Lecture Notes 1: Matrix Algebra

Part B: Determinants and Inverses

Peter J. Hammond

revised 2020 September 14th

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 1 of 87

Outline
Special Matrices
Square, Symmetric, and Diagonal Matrices
The Identity Matrix
The Inverse Matrix
Partitioned Matrices
Permutations and Their Signs
Permutations
Transpositions
Signs of Permutations
The Product Rule for the Signs of Permutations
Determinants: Introduction
Determinants of Order 2
Determinants of Order 3
The Determinant Function
Permutation and Transposition Matrices
Triangular Matrices
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 2 of 87
Square Matrices
A square matrix has an equal number of rows and columns,
this number being called its dimension.
The (principal, or main) diagonal
of a square matrix A = (aij )n×n of dimension n
is the list (aii )ni=1 = (a11 , a22 , . . . , ann ) of its n diagonal elements.
The other elements aij with i 6= j are the off-diagonal elements.

A square matrix is often expressed in the form

 
a11 a12 . . . a1n
a21 a22 . . . a2n 
A= .
 
.. . . . 
 .. . . .. 
an1 an2 . . . ann

with some extra dots along the diagonal.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 3 of 87

Symmetric Matrices

Definition
A square matrix A is symmetric just in case
it is equal to its transpose — i.e., if A> = A.

Example
The product of two symmetric matrices need not be symmetric.
Using again our example of non-commuting 2 × 2 matrices,
here are two examples
where the product of two symmetric matrices is asymmetric:
    
0 1 0 0 0 1
I = ;
1 0 0 1 0 0
    
0 0 0 1 0 0
I =
0 1 1 0 1 0

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 4 of 87

Two Exercises with Symmetric Matrices

Exercise
Let x be a column n-vector.
1. Find the dimensions of x> x and of xx> .
2. Show that one is a non-negative number
which is positive unless x = 0,
and that the other is an n × n symmetric matrix.

Exercise
Let A be an m × n-matrix.
1. Find the dimensions of A> A and of AA> .
2. Show that both A> A and AA> are symmetric matrices.
3. Show that m = n is a necessary condition for A> A = AA> .
4. Show that m = n with A symmetric
is a sufficient condition for A> A = AA> .

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 5 of 87

Diagonal Matrices
A square matrix A = (aij )n×n is diagonal
just in case all of its off diagonal elements are 0
— i.e., i 6= j =⇒ aij = 0.
A diagonal matrix of dimension n can be written in the form
 
d1 0 0 ... 0
0 d2 0 . . . 0 
 
D=0 0 d3 . . . 0   = diag(d1 , d2 , d3 , . . . , dn ) = diag d

 .. .. .. . . .. 
. . . . .
0 0 0 . . . dn

where the n-vector d = (d1 , d2 , d3 , . . . , dn ) = (di )ni=1

consists of the diagonal elements of D.
Note that diag d = (dij )n×n where each dij = δij dii = δij djj .

Obviously, any diagonal matrix is symmetric.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 6 of 87
Multiplying by Diagonal Matrices
Example
Let D be a diagonal matrix of dimension n.
Suppose that A and B are m × n and n × m matrices, respectively.
Then E := AD and F := DB are well defined matrices
of dimensions m × n and n × m, respectively.
By the law of matrix multiplication, their elements are
Xn Xn
eij = aik δkj djj = aij djj and fij = δik dii bkj = dii bij
k=1 k=1

Thus, post-multiplying A by D is the column operation

of simultaneously multiplying every column aj of A
by its matching diagonal element djj .
Similarly, pre-multiplying B by D is the row operation
of simultaneously multiplying every row b> i of B
by its matching diagonal element dii .
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 7 of 87
Two Exercises with Diagonal Matrices
Exercise
Let D be a diagonal matrix of dimension n.
Give conditions that are both necessary and sufficient
for each of the following:
1. AD = A for every m × n matrix A;
2. DB = B for every n × m matrix B.

Exercise
Let D be a diagonal matrix of dimension n,
and C any n × n matrix.
An earlier example shows that
one can have CD 6= DC even if n = 2.
1. Show that C being diagonal
is a sufficient condition for CD = DC.
2. Is this condition necessary?
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 8 of 87
Outline
Special Matrices
Square, Symmetric, and Diagonal Matrices
The Identity Matrix
The Inverse Matrix
Partitioned Matrices
Permutations and Their Signs
Permutations
Transpositions
Signs of Permutations
The Product Rule for the Signs of Permutations
Determinants: Introduction
Determinants of Order 2
Determinants of Order 3
The Determinant Function
Permutation and Transposition Matrices
Triangular Matrices
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 9 of 87
The Identity Matrix

The identity matrix of dimension n is the diagonal matrix

In = diag(1, 1, . . . , 1)

whose n diagonal elements are all equal to 1.

Equivalently, it is the n × n-matrix A = (aij )n×n
whose elements are all given by aij = δij
for the Kronecker delta function (i, j) 7→ δij
defined on {1, 2, . . . , n}2 .
Exercise
Given any m × n matrix A, verify that Im A = AIn = A.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 10 of 87

Uniqueness of the Identity Matrix

Exercise
Suppose that the two n × n matrices X and Y respectively satisfy:
1. AX = A for every m × n matrix A;
2. YB = B for every n × m matrix B.

Prove that X = Y = In .
(Hint: Consider each of the mn different cases where A (resp. B)
has exactly one non-zero element that is equal to 1.)
The results of the last two exercises together serve to prove:
Theorem
The identity matrix In is the unique n × n-matrix such that:
I In B = B for each n × m matrix B;
I AIn = A for each m × n matrix A.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 11 of 87

How the Identity Matrix Earns its Name

Remark
The identity matrix In earns its name because it represents
a multiplicative identity on the “algebra” of all n × n matrices.
That is, In is the unique n × n-matrix with the property
that In A = AIn = A for every n × n-matrix A.

