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 R2 :
(1) Area is always non-negative: area(S) ≥ 0 for all subsets S ⊂ R2 such that
it makes sense to speak about their area;
(2) Area is additive: area(S1 ∪ S2 ) = areaS1 + areaS2 , if the intersection
S1 ∩ S2 is empty.
A translation of the space Rn is the map Ta : Rn → Rn that takes every
point x ∈ Rn to x + a, where a ∈ Rn is a fixed vector. For a subset S ⊂ Rn , a
translation Ta “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 Rn .
The following properties hold for areas in R2 :
(3) Area is invariant under translations: areaTa (S) = areaS, for all vectors
a ∈ R2 .
(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 Rn .
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Let a = (a1 , a2 ), b = (b1 , b2 ) be vectors in R2 . The parallelogram on a, b
with basepoint O ∈ R2 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
>
b½½
½
C
½
½
O
½
-½
½
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 R2 : 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):
>
b½½
>
b½½
½
½
a
>
b½½
½
-½
a
½
-½
½
½
a
½
-½
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 0 , shift
it by the vector a and attach it back as the triangle ACC 0 :
B
O
a
C
½
½ ³³
³
½³
½³
-³
>
b½½
³
³³
½³³
½³
³
B0
1
³³
C0
³
1
³³
A
2
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
areaΠ(ka, b)= areaΠ(a, kb)= |k| · areaΠ(a, b)
for any real number k ∈ R.
Proof Omitted.
Theorem 1 Suppose a= (a1 , a2 ), b= (b1 , b2 ). Then
¯
µ
¯
a1
¯
areaΠ(a, b) = |det(a, b)| = ¯det
b1
¶¯
a2 ¯¯
.
b2 ¯
(1)
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
µ
¶
a1 a2
areaΠ(a, b)= areaΠ
b1 µ b2
µ
¶
¶
a1
a2
a1
0
= areaΠ
= areaΠ
b1
0 b2 − ab11 a2
µ0 b2 −
¶ a1 a 2
1 0
= |a1 | |b2 − ab11 a2 | areaΠ
= |a1 b2 − b1 a2 | · areaΠ(e1 , e2 ) = |a1 b2 − b1 a2 | ,
0 1
where e1 = (1, 0) and e2 = (0, 1), as desired.
N.B. In the theorem we chose the unit of area so that areaΠ(e1 , e2 ) = 1.
Example Find the area of the parallelogram built on vectors a = (−2, 5) and
b = (1, 1). Solution: we have
¯
¯
¯ −2 5 ¯
¯ = −7 .
∆ = ¯¯
1 1 ¯
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
¯
¯
¯
¯
1 ¯ 3 2 ¯
= 1.
area(ABC) = 2 ¯
2 2 ¯
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Volumes and determinants
All the above results can be generalized to higher dimensions. Consider vectors
a1 , . . . , an in Rn . The parallelepiped on a1 , . . . , an with basepoint O ∈ Rn is
the set of points
x = O+t1 a1 + . . . + tn an
where 0 ≤ t1 ≤ 1 for all i = 1, . . . , n .
Denote it Π(a1 , . . . , an ). 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
¯
¯
¯ a11 . . . a1n ¯
¯
¯
(2)
volΠ(a1 , . . . , an ) = ¯¯ . . . . . . . . . ¯¯
¯ an1 . . . ann ¯
where a1 = (a11 , . . . , a1n ), . . . , an = (an1 , . . . , ann ). 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(e1 , . . . , en ) is set to 1.
Example Find the oriented volume of the parallelepiped built on a1 = (2, 1, 0),
a2 = (0, 3, 11) and a3 = (1, 2, 7) in R3 . Solution:
¯
¯
¯ 2 1 0 ¯
¯
¯
volΠ(a1 , a2 , a3 ) = ¯¯ 0 3 11 ¯¯= 2 · (−1) − 1 · (−11) = 9 .
¯ 1 2 7 ¯
So the volume equals 9.
Areas and volumes in Euclidean space
Recall that on Rn one can define the scalar product of vectors by the formula
(a, b)= a1 b1 + . . . + an bn
(see Problem 5 in week 1. We can also write a · b). It follows that the ‘standard
basis vectors’ e1 , . . . , en satisfy (ei , ej ) = 0 if i 6= j and (ei , ei ) = 1. The length
of a vector is defined as
p
|a| = (a, a)
and the angle between two vectors is defined by the equality
(a, b) = |a| |b| cos α
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from where we can find cos α if (a, b), (a, a), (b, b) are known.
The unit cube Π(e1 , . . . , en ) plays above a distinguished role. We shall see
that any unit cube in Rn will have unit volume. Consider an example (for n = 2
we continue to use area instead of vol).
Example Let g1 = (cos α, sin α), g2 = (− sin α, cos α) in R2 . We can immediately
see that |g1 | = |g2 |= 1 and g1 · g2 = 0, so Π(g1 , g2 ) is a unit square. We have
¯
¯
¯ cos α sin α ¯
¯= cos2 α + sin2 α = 1.
¯
areaΠ(g1 , g2 ) = ¯
− sin α cos α ¯
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


