plain $1 \times 1$ square.

WHY. Every other arrow is built by stretching and adding these two, so if we know the area of this square we can reach the area of any box. We simply declare its area to be $1$ — that is our unit of measurement, nothing to prove.

PICTURE. The blue and yellow arrows sit on the axes; the green square between them is our ruler.

Figure — Geometric interpretation — signed volume

Step 2 — Stretching one edge stretches the area (multilinearity, part 1)

WHAT. Keep $b$ fixed. Replace $a$ by $c a$ (same direction, $c$ times as long). The box's area is multiplied by exactly $c$ .

WHY. The box has a base and a height. Stretching one edge by $c$ leaves the height the same but multiplies the base by $c$ — so area scales by $c$ . This is the only sane way "area" could react to stretching, so we demand it.

PICTURE. Same yellow $b$ , but the blue edge is doubled; the shaded box is visibly twice as big.

Step 3 — Adding along one edge adds the areas (multilinearity, part 2)

WHAT. Split the first arrow into two pieces $a + a^{'}$ . The box on the sum equals the two smaller boxes stacked.

WHY. Because area is additive along a shared edge: glue two parallelograms that share the yellow side, and the total area is the sum. Together with Step 2, this is called being multilinear — linear in each edge separately.

PICTURE. The box over $a + a^{'}$ is shown split by a dashed line into a box over $a$ and a box over $a^{'}$ .

Step 4 — Equal edges = flat box = zero (alternating)

WHAT. If both arrows are the same ( $a = b$ ), the box collapses onto a single line. A line has no area: $V (a, a) = 0$ .

WHY. There is no "width" to sweep out when the two edges lie on top of each other — the parallelogram is squashed flat. We call this the alternating rule.

PICTURE. Two overlapping arrows and a red highlighted line where the box should be — width zero.

Step 5 — Swapping the two arrows flips the sign

WHAT. From Rule 2 we can derive a hidden consequence: swapping the order of the arrows negates the area. So $V (e_{2}, e_{1}) = - 1$ while $V (e_{1}, e_{2}) = + 1$ .

WHY. Feed the equal-edge rule the arrow $a + b$ into both slots: $0 = V (a + b, a + b)$ Now use multilinearity (Steps 2–3) to blow this open into four boxes: $0 = = 0 V (a, a) + V (a, b) + V (b, a) + = 0 V (b, b)$ The two underbraced terms vanish by Rule 2, leaving $V (a, b) = - V (b, a)$ . Swapping = sign flip. This is where the "signed" in signed area comes from.

PICTURE. Same box drawn twice; the little curved arrow of turning goes counterclockwise (green, $+$ ) on the left and clockwise (red, $-$ ) on the right.

Step 6 — Write any arrow in the basic arrows, then expand

WHAT. Any arrow is its right-part plus its up-part: $a = a_{1} e_{1} + a_{2} e_{2}, b = b_{1} e_{1} + b_{2} e_{2} .$ Feed these into $V$ and use multilinearity (Steps 2–3) to pull every scalar out. Four terms fall out: $V (a, b) = a_{1} b_{1} V (e_{1}, e_{1}) + a_{1} b_{2} V (e_{1}, e_{2}) + a_{2} b_{1} V (e_{2}, e_{1}) + a_{2} b_{2} V (e_{2}, e_{2}) .$

WHY. We already know each of these four "corner" areas from Steps 1, 4, 5. Substituting decides everything.

PICTURE. The four terms shown as four tiny tiles; two are stamped $0$ (equal edges), two carry $+ 1$ and $- 1$ .

Substitute the known corner values:

$V (a, b) = a_{1} b_{1} Rule 2 (0) + a_{1} b_{2} Step 1 (+ 1) + a_{2} b_{1} Step 5 (- 1) + a_{2} b_{2} Rule 2 (0) .$

WHAT. The formula $a_{1} b_{2} - a_{2} b_{1}$ already handles every quadrant and every degenerate case automatically. Let's check each so no scenario surprises you.

WHY. A formula you trust is one you have tested at its edges.

PICTURE. Four mini-boxes: counterclockwise turn ( $+$ ), clockwise turn ( $-$ ), collinear arrows (flat, $0$ ), and a stretched-but-same-turn box (bigger number, same sign).

$b$ is counterclockwise from $a$ (turn left): $a_{1} b_{2} - a_{2} b_{1} > 0$ . Orientation preserved. Check: $a = (1, 0), b = (0, 1) \Rightarrow 1 \cdot 1 - 0 \cdot 0 = + 1$ .
$b$ is clockwise from $a$ (turn right): value $< 0$ . Orientation reversed. Check: swap them, $a = (0, 1), b = (1, 0) \Rightarrow 0 \cdot 0 - 1 \cdot 1 = - 1$ .
$a$ and $b$ point the same (or opposite) way — one is a multiple of the other: value $= 0$ . Flat box. Check: $a = (2, 4), b = (1, 2) \Rightarrow 2 \cdot 2 - 4 \cdot 1 = 0$ .
One arrow is the zero arrow $(0, 0)$ : value $= 0$ . No box at all. Check: $b = (0, 0) \Rightarrow a_{1} \cdot 0 - a_{2} \cdot 0 = 0$ .
Stretch $a$ by $c$ (Step 2): value scales by $c$ , sign unchanged if $c > 0$ , flips if $c < 0$ . Check: $a = (2, 0), b = (0, 1) \Rightarrow 2$ , exactly $2 \times$ the unit case.

The one-picture summary

Every rule collapses into one frame: two arrows, the box between them, the swept turn that fixes the sign, and the formula reading off the two "cross" products $a_{1} b_{2}$ (grows the box) and $a_{2} b_{1}$ (subtracts the overlap).

Recall Feynman retelling — the whole walkthrough in plain words

I drew two little arrows on graph paper. First I said the plain unit square between the right-arrow and up-arrow counts as area $1$ — that's just my ruler. Then I noticed three fair rules any area must obey: stretch an edge by 3 and the area triples; glue two boxes along a shared side and areas add; and if both arrows are identical the box is squashed flat so its area is $0$ . That last rule secretly forces a sign: swapping the two arrows flips $+$ to $-$ , which is how we remember whether we turned left or right. Finally I wrote each arrow as "so-much-right plus so-much-up", multiplied everything out, and three of my rules erased the useless pieces — leaving exactly $a_{1} b_{2} - a_{2} b_{1}$ . That's the determinant. I never assumed the determinant; the drawing rules built it for me.

Connections

Geometric interpretation — signed volume — the parent result this page derives.
Determinant — definition and cofactor expansion
Cross product and scalar triple product
Linear independence and rank
Orientation and handedness of bases
Change of variables and the Jacobian