Typical notation suppresses the subscript n in In

that indicates the dimension of the identity matrix.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 12 of 87

Outline
Special Matrices
Square, Symmetric, and Diagonal Matrices
The Identity Matrix
The Inverse Matrix
Partitioned Matrices
Permutations and Their Signs
Permutations
Transpositions
Signs of Permutations
The Product Rule for the Signs of Permutations
Determinants: Introduction
Determinants of Order 2
Determinants of Order 3
The Determinant Function
Permutation and Transposition Matrices
Triangular Matrices
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 13 of 87
Left and Right Inverse Matrices

Definition
Let A denote any n × n matrix.
1. The n × n matrix X is a left inverse of A
just in case XA = In .
2. The n × n matrix Y is a right inverse of A
just in case AY = In .
3. The n × n matrix Z is an inverse of A
just in case it is both a left and a right inverse
— i.e., ZA = AZ = In .

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 14 of 87

The Unique Inverse Matrix
Theorem
Suppose that the n × n matrix A has both a left and a right inverse.
Then both left and right inverses are unique,
and both are equal to a unique inverse matrix denoted by A−1 .
Proof.
If XA = AY = I, then XAY = XI = X and XAY = IY = Y,
implying that X = XAY = Y.
Now, if X̃ is any alternative left inverse,
then X̃A = I and so X̃ = X̃AY = Y = X.
Similarly, if Ỹ is any alternative right inverse,
then AỸ = I and so Ỹ = XAỸ = X = Y.
It follows that X̃ = X = Y = Ỹ, so we can define A−1
as the unique common value of all these four matrices.
Big question: when does the inverse exist?
Answer: if and only if the determinant is non-zero.
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 15 of 87
Rule for Inverting Products

Theorem
Suppose that A and B are two invertible n × n matrices.
Then the inverse of the matrix product AB exists,
and is the reverse product B−1 A−1 of the inverses.

Proof.
Using the associative law for matrix multiplication repeatedly gives:

(B−1 A−1 )(AB) = B−1 (A−1 A)B = B−1 (I)B = B−1 (IB) = B−1 B = I

and

(AB)(B−1 A−1 ) = A(BB−1 )A−1 = A(I)A−1 = (AI)A−1 = AA−1 = I.

These equations confirm that X := B−1 A−1 is the unique matrix

satisfying the double equality (AB)X = X(AB) = I.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 16 of 87

Rule for Inverting Chain Products and Transposes
Exercise
Prove that, if A, B and C are three invertible n × n matrices,
then (ABC)−1 = C−1 B−1 A−1 .
Then use mathematical induction
to extend the rule for inverting any product BC
in order to find the inverse of the product A1 A2 · · · Ak
of any finite chain of invertible n × n matrices.
Theorem
Suppose that A is an invertible n × n matrix.
Then the inverse (A> )−1 of its transpose
is (A−1 )> , the transpose of its inverse.
Proof.
By the rule for transposing products, one has

A> (A−1 )> = (A−1 A)> = I> = I

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 17 of 87

Orthogonal and Orthonormal Sets of Vectors

Definition
A set of k vectors {x1 , x2 , . . . , xk } ⊂ Rn is said to be:
I pairwise orthogonal just in case xi · xj = 0 whenever j 6= i;
I orthonormal just in case, in addition, each kxi k = 1
— i.e., all k elements of the set are vectors of unit length.
The set of k vectors {x1 , x2 , . . . , xk } ⊂ Rn is orthonormal
just in case xi · xj = δij for all pairs i, j ∈ {1, 2, . . . , k}.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 18 of 87

Orthogonal Matrices

Definition
Any n × n matrix is orthogonal
just in case its n columns form an orthonormal set.

Theorem
Given any n × n matrix P, the following are equivalent:
1. P is orthogonal;
2. PP> = P> P = I;
3. P−1 = P> ;
4. P> is orthogonal.
The proof follows from the definitions, and is left as an exercise.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 19 of 87

Outline
Special Matrices
Square, Symmetric, and Diagonal Matrices
The Identity Matrix
The Inverse Matrix
Partitioned Matrices
Permutations and Their Signs
Permutations
Transpositions
Signs of Permutations
The Product Rule for the Signs of Permutations
Determinants: Introduction
Determinants of Order 2
Determinants of Order 3
The Determinant Function
Permutation and Transposition Matrices
Triangular Matrices
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 20 of 87
Partitioned Matrices: Definition
A partitioned matrix is a rectangular array of different matrices.
Example
Consider the (m + `) × (n + k) matrix
   
A B Am×n Bm×k
=
C D C`×n D`×k

where, as indicated, the four submatrices A, B, C, D

are of dimension m × n, m × k, ` × n and ` × k respectively.
Note: Here matrix D may not be diagonal, or even square.
For any scalar α ∈ R,
the scalar multiple of a partitioned matrix is
   
A B αA αB
α =
C D αC αD

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 21 of 87

Partitioned Matrices: Addition

Suppose the two partitioned matrices

   
A B E F
and
C D G H

have the property that the following four pairs

of corresponding matrices have equal dimensions:
(i) A and E; (ii) B and F; (iii) C and G; (iv) D and H.
Then the sum of the two matrices is
     
A B E F A+E B+F
+ =
C D G H C+G D+H

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 22 of 87

Partitioned Matrices: Multiplication

Suppose that the two partitioned matrices

   
A B E F
and
C D G H

along with all the relevant pairs of their sub-matrices,

are compatible for multiplication.
Then their product is defined as
    
A B E F AE + BG AF + BH
=
C D G H CE + DG CF + DH

This extends the usual multiplication rule for matrices:

multiply the rows of sub-matrices in the first partitioned matrix
by the columns of sub-matrices in the second partitioned matrix.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 23 of 87

Transposes and Some Special Matrices

The rule for transposing a partitioned matrix is

>  >
C>
 
A B A
=
C D B> D >

So the original matrix is symmetric

iff A = A> , D = D> , and B = C> ⇐⇒ C = B> .
It is diagonal iff A, D are both diagonal,
while also B = 0 and C = 0.
The identity matrix is diagonal with A = I, D = I,
possibly identity matrices of different dimensions.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 24 of 87

Partitioned Matrices: Inverses, I
For an (m + n) × (m + n) partitioned matrix to have an inverse,
the equation
      
A B E F AE + BG AF + BH Im 0m×n
= =
C D G H CE + DG CF + DH 0n×m In
should have a solution for the matrices E, F, G, H, given A, B, C, D.
Assuming that the m × m matrix A has an inverse, we can:
1. construct new first m equations
by premultiplying the old ones by A−1 ;
2. construct new second n equations by:
I premultiplying the new first m equations
by the n × m matrix C;
I then subtracting this product from the old second n equations.
The result is
  −1
A−1 B
  