(a1 , a1 ) . . . (a1 , ak )
G(a1 , . . . , ak ) =  . . .
...
...  .
(ak , a1 ) . . . (ak , ak )
(3)
where k ≤ n, is called the Gram matrix of the system of vectors a1 , . . . , an and
its determinant, the Gram determinant.
Theorem 2 The Gram determinant of a1 , . . . , an is the square of the volume of
the parallelepiped Π(a1 , . . . , an ).
Proof Indeed, consider

a11 . . .
T

AA = . . . . . .
an1 . .
.
the n × n matrix A with rows ai . Consider

  
a1n
a11 . . . an1
a1




...
. . . . . . . . . = . . .  ( aT
. . . aT
1
n )
ann
a1n . . .  ann
an
T
a1 aT
.
.
.
a
a
1 n
1
=  ...
...
. . .  = G(a1 , . . . , an ) .
an aT
. . . an aT
1
n
Hence
det G(a1 , . . . , an )= det(AAT ) = det A det AT
³
´2
2
= (det A) = volΠ(a1 , . . . , an ) .
Corollary If gi are arbitrary mutually orthogonal unit vectors (i.e., Π(g1 , . . . , gn )
is a unit cube), then volΠ(g1 , . . . , gn )= 1.
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One of the advantages of expressing volumes via the Gram determinants is
that it allows us to consider easily parallelepipeds in Rn of dimensions less than
n. More precisely, if we are given k vectors a1 , . . . , ak , then we can consider a
k-dimensional parallelepiped Π(a1 , . . . , ak ) contained in the k-dimensional space
spanned by a1 , . . . , ak . The formula
³
´2
volΠ(a1 , . . . , ak ) = det G(a1 , . . . , ak )
is applicable, with the scalar products calculated in the ambient space Rn .
Example Find the area of the parallelogram built on a= (1, −1, 2) and b=
(2, 0, 3). Solution: the Gram determinant is
¯
¯
¯ 6 8 ¯
¯
¯
¯ 8 13 ¯ = 14 .
√
Hence the area is 14.
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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 R2 ; make a sketch;
(b) volΠ(a, b, c) if a= (3, 2, −1), b= (2, 2, 5), c= (0, 0, 1) in R3 ;
(c) areaΠ(a, b) if a= (1, −1, 2, 3), b= (0, 3, 1, 2) in R4 . (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 R2 is given by the
formula
¯
¯
¯ a1 a2 1 ¯
¯
¯
area(ABC) = 12 ¯¯ b1 b2 1 ¯¯
¯ c1 c2 1 ¯
where A = (a1 , a2 ), B = (b1 , b2 ), C = (c1 , c2 ).
Problem 4 (See Problem 5 from week 1)
(a) Show that the so-called triple or mixed product (a, b, c) in R3 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
R3 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
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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.
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