# Determinants, Areas and Volumes

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1 Determinants, Areas and Volumes Theodore Voronov Part 2 Areas and Volumes The area of a two-dimensional object such as a region of the plane and the volume of a three-dimensional object such as a solid body in space, as well the length of an interval of the real line, are all particular cases of a very general notion of measure. General measure theory is a part of analysis. Here we shall focus on the geometrical idea of measure and its relation with the algebraic notion of determinant. Why the area of a parallelogram is represented by a determinant Although, from practice we know very well what is the area of simple geometrical figures such as, for example, a rectangle or a disk, it is not easy to give a rigorous general definition of area. However, some properties of area should be clear; consider subsets of the plane R 2 : (1) Area is always non-negative: area(s) 0 for all subsets S R 2 such that it makes sense to speak about their area; (2) Area is additive: area(s 1 S 2 ) = areas 1 + areas 2, if the intersection S 1 S 2 is empty. A translation of the space R n is the map T a : R n R n that takes every point x R n to x + a, where a R n is a fixed vector. For a subset S R n, a translation T a shifts all points of S along a, i.e., each point x S is mapped to x + a, and S moves rigidly to its new location in R n. The following properties hold for areas in R 2 : (3) Area is invariant under translations: areat a (S) = areas, for all vectors a R 2. (4) For a one-dimensional object, such as a line segment, the area should vanish. The plan now is as follows: using conditions (1) (4) we shall establish a deep link between the notion of area and the theory of determinants. To this end, we consider the area of a simple polygon, a parallelogram. Later our considerations will be generalized to R n. 1

2 Let a = (a 1, a 2 ), b = (b 1, b 2 ) be vectors in R 2. The parallelogram on a, b with basepoint O R 2 is the set of points of the form x = O + ta + sb where 0 t, s 1. One can easily see that it is the plane region bounded by the two pairs of parallel straight line segments: OA, BC and OB, AC where C = O + a + b. b B C O a A What is the area of it? From property (3) it follows that the area does not depend on the location of our parallelogram in R 2 : by a translation the basepoint O can be made an arbitrary point of the plane without changing the area. Let us assume that O = 0 is the point (0, 0). Denote the parallelogram by Π(a, b). Then areaπ(a, b) is a function of vectors a, b. Proposition 1 The function areaπ(a, b) has the following properties: (1) areaπ(na, b)= areaπ(a, nb)= n areaπ(a, b) for any n Z; (2) areaπ(a, b + ka)= areaπ(a + kb, b)= areaπ(a, b) for any k R. Proof Suppose we replace a by na for a positive integer n. Then Π(na, b) is the union of n copies of the parallelogram Π(a, b): b b a b a a From the additivity of area it follows that areaπ(na, b)= n Π(a, b). The same is true if we replace b by nb, for positive n. For n = 0, Π(0, b) or Π(a, 0) are just segments, therefore have zero area, by (4). Notice also that Π( a, b) and Π(a, b) differ by a shift, so have the same area. Hence, areaπ(na, b)= areaπ(a, nb)= n areaπ(a, b) holds in general. To prove the second relation, we again use the additivity of area: it is clear that to obtain Π(a, b + ka), one has to cut from Π(a, b) the triangle OBB, shift it by the vector a and attach it back as the triangle ACC : B B C C b O a A 2