Im E F A 0m×n
=
0n×m D − CA−1 B G H −CA−1 In
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 25 of 87
Partitioned Matrices: Inverses, II
For the next step,
assume the n × n matrix X := D − CA−1 B
also has an inverse X−1 = (D − CA−1 B)−1 .
  −1
A−1 B
  
Im E F A 0m×n
Given = ,
0n×m D − CA−1 B G H −CA−1 In
we first premultiply the last n equations by X−1 to get

A−1 B A−1
    
Im E F 0m×n
=
0n×m In G H −X−1 CA−1 X−1

Next, we subtract A−1 B times the last n equations

from the first m equations to obtain
    
E F Im 0m×n E F
=
G H 0n×m In G H
A + A BX CA−1 −A−1 BX−1
 −1 −1 −1

=
−X−1 CA−1 X−1
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 26 of 87
Final Exercises

Exercise
1. Assume that A−1 and X−1 = (D − CA−1 B)−1 exist.
A + A−1 BX−1 CA−1 −A−1 BX−1
 −1 
Given Z := ,
−X−1 CA−1 X−1
use direct multiplication twice in order to verify that
     
A B A B Im 0m×n
Z=Z =
C D C D 0n×m In

2. Let A be any invertible m × m matrix.

 
A b
Show that the bordered (m + 1) × (m + 1) matrix
c> d
is invertible provided that d 6= c> A−1 b,
and find its inverse in this case.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 27 of 87

Partitioned Matrices: Extension
Exercise
Suppose that the two partitioned matrices

A = (Aij )k×` and B = (Bij )k×`

are both k × ` arrays of respective mi × nj matrices Aij , Bij ,

for i = 1, 2, . . . , k and j = 1, 2, . . . , `.
1. Under what conditions can the product AB
be defined as a k × ` array of matrices?
2. Under what conditions can the product BA
be defined as a k × ` array of matrices?
3. When either AB or BA can be so defined,
give a formula for its product, using summation notation.
4. Express A> as a partitioned matrix.
5. Under what conditions is the matrix A symmetric?

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 28 of 87

Outline
Special Matrices
Square, Symmetric, and Diagonal Matrices
The Identity Matrix
The Inverse Matrix
Partitioned Matrices
Permutations and Their Signs
Permutations
Transpositions
Signs of Permutations
The Product Rule for the Signs of Permutations
Determinants: Introduction
Determinants of Order 2
Determinants of Order 3
The Determinant Function
Permutation and Transposition Matrices
Triangular Matrices
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 29 of 87
Permutations
Definition
Given Nn = {1, . . . , n} for any n ∈ N with n ≥ 2,
a permutation of Nn is a bijective mapping Nn 3 i 7→ π(i) ∈ Nn .
That is, the mapping Nn 3 i 7→ π(i) ∈ Nn is both:
1. a surjection, or mapping of Nn onto Nn ,
in the sense that the range set satisfies
π(Nn ) := {j ∈ Nn | ∃i ∈ Nn : j = π(i)} = Nn ;
2. an injection, or a one to one mapping,
in the sense that π(i) = π(j) =⇒ i = j or,
equivalently, i 6= j =⇒ π(i) 6= π(j).

Exercise
Prove that the mapping Nn 3 i 7→ f (i) ∈ Nn is a bijection,
and so a permutation, if and only if
its range set f (Nn ) := {j ∈ Nn | ∃i ∈ Nn : j = f (i)}
has cardinality #f (Nn ) = #Nn = n.
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 30 of 87
Products of Permutations
Definition
The product π ◦ ρ of two permutations π, ρ ∈ Πn
is the composition mapping Nn 3 i 7→ (π ◦ ρ)(i) := π[ρ(i)] ∈ Nn .
Exercise
Prove that the product π ◦ ρ of any two permutations π, ρ ∈ Πn
is a permutation.
Hint: Show that #(π ◦ ρ)(Nn ) = #ρ(Nn ) = #Nn = n.
Example
1. If you shuffle a pack of 52 playing cards once,
without dropping any on the floor,
the result will be a permutation π of the cards.
2. If you shuffle the same pack a second time,
the result will be a new permutation ρ of the shuffled cards.
3. Overall, the result of shuffling the cards twice
will be the single permutation ρ ◦ π.
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 31 of 87
Finite Permutation Groups
Definition
Given any n ∈ N, the family Πn of all permutations of Nn includes:
I the identity permutation ι defined by ι(h) = h for all h ∈ Nn ;
I because the mapping Nn 3 i 7→ f (i) ∈ Nn is bijective,
for each π ∈ Πn , a unique inverse permutation π −1 ∈ Πn
satisfying π −1 ◦ π = π ◦ π −1 = ι.
Definition
The associative law for functions says that,
given any three functions h : X → Y , g : Y → Z and f : Z → W ,
the composite function f ◦ g ◦ h : X → W satisfies

(f ◦ g ◦ h)(x) ≡ f (g (h(x))) ≡ [(f ◦ g ) ◦ h](x) ≡ [f ◦ (g ◦ h)](x)

Exercise
Given any n ∈ N, show that (Πn , π, ι) is an algebraic group
— i.e., the group operation (π, ρ) 7→ π ◦ ρ is well-defined,
associative, with ι as the unit, and an inverse π −1 for every π ∈ Πn .
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 32 of 87
Outline
Special Matrices
Square, Symmetric, and Diagonal Matrices
The Identity Matrix
The Inverse Matrix
Partitioned Matrices
Permutations and Their Signs
Permutations
Transpositions
Signs of Permutations
The Product Rule for the Signs of Permutations
Determinants: Introduction
Determinants of Order 2
Determinants of Order 3
The Determinant Function
Permutation and Transposition Matrices
Triangular Matrices
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 33 of 87
Transpositions

Definition
For each disjoint pair k, ` ∈ {1, 2, . . . , n},
the transposition mapping i 7→ τk` (i) on {1, 2, . . . , n}
is the permutation defined by

` if i = k;

τk` (i) := k if i = `;

i otherwise;


That is, τk` transposes the order of k and `,

leaving all i 6∈ {k, `} unchanged.
Evidently τk` = τ`k and τk` ◦ τ`k = ι, the identity permutation,
and so τ ◦ τ = ι for every transposition τ .

