Definition D1
For A = (a b ; c d), set u = A(1,0) = (a, c), v = A(0,1) = (b, d).
Proof.
(1)Computing by the matrix action, the components of A(1,0) are a·1 + b·0 = a and c·1 + d·0 = c.A1, A25
(2)In the same way A(0,1) = (b, d). That is, u and v are the two columns of A.A1, A25
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Definition D2
Set det A = ad − bc. Write this value as D.
Proof.
(1)D is a number determined by the four entries of A alone.convention
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Definition D3
The quadrilateral with vertices 0, u, u+v, v is called the image parallelogram of A.
Proof.
(1)The vertices (0,0), (1,0), (1,1), (0,1) of the unit square go to 0, u, u+v, v respectively.D1, A1
(2)The side from 0 to u and the side from v to u+v are both u, and the other two sides are both v, so it is a parallelogram.(1), convention
(3)Let θ be the angle between u and v. Below, the case 0 < θ < π is treated first.(2), convention
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Lemma L1
The area of a parallelogram is the product of base and height.
Proof.
(1)A diagonal forms two triangles T₁ = (0, u, u+v) and T₂ = (0, u+v, v).D3, A12
(2)The three sides of T₁ are |u|, |v|, |u+v|, and the three sides of T₂ are also |u+v|, |u|, |v|, so the two triangles are congruent.(1), A13, A11
(3)Being congruent, the two triangles have equal area.(2), A10
(4)The two triangles do not overlap and fill the parallelogram, so the area is their sum.(1), A9
(5)Since a triangle's area is ½ × base × height, the parallelogram's area is 2 × ½ × base × height = base × height.(3), (4), A8, A21, A25
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Lemma L2
|u|²|v|² − (u·v)² = D², where D = ad − bc.
Proof.
(1)|u|² = a² + c², |v|² = b² + d²D1, A4, A16
(2)u·v = ab + cdD1, A3
(3)(a²+c²)(b²+d²) = a²b² + a²d² + c²b² + c²d²(1), A19, A21
(4)(ab+cd)² = a²b² + 2abcd + c²d²(2), A19, A21, A22, A23
(5)Subtracting the two expressions gives a²d² − 2abcd + c²b²(3), (4), A20, A22, A23, A24
(6)Meanwhile (ad − bc)² = a²d² − 2abcd + b²c²D2, A19, A20, A21, A22, A23, A24
(7)Since c²b² = b²c², |u|²|v|² − (u·v)² = D²(1), (5), (6), A21, E
(8)Up to this line only the four arithmetic operations were used. Area has not yet appeared.(7)
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Theorem T
The area of the image parallelogram of A is |det A|.
Proof.
(1)First take the case 0 < θ < π, that is, u and v not on one line.D3
(2)Taking the base as |u|, the height is |v| sin θ, so S = |u||v| sin θ(1), L1, A7
(3)S² = |u|²|v|² sin²θ = |u|²|v|²(1 − cos²θ)(2), A5, A21, A24
(4)|u|²|v|² cos²θ = (|u||v| cos θ)² = (u·v)²A2, A21
(5)Therefore S² = |u|²|v|² − (u·v)²(3), (4), A20, A25
(6)By L2, S² = D²(5), L2
(7)Since |u| > 0, |v| > 0, sin θ > 0, we have S > 0, and taking square roots of both sides gives S = |D|(2), (6), A17, A6, A18, A15, A26, E
(8)Next, the case where u and v lie on one line: if v = t u, then b = ta, d = tc, so D = a(tc) − (ta)c = 0D1, D2, A21, A24, N
(9)The image is then a segment, so the area is 0 and |D| = 0. The conclusion still holds.(8), A14, A15
(10)If u = 0, then a = c = 0; if v = 0, then b = d = 0. Either way D = 0 and the image is a point or a segment.D1, D2, A25, A24, A14
(11)Therefore in every case the area of the image parallelogram is |det A|.(7), (9), (10)
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Corollary C1
Swapping the two columns flips the sign of the determinant, but the image parallelogram is unchanged.
Proof.
(1)Setting A′ = (b a ; d c) gives det A′ = bc − ad = −(ad − bc) = −DD2, A20, A22, A24, N
(2)The two columns of A′ are v, u, so the vertices of its image parallelogram are 0, v, v+u, u.D1, D3
(3)Since u + v = v + u, the figure in (2) is the same as the image parallelogram of A.(2), A22
(4)The area is |−D| = |D|, the same.(1), (3), A26, T
(5)Therefore the sign records not area but the order of the two columns. The moment |det A| is taken, that information is discarded.(1), (4)
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Corollary C2
If a linear map scales the area of every figure by the same ratio, then |det(AB)| = |det A|·|det B|.
Proof.
(1)Theorem T concerns the single unit square. Assume the scaling ratio is the same regardless of the figure.assumption, T
(2)AB means applying B first, then A.A27
(3)The unit square becomes, under B, a figure of area |det B|, which under A is scaled by |det A|.(1), (2), T
(4)The result is the image parallelogram of AB, so its area is |det(AB)|.(2), T
(5)Therefore |det(AB)| = |det A|·|det B|(3), (4)
(6)Proof 179 verifies this identity, sign included, by entrywise computation. Area tells why the identity is natural.(5)
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