3 Clearly, due to additivity and invariance under translations, the area will not change, and we arrive at areaπ(a, b + ka)= areaπ(a, b) as claimed. In fact, the first assertion in Proposition 1 is valid in a stronger form. Proposition 2 The area of a parallelogram satisfies for any real number k R. Proof Omitted. areaπ(ka, b)= areaπ(a, kb)= k areaπ(a, b) Theorem 1 Suppose a= (a 1, a 2 ), b= (b 1, b 2 ). Then ( ) areaπ(a, b) = det(a, b) = det a1 a 2. (1) b 1 b 2 Proof Indeed, by Propositions 1 and 2, areaπ(a, b) satisfies almost the same properties as the determinant det(a, b). They can be used to calculate areaπ(a, b); it is similar to using row operations for calculating determinants. We have ( ) a1 a 2 ( a1 a = areaπ 2 = a 1 b 2 b 1 a 1 a 2 areaπ where e 1 = (1, 0) and e 2 = (0, 1), as desired. areaπ(a, b)= areaπ ) b 1 ( b 2 ) a1 0 0 b 2 b = areaπ 1 a 1 a 2 0 b 2 b 1 ( ) a 1 a = a b 2 b 1 a 2 areaπ(e 1, e 2 ) = a 1 b 2 b 1 a 2, N.B. In the theorem we chose the unit of area so that areaπ(e 1, e 2 ) = 1. Example Find the area of the parallelogram built on vectors a = ( 2, 5) and b = (1, 1). Solution: we have = = 7. Hence areaπ(a, b) = 7 = 7. Example Find the area of the triangle ABC if A = (3, 2), B = (4, 2), C = (1, 0). Solution: it is half of the area of the parallelogram built on CA = A C = (2, 2), CB = B C = (3, 2). Hence area(abc) = = 1. 3

4 Volumes and determinants All the above results can be generalized to higher dimensions. Consider vectors a 1,..., a n in R n. The parallelepiped on a 1,..., a n with basepoint O R n is the set of points x = O+t 1 a t n a n where 0 t 1 1 for all i = 1,..., n. Denote it Π(a 1,..., a n ). In the sequel nothing depends on the basepoint, so we shall suppress any mentioning of it. Instead of deducing a formula for the volume of a parallelepiped similar to (1) from general properties of volumes such as additivity, as we did above for area, it is convenient to set by definition a a 1n volπ(a 1,..., a n ) = (2) a n1... a nn where a 1 = (a 11,..., a 1n ),..., a n = (a n1,..., a nn ). Note that if this is negative we take the absolute value. This definition implies that the unit of volume is such that the volume of Π(e 1,..., e n ) is set to 1. Example Find the oriented volume of the parallelepiped built on a 1 = (2, 1, 0), a 2 = (0, 3, 11) and a 3 = (1, 2, 7) in R 3. Solution: volπ(a 1, a 2, a 3 ) = = 2 ( 1) 1 ( 11) = So the volume equals 9. Areas and volumes in Euclidean space Recall that on R n one can define the scalar product of vectors by the formula (a, b)= a 1 b a n b n (see Problem 5 in week 1. We can also write a b). It follows that the standard basis vectors e 1,..., e n satisfy (e i, e j ) = 0 if i j and (e i, e i ) = 1. The length of a vector is defined as a = (a, a) and the angle between two vectors is defined by the equality (a, b) = a b cos α 4

5 from where we can find cos α if (a, b), (a, a), (b, b) are known. The unit cube Π(e 1,..., e n ) plays above a distinguished role. We shall see that any unit cube in R n will have unit volume. Consider an example (for n = 2 we continue to use area instead of vol). Example Let g 1 = (cos α, sin α), g 2 = ( sin α, cos α) in R 2. We can immediately see that g 1 = g 2 = 1 and g 1 g 2 = 0, so Π(g 1, g 2 ) is a unit square. We have areaπ(g 1, g 2 ) = cos α sin α sin α cos α = cos2 α + sin 2 α = 1. There is a way of expressing the volume of a parallelepiped entirely in terms of intrinsic geometric information: lengths of vectors and angles between them, rather than their coordinates as in the previous formulas. Definition G(a 1,..., a k ) = (a 1, a 1 )... (a 1, a k ) (3) (a k, a 1 )... (a k, a k ) where k n, is called the Gram matrix of the system of vectors a 1,..., a n and its determinant, the Gram determinant. Theorem 2 The Gram determinant of a 1,..., a n is the square of the volume of the parallelepiped Π(a 1,..., a n ). Proof Indeed, consider the n n matrix A with rows a i. Consider AA T = a a 1n a a n = a 1... ( a T 1... a T n ) a n1... a nn a 1n... a nn a n Hence = a 1a T 1... a 1 a T n a n a T 1... a n a T n = G(a 1,..., a n ). det G(a 1,..., a n )= det(aa T ) = det A det A ( T 2. = (det A) 2 = volπ(a 1,..., a n )) Corollary If g i are arbitrary mutually orthogonal unit vectors (i.e., Π(g 1,..., g n ) is a unit cube), then volπ(g 1,..., g n )= 1. 5