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 34 of 87

Transposition is Not Commutative

Any (j1 , j2 , . . . , jn ) ∈ Nnn whose components are all different

corresponds to a unique permutation, denoted by π j1 j2 ... jn ∈ Πn ,
that satisfies π(i) = ji for all i ∈ Nnn .
Example
Two transpositions defined
on a set containing more than two elements
may not commute because, for example,

τ12 ◦ τ23 = π 231 6= τ23 ◦ τ12 = π 312

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 35 of 87

Permutations are Products of Transpositions

Theorem
Any permutation π ∈ Πn on Nn := {1, 2, . . . , n}
is the product of at most n − 1 transpositions.
We will prove the result by induction on n.
As the induction hypothesis,
suppose the result holds for permutations on Nn−1 .
Any permutation π on N2 := {1, 2} is either the identity,
or the transposition τ12 , so the result holds for n = 2.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 36 of 87

Proof of Induction Step
For general n, let j := π −1 (n) denote the element
that π moves to the end.
By construction, the permutation π ◦ τjn
must satisfy π ◦ τjn (n) = π(τjn (n)) = π(j) = n.
So the restriction π̃ of π ◦ τjn to Nn−1 is a permutation on Nn−1 .
By the induction hypothesis, for all k ∈ Nn−1 ,
there exist transpositions τ 1 , τ 2 , . . . , τ q
such that π̃(k) = (π ◦ τjn )(k) = (τ 1 ◦ τ 2 ◦ . . . ◦ τ q )(k)
where q ≤ n − 2 is the number of transpositions in the product.
For p = 1, . . . , q, because τ p interchanges only elements of Nn−1 ,
one can extend its domain to include n by letting τ p (n) = n.
Then (π ◦ τjn )(k) = (τ 1 ◦ τ 2 ◦ . . . ◦ τ q )(k) for k = n as well,
so π = (π ◦ τjn ) ◦ τjn−1 = τ 1 ◦ τ 2 ◦ . . . ◦ τ q ◦ τjn−1 .
Hence π is the product of at most q + 1 ≤ n − 1 transpositions.
This completes the proof by induction on n.
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 37 of 87
Adjacency Transpositions and Their Products, I

Definition
For each k ∈ {1, 2, . . . , n − 1}, the transposition τk,k+1
of element k with its successor is an adjacency transposition.

Definition
For each pair k, ` ∈ Nn with k < `, define:
1. π k%` := τ`−1,` ◦ τ`−2,`−1 ◦ . . . ◦ τk,k+1 ∈ Πn
as the composition of ` − k
successive adjacency transpositions in order,
starting with τk,k+1 and ending with τ`−1,` ;
2. π `&k := τk,k+1 ◦ τk+1,k+2 ◦ . . . ◦ τ`−1,` ∈ Πn
as the composition of the same ` − k
successive adjacency transpositions in reverse order.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 38 of 87

Adjacency Transpositions and Their Products, II
Exercise
For each pair k, ` ∈ Nn with k < `, prove that:

i
 if i < k or i > `;
I π k%` (i) := i − 1 if k < i ≤ `;

` if i = k.


I π k%k = π k&k = ι
I π k%` and π `&k are inverses
I π k%` = π 1,2,...,k−1,k+1,...,`−1,`,k,`+1,...,n
I π `&k = π 1,2,...,k−1,`,k,k+1,...,`−2,`−1,`+1,...,n

1. Note that π k%` moves k up to the `th position,

while moving each element between k + 1 and ` down by one.
2. By contrast, π `&k moves ` down to the kth position,
while moving each element between k and ` − 1 up by one.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 39 of 87

Reduction to the Product of Adjacency Transpositions
Lemma
For each pair k, ` ∈ Nn with k < `, the transposition τk`
equals both π `−1&k ◦ π k%` and π k+1%` ◦ π `&k ,
the compositions of 2(` − k) − 1 adjacency transpositions.

Proof.
1. As noted, π k%` moves k up to the `th position,
while moving each element between k + 1 and ` down by one.
Then π `−1&k moves `, which π k%` left in position ` − 1,
down to the k position, and moves k + 1, k + 2, . . . , ` − 1
up by one, back to their original positions.
This proves that π `−1&k ◦ π k%` = τk` .
It also expresses τk` as the composition of
(` − k) + (` − 1 − k) = 2(` − k) − 1 adjacency transpositions.
2. The proof that π k+1%` ◦ π `&k = τk` is similar;
details are left as an exercise.
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 40 of 87
Outline
Special Matrices
Square, Symmetric, and Diagonal Matrices
The Identity Matrix
The Inverse Matrix
Partitioned Matrices
Permutations and Their Signs
Permutations
Transpositions
Signs of Permutations
The Product Rule for the Signs of Permutations
Determinants: Introduction
Determinants of Order 2
Determinants of Order 3
The Determinant Function
Permutation and Transposition Matrices
Triangular Matrices
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 41 of 87
The Inversions of a Permutation
Definition
1. Let Nn,2 = {S ⊆ Nn | #S = 2} denote
the set of all (unordered) pair subsets of Nn .
2. Obviously, if {i, j} ∈ Nn,2 , then i 6= j.
3. Given any pair {i, j} ∈ Nn,2 , define
i ∨ j := max{i, j} and i ∧ j := min{i, j}

For all {i, j} ∈ Nn,2 , because i 6= j, one has i ∨ j > i ∧ j.

4. Given any permutation π ∈ Πn ,
the pair {i, j} ∈ Nn,2 is an inversion of π
just in case π “reorders” {i, j}
in the sense that π(i ∨ j) < π(i ∧ j).
5. Denote the set of inversions of π by
N(π) := {{i, j} ∈ Nn,2 | π(i ∨ j) < π(i ∧ j)}

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 42 of 87

The Sign of a Permutation

Definition
1. Given any permutation π : Nn → Nn ,
let n(π) := #N(π) ∈ N ∪ {0}
denote the number of its inversions.
2. A permutation π : Nn → Nn is either even or odd
according as n(π) is an even or odd number.
3. The sign or signature of a permutation π,
is defined as sgn(π) := (−1)n(π) , which is:
(i) +1 if π is even; (ii) −1 if π is odd.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 43 of 87

The Sign of an Adjacency Transposition
Theorem
For each k ∈ Nn−1 , if π is the adjacency transposition τk,k+1 ,
then N(π) = {{k, k + 1}}, so n(π) = 1 and sgn(π) = −1.

Proof.
If π is the adjacency transposition τk,k+1 , then

i
 if i 6∈ {k, k + 1}
π(i) = k + 1 if i = k

k if i = k + 1


It is evident that {k, k + 1} is an inversion.

Also π(i) ≤ i for all i 6= k, and π(j) ≥ j for all j 6= k + 1.
So if i < j, then π(i) ≤ i < j ≤ π(j) unless i = k and j = k + 1,
and so π(i) > π(j) only if (i, j) = (k, k + 1).
Hence N(π) = {{k, k + 1}}, implying that n(π) = 1.
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 44 of 87
A Multi-Part Exercise
Exercise
Show that:
1. For each permutation π ∈ Πn , one has

N(π) := {{i, j} ∈ Nn,2 | (i − j)[π(i) − π(j)] < 0}

 
π(i) − π(j)
= {i, j} ∈ Nn,2 | <0
i −j

2. n(π) = 0 ⇐⇒ π = ι, the identity permutation;

3. n(π) ≤ 12 n(n − 1), with equality
if and only if π is the reversal permutation
defined by π(i) = n − i + 1 for all i ∈ Nn — i.e.,

(π(1), π(2), . . . , π(n − 1), π(n)) = (n, n − 1, . . . , 2, 1)

Hint: Consider the number

of ordered pairs (i, j) ∈ Nn × Nn that satisfy i < j.
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 45 of 87
Outline
Special Matrices
Square, Symmetric, and Diagonal Matrices
The Identity Matrix
The Inverse Matrix
Partitioned Matrices
Permutations and Their Signs
Permutations
Transpositions
Signs of Permutations
The Product Rule for the Signs of Permutations
Determinants: Introduction
Determinants of Order 2
Determinants of Order 3
The Determinant Function
Permutation and Transposition Matrices
Triangular Matrices
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 46 of 87
Double Products

Let X = hxij i(i,j)∈Nn ×Nn denote an n × n matrix.

We introduce the notation
Yn Yn Yn−1 Yn Yn
xij := xij := xij
i>j i=1 j=1 j=1 i=j+1

for the product of all(the elements in the lower triangular matrix L

xij if i > j
with elements `ij :=
0 if i ≤ j
In case the matrix X is symmetric, one has
Yn Yn Yn
xij = xji = xij
i>j i>j i<j
Qn Q
This can be rewritten as i>j xij = {i,j}∈Nn,2 xij ,
which is the product over all unordered pairs of elements in Nn .

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 47 of 87

Preliminary Example and Definition
Example
For every n ∈ N, define the double product
Y Yn Yn
Pn,2 := |i − j| = |i − j| = |i − j|
{i,j}∈Nn,2 i>j i<j

Then one has

Pn,2 = (n − 1) (n − 2)2 (n − 3)3 · · · 3n−3 2n−2 1n−1
Qn−1 n−k
= k=1 k Qn−1
= (n − 1)! (n − 2)! (n − 3)! · · · 3! 2! = k=1 k!

Definition
For every permutation π ∈ Πn , define the symmetric matrix Xπ
 π(i) − π(j) if i 6= j

π
so that xij := i −j
1 if i = j


University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 48 of 87

Basic Lemma
Lemma
π
Q
For every permutation π ∈ Πn , one has sgn(π) = {i,j}∈Nn,2 xij .

Proof.
I Because π is a permutation,
the mapping Nn,2 3 {i, j} 7→ {π(i), π(j)} ∈ Nn,2
has inverse Nn,2 3 {i, j} 7→ {π −1 (i), π −1 (j)} ∈ Nn,2 .
In fact it is a bijection between Nn,2 and itself.
I Hence Pn,2 := {i,j}∈N |i − j| = {i,j}∈N |π(i) − π(j)| .
Q Q
n,2 n,2

|π(i) − π(j)| Q
I So = {i,j}∈Nn,2 |xijπ | = 1.
Q
{i,j}∈Nn,2
|i − j|
I Also xijπ = ∓1 according as {i, j} is or is not a reversal of π.
I It follows that
π n(π) π n(π) = sgn(π)
Q Q
{i,j}∈Nn,2 xij = (−1) {i,j}∈Nn,2 |xij | = (−1)

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 49 of 87

The Product Rule for Signs of Permutations
Theorem
For all permutations ρ, π ∈ Πn one has sgn(ρ ◦ π) = sgn(ρ) sgn(π).
Proof.
The basic lemma implies that

sgn(ρ ◦ π) Q ρ(π(i)) − ρ(π(j)) Q k −`

=
sgn(π) {i,j}∈Nn,2 i −j {k,`}∈Nn,2 π(k) − π(`)
Q ρ(π(i)) − ρ(π(j)) Q i −j
=
{i,j}∈Nn,2 i −j {i,j}∈Nn,2 π(i) − π(j)
Q
After cancelling the product {i,j}∈Nn,2 (i − j)
and then replacing π(i) by k and π(j) by `,
because π and ρ are permutations, one obtains

sgn(ρ ◦ π) Y ρ(k) − ρ(`)

= = sgn(ρ)
sgn(π) k −`
{k,`}∈Nn,2

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 50 of 87

The Sign of the Inverse Permutations

Corollary
Given any permutation π ∈ Πn , one has sgn(π −1 ) = sgn(π).

Proof.
Because the identity permutation satisfies ι = π ◦ π −1 ,
the product rule implies that

1 = sgn(ι) = sgn(π ◦ π −1 ) = sgn(π) sgn(π −1 )

Because sgn(π), sgn(π −1 ) ∈ {−1, 1},

they must both have the same sign, and the result follows.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 51 of 87

Outline
Special Matrices
Square, Symmetric, and Diagonal Matrices
The Identity Matrix
The Inverse Matrix
Partitioned Matrices
Permutations and Their Signs
Permutations
Transpositions
Signs of Permutations
The Product Rule for the Signs of Permutations
Determinants: Introduction
Determinants of Order 2
Determinants of Order 3
The Determinant Function
Permutation and Transposition Matrices
Triangular Matrices
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 52 of 87
Determinants of Order 2: Definition
Consider again the pair of linear equations

a11 x1 + a12 x2 = b1
a21 x1 + a12 x2 = b2

with its associated coefficient matrix

 
a a
A = 11 12
a21 a22

Let us define the number D := a11 a22 − a21 a12 .