6 One of the advantages of expressing volumes via the Gram determinants is that it allows us to consider easily parallelepipeds in R n of dimensions less than n. More precisely, if we are given k vectors a 1,..., a k, then we can consider a k-dimensional parallelepiped Π(a 1,..., a k ) contained in the k-dimensional space spanned by a 1,..., a k. The formula ( volπ(a 1,..., a k )) 2 = det G(a1,..., a k ) is applicable, with the scalar products calculated in the ambient space R n. Example Find the area of the parallelogram built on a= (1, 1, 2) and b= (2, 0, 3). Solution: the Gram determinant is = 14. Hence the area is 14. 6

7 MT1000 Project 4 Groupwork Week 2 Problem 1 Find the areas and volumes: (a) areaπ(a, b) if a= ( cos α, sin α), b= ( sin α, cos α) in R 2 ; make a sketch; (b) volπ(a, b, c) if a= (3, 2, 1), b= (2, 2, 5), c= (0, 0, 1) in R 3 ; (c) areaπ(a, b) if a= (1, 1, 2, 3), b= (0, 3, 1, 2) in R 4. (Use the Gram determinant.) Problem 2 Verify by a direct calculation that the Gram determinant det G(a, b) vanishes if one of the vectors a, b is a scalar multiple of the other. (Geometrically that means that they are in the same line.) What can be said about the area of the parallelogram Π(a, b)? Problem 3 Show that the oriented area of a triangle ABC in R 2 is given by the formula a 1 a 2 1 area(abc) = 1 2 b 1 b 2 1 c 1 c 2 1 where A = (a 1, a 2 ), B = (b 1, b 2 ), C = (c 1, c 2 ). Problem 4 (See Problem 5 from week 1) (a) Show that the so-called triple or mixed product (a, b, c) in R 3 defined as (a, b, c) = a (b c) = (a b) c is the volume of Π(a, b, c). (b) Let n be a unit vector perpendicular to a and b. Show that det G(n, a, b)= det G(a, b) and deduce that volπ(n, a, b)= areaπ(a, b). (c) Use parts (a) and (b) to prove that the length of the vector product a b in R 3 is the area of the parallelogram Π(a, b). Problem 5 Show that a b = a b sin α where α is the angle between a and b. Hint: use the result of the previous problem and express the area by the Gram determinant. You may assume that a b = a b cos α. Project Report You should write a report on the topics covered in this project. The report should include a description (in your own words) of the defining properties of a determinant and the key properties that follow from this (you do not need to include properties of matrices and vectors). The link between determinants and areas and volumes should be explained and also the role of the Gram determinant in calculating volumes (do not include the proofs of Proposition 1 and Theorems 1 and 2). The solutions to all groupwork problems should be included in an 7

8 appropriate place. Even though you have worked as a group on this project, the report should be all your own work. Please hand in your report (with your name and group number on the front) to the Student Support Office in Lamb by 1pm on Friday 16th December. There are 25 marks for this project. (a) 10 marks for the solutions to the mathematical problems. (b) 10 marks for clearly and correctly explaining the key ideas in the lecture notes in your own words. (c) half the average mark for your group for (a) out of 5. Any student who does not attend a group session, without good reason, will get 50% of the group mark (c). Any student who does not attend both group sessions, without good reason, will get 0 for the group mark. Please notify us as soon as possible if you miss a session and fill in a Self Certification Form available from the Student Support Office. 8

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