6 0,
We saw earlier that, provided that D =
the two simultaneous equations have a unique solution given by
1 1
x1 = (b1 a22 − b2 a12 ), x2 = (b2 a11 − b1 a21 )
D D
The number D is called the determinant of the matrix A.
It is denoted by either det(A), or more concisely, by |A|.
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 53 of 87
Determinants of Order 2: Simple Rule
Thus, for any 2 × 2 matrix A, its determinant D is

a11 a12
|A| = = a11 a22 − a21 a12
a21 a22

For this special case of order 2 determinants, a simple rule is:

1. multiply the diagonal elements together;
2. multiply the off-diagonal elements together;
3. subtract the product of the off-diagonal elements
from the product of the diagonal elements.

Exercise
Show that the determinant satisfies
1 0 0 1
|A| = a11 a22 + a21 a12
0 1 1 0

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 54 of 87

Transposing the Rows or Columns

Example    
a b 0 1
Consider the two 2 × 2 matrices A = , T= .
c d 1 0
Note that T is orthogonal.
   
b a c d
Also, one has AT = and TA = .
d c a b
Here T is a transposition matrix which interchanges:
(i) the columns of A in AT; (ii) the rows of A in TA.
Evidently |T| = −1 and |TA| = |AT| = (bc − ad) = −|A|.
So interchanging the two rows or columns of A
changes the sign of |A|.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 55 of 87

Sign Adjusted Transpositions

Example
Next, consider the following three 2 × 2 matrices:
     
a b 0 1 0 −1
A= , T= , T̂ =
c d 1 0 1 0

Note that, like T, the matrix T̂ is orthogonal.

   
b −a −c −d
Here one has AT̂ = and T̂A = .
d −c a b
Evidently |T̂| = 1 and |T̂A| = |AT̂| = (ad − bc) = |A|.
 
> 0 1
The same is true of its transpose (and inverse) T̂ = .
−1 0
This key property makes both T̂ and T̂>
sign adjusted versions of the transposition matrix T.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 56 of 87

Cramer’s Rule in the 2 × 2 Case
Using determinant notation, the solution to the equations
a11 x1 + a12 x2 = b1
a21 x1 + a12 x2 = b2
can be written in the alternative form
1 b1 a12 1 a11 b1
x1 = , x2 =
D b2 a22 D a21 b2
This accords with Cramer’s rule,
which says that the solution to Ax = b is the vector x = (xi )ni=1
each of whose components xi is the fraction with:
1. denominator equal to the determinant D
of the coefficient matrix A (provided, of course, that D 6= 0);
2. numerator equal to the determinant of the matrix [A−i /b]
formed from A by excluding its ith column,
then replacing it with the b vector of right-hand side elements,
while keeping all the columns in their original order.
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 57 of 87
Outline
Special Matrices
Square, Symmetric, and Diagonal Matrices
The Identity Matrix
The Inverse Matrix
Partitioned Matrices
Permutations and Their Signs
Permutations
Transpositions
Signs of Permutations
The Product Rule for the Signs of Permutations
Determinants: Introduction
Determinants of Order 2
Determinants of Order 3
The Determinant Function
Permutation and Transposition Matrices
Triangular Matrices
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 58 of 87
Determinants of Order 3: Definition
Determinants of order 3 can be calculated
from those of order 2 according to the formula

a22 a23 a a a a
|A| = a11 − a12 21 23 + a13 21 22
a32 a33 a31 a33 a31 a32
X3
= (−1)1+j a1j |C1j |
j=1

where, for j = 1, 2, 3, the 2 × 2 matrix C1j is the (1, j)-cofactor

obtained by removing both row 1 and column j from the matrix A.
The result is the following sum

|A| = a11 a22 a33 − a11 a23 a32 + a12 a23 a31
− a12 a21 a33 + a13 a21 a32 − a13 a22 a31

of 3! = 6 terms, each the product of 3 elements chosen

so that each row and each column is represented just once.
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 59 of 87
Determinants of Order 3: Cofactor Expansion
The determinant expansion

|A| = a11 a22 a33 − a11 a23 a32 + a12 a23 a31
− a12 a21 a33 + a13 a21 a32 − a13 a22 a31

is very symmetric, suggesting (correctly)

that the cofactor expansion along the first row (a11 , a12 , a13 )
X3
|A| = (−1)1+j a1j |C1j |
j=1

gives the same answer as the other cofactor expansions

X3 X3
|A| = (−1)r +j arj |Crj | = (−1)i+s ais |Cis |
j=1 i=1

along, respectively:
I the r th row (ar 1 , ar 2 , ar 3 )
I the sth column (a1s , a2s , a3s )
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 60 of 87
Determinants of Order 3: Alternative Expressions
One way of condensing the notation

|A| = a11 a22 a33 − a11 a23 a32 + a12 a23 a31
− a12 a21 a33 + a13 a21 a32 − a13 a22 a31

is to reduce it to |A| = π∈Π3 sgn(π) 3i=1 aiπ(i)

P Q
for the sign function Π3 3 π 7→ sgn(π) ∈ {−1, +1}.
The six values of sgn(π) can be read off as
sgn(π 123 ) = +1; sgn(π 132 ) = −1; sgn(π 231 ) = +1;
sgn(π 213 ) = −1; sgn(π 312 ) = +1; sgn(π 321 ) = −1.

Exercise
Verify these values for each of the six π ∈ Π3 by:
1. calculating the number of inversions directly;
2. expressing each π as the product of transpositions,
and then counting these.
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 61 of 87
Sarrus’s Rule: Diagram
An alternative way to evaluate determinants only of order 3
is to add two new columns
that repeat the first and second columns:

a11 a12 a13 a11 a12

a21 a22 a23 a21 a22
a31 a32 a33 a31 a32

Then add lines/arrows going up to the right or down to the right,

as shown below
a11 a12 a13 a11 a12
& %
& %
& %
a21 a22 a23 a21 a22
% %
& %
& &
a31 a32 a33 a31 a32

Note that some pairs of arrows in the middle cross each other.
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 62 of 87
Sarrus’s Rule Defined

Now:
1. multiply along the three lines falling to the right,
then sum these three products, to obtain

a11 a22 a33 + a12 a23 a31 + a13 a21 a32

2. multiply along the three lines rising to the right,

then sum these three products, giving the sum a minus sign,
to obtain

−a11 a23 a32 − a12 a21 a33 − a13 a22 a31

The sum of all six terms exactly equals the earlier formula for |A|.
Note that this method, known as Sarrus’s rule,
does not generalize to determinants of order higher than 3.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 63 of 87

Outline
Special Matrices
Square, Symmetric, and Diagonal Matrices
The Identity Matrix
The Inverse Matrix
Partitioned Matrices
Permutations and Their Signs
Permutations
Transpositions
Signs of Permutations
The Product Rule for the Signs of Permutations
Determinants: Introduction
Determinants of Order 2
Determinants of Order 3
The Determinant Function
Permutation and Transposition Matrices
Triangular Matrices
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 64 of 87
The Determinant Mapping

Let Dn denote the domain Rn×n of n × n matrices.

Definition
For all n ∈ N, the determinant mapping
X Yn
Dn 3 A 7→ |A| := sgn(π) aiπ(i) ∈ R
π∈Πn i=1

specifies the determinant |A| of each n × n matrix A

as a function of its n row vectors (a> n
i )i=1 .

Here the multiplier sgn(π) attached to each product of n terms

can be regarded as the sign adjustment
associated with the permutation π ∈ Πn .

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 65 of 87

Row Mappings

For a general natural number n ∈ N, consider any row mapping

 
Dn 3 A 7→ D(A) = D ha> n
i i=1 ∈ R
i

defined on the domain Dn of n × n matrices A

with row vectors ha> n
i ii=1 .

Notation: For each fixed r ∈ Nn , let D(A/b> r )

denote the new value D(a> 1 , . . . , a> , b> , a> , . . . , a> )
r −1 r r +1 n
of the row mapping D after the r th row a> r of the matrix A
has been replaced by the new row vector b> r ∈R .
n

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 66 of 87

Row Multilinearity

Definition
The function Dn 3 A 7→ D(A) of the n rows ha> n
i ii=1 of A
is (row) multilinear just in case,
for each row number i ∈ {1, 2, . . . , n},
each pair b> > n
i , ci ∈ R of new versions of row i,
and each pair of scalars λ, µ ∈ R, one has

D(A−i /λb> > > >

i + µci ) = λD(A−i /bi ) + µD(A−i /ci )

Formally, the mapping Rn 3 a> >

i 7→ D(A−i /ai ) ∈ R
is required to be linear, for fixed each row i ∈ Nn .
That is, D is a linear function of the ith row vector a>
i on its own,
when all the other rows a> h (h 6
= i) are fixed.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 67 of 87

Determinants are Row Multilinear
Theorem
For all n ∈ N, the determinant mapping
X Yn
Dn 3 A 7→ |A| := sgn(π) aiπ(i) ∈ R
π∈Πn i=1

is a row multilinear function of its n row vectors (a> n

i )i=1 .

Proof.
For each fixed row r ∈ N, we have

det(A−i /λb>
r + µcr )
>
P Q
= π∈Πn sgn(π) (λbr π(r ) + µcr π(r ) ) i6=r aiπ(i)
P h Q Q i
= π∈Πn sgn(π) λb r π(r ) a
i6=r iπ(i) + µc r π(r ) a
i6=r iπ(i)
= λ det(A−i /b> >
r ) + µ det(A−i /cr )

as required for multilinearity.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 68 of 87
Outline
Special Matrices
Square, Symmetric, and Diagonal Matrices
The Identity Matrix
The Inverse Matrix
Partitioned Matrices
Permutations and Their Signs
Permutations
Transpositions
Signs of Permutations
The Product Rule for the Signs of Permutations
Determinants: Introduction
Determinants of Order 2
Determinants of Order 3
The Determinant Function
Permutation and Transposition Matrices
Triangular Matrices
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 69 of 87
Permutation Matrices: Definition
Definition
Given any permutation π ∈ Πn on {1, 2, . . . , n},
define Pπ as the n × n permutation matrix
π
whose elements satisfy pπ(i),j π =δ
= δi,j or equivalently pi,j π −1 (i),j .

That is, the rows of the identity matrix In are permuted

so that for each i = 1, 2, . . . , n,
its ith row vector is moved to become row π(i) of Pπ .
Lemma
−1
For each permutation matrix Pπ one has (Pπ )> = Pπ .
Proof.
Because π is a permutation, i = π(j) ⇐⇒ j = π −1 (i).
Then the definitions imply that for all (i, j) ∈ N2n one has
−1
(Pπ )> π
i,j = pj,i = δπ(j),i = δπ −1 (i),j = p
π
(i, j)

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 70 of 87

Permutation Matrices: Examples
Example
There are two 2 × 2 permutation matrices, which are given by:
 
12 21 0 1
P = I2 ; P = .
1 0

Their signs are respectively +1 and −1.

There are 3! = 6 permutation matrices in 3 dimensions given by:
     
1 0 0 1 0 0 0 1 0
P123 = 0 1 0 ; P132 = 0 0 1 ; P213 = 1 0 0 ;
0 0 1 0 1 0 0 0 1
     
0 1 0 0 0 1 0 0 1
P231 = 0 0 1 ; P312 = 1 0 0 ; P321 = 0 1 0 .
1 0 0 0 1 0 1 0 0
Their signs are respectively +1, −1, −1, +1, +1 and −1.
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 71 of 87
Multiplying a Matrix by a Permutation Matrix
Lemma
Given any n × n matrix A, for each permutation π ∈ Πn
the corresponding permutation matrix Pπ satisfies

(Pπ A)π(i),j = aij = (APπ )i,π(j)

Proof.
For each pair (i, j) ∈ N2n , one has
Xn Xn
(Pπ A)π(i),j = π
pπ(i),k akj = δik akj = aij
k=1 k=1

and also
Xn Xn
(APπ )i,π(j) = π
aik pk,π(j) = aik δkj = aij
k=1 k=1
   
premultiplying π rows
So A by P applies π to A’s .
postmultiplying columns
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 72 of 87
Multiplying Permutation Matrices
Theorem
Given the composition π ◦ ρ of two permutations π, ρ ∈ Πn ,
the associated permutation matrices satisfy Pπ Pρ = Pπ◦ρ .

Proof.
For each pair (i, j) ∈ N2n , one has
π ρ
Pn Pn
(Pπ Pρ )ij = k=1 pik pkj = k=1 δπ −1 (i),k δρ−1 (k),j
Pn
= k=1 δ(ρ−1 ◦π −1 )(i),ρ−1 (k) δρ−1 (k),j
Pn π◦ρ
= `=1 δ(π◦ρ)−1 (i),` δ`,j = δ(π◦ρ)−1 (i),j = pij

Corollary
1 2 q
If π = π 1 ◦ π 2 ◦ · · · ◦ π q , then Pπ = Pπ Pπ · · · Pπ .

Proof.
By induction on q, using the result of the Theorem.
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 73 of 87
Any Permutation Matrix Is Orthogonal

Proposition
Any permutation matrix Pπ satisfies Pπ (Pπ )> = (Pπ )> Pπ = In ,
so is orthogonal.

Proof.
For each pair (i, j) ∈ N2n , one has
Pn Pn
[Pπ (Pπ )> ]ij = π π
k=1 pik pjk = k=1 δπ −1 (i),k δπ −1 (j),k
= δπ−1 (i),π−1 (j) = δij

and also
Pn Pn
[(Pπ )> Pπ ]ij = π π
k=1 pki pkj = k=1 δπ −1 (k),i δπ −1 (k),j
Pn
= `=1 δ`,i δ`,j = δij

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 74 of 87

Transposition Matrices
A special case of a permutation matrix
is a transposition Trs of rows r and s.
As the matrix I with rows r and s transposed, it satisfies

δij
 if i 6∈ {r , s}
(Trs )ij = δsj if i = r

δrj if i = s


Exercise
Let A be any n × n matrix. Prove that:
1) any transposition matrix Trs is symmetric and orthogonal;
2) Trs = Tsr ; 3) Trs Tsr = Tsr Trs = I;
4) Trs A is A with rows r and s interchanged;
5) ATrs is A with columns r and s interchanged.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 75 of 87

Determinants with Permuted Rows: Theorem

Theorem
Given any n × n matrix A and any permutation π ∈ Nn ,
one has |Pπ A| = |APπ | = sgn(π) |A|.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 76 of 87

Determinants with Permuted Rows: Proof
Proof.
The expansion formula for determinants gives
X Yn
|Pπ A| = sgn(ρ) (Pπ A)i,ρ(i)
ρ∈Πn i=1

But for each i ∈ Nn , ρ ∈ Πn , one has (Pπ A)i,ρ(i) = aπ−1 (i),ρ(i) , so

Qn
|Pπ A| =
P
ρ∈Πn sgn(ρ) a −1
P i=1 π (i),ρ(i) Q
= [1/ sgn(π)] π◦ρ∈Πn sgn(π ◦ ρ) ni=1 ai,(π◦ρ)(i)
= sgn(π) σ∈Πn sgn(σ) ni=1 ai,σ(i) = sgn(π) |A|
P Q

because sgn(π ◦ ρ) = sgn(π) sgn(ρ) and 1/ sgn(π) = sgn(π),

whereas there is an obvious bijection Πn 3 ρ ↔ π ◦ ρ = σ ∈ Πn
on the set of permutations Πn .
The proof that |APπ | = sgn(π) |A| is sufficiently similar
to be left as an exercise.
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 77 of 87
The Alternation Rule for Determinants

Corollary
Given any n × n matrix A
and any transposition τrs with associated transposition matrix Trs ,
one has |Trs A| = |ATrs | = −|A|.

Proof.
Apply the previous theorem in the special case
when π = τrs and so Pπ = Trs .
Then, because sgn(π) = sgn(τrs ) = −1,
the equality |Pπ A| = sgn(π) |A| implies that |Trs A| = −|A|.
We have shown that, for any n × n matrix A, given any:
1. permutation π ∈ Nn , one has |Pπ A| = |APπ | = sgn(π) |A|;
2. transposition τrs , one has |Trs A| = |ATrs | = −|A|.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 78 of 87

Sign Adjusted Transpositions

We define the sign adjusted transposition matrix T̂rs

as either one of the two matrices that:
(i) swaps rows or columns r and s;
(ii) then multiplies one, but only one,
of the two swapped rows or columns by −1.
As the matrix I with rows r and s transposed,
and then one sign changed, it satisfies

δij
 if i 6∈ {r , s}
(Trs )ij = αs δsj if i = r

αr δrj if i = s


where αr , αs ∈ {−1, +1} with αr = −αs .

It evidently satisfies |T̂rs A| = |AT̂rs | = |A|.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 79 of 87

Sign Adjusted Permutations
Given any permutation matrix P,
there is a unique permutation π such that P = Pπ .
Suppose that π = τr1 s1 ◦ · · · ◦ τr` s` is any one of the several ways
in which the permutation π can be decomposed
into a composition of transpositions.
Then P = `k=1 Trk sk and |PA| = (−1)` |A| for any A.
Q

Definition
Say that P̂ is a sign adjusted version of P = Pπ
just in case it can be expressed as the product P̂ = `k=1 T̂rk sk
Q

of sign adjusted transpositions satisfying P = `k=1 Trk sk .

Q

Then it is easy to prove by induction on `

that for every n × n matrix A one has |P̂A| = |AP̂| = |A|.
Recall that all the elements of a permutation matrix P are 0 or 1.
A sign adjustment of P involves changing some of the 1 elements
into −1 elements, while leaving all the 0 elements unchanged.
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 80 of 87
Outline
Special Matrices
Square, Symmetric, and Diagonal Matrices
The Identity Matrix
The Inverse Matrix
Partitioned Matrices
Permutations and Their Signs
Permutations
Transpositions
Signs of Permutations
The Product Rule for the Signs of Permutations
Determinants: Introduction
Determinants of Order 2
Determinants of Order 3
The Determinant Function
Permutation and Transposition Matrices
Triangular Matrices
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 81 of 87
Triangular Matrices: Definition

Definition
A square matrix is upper (resp. lower) triangular
if all its non-zero off diagonal elements are above and to the right
(resp. below and to the left) of the diagonal
— i.e., in the upper (resp. lower) triangle
bounded by the principal diagonal.
I The elements of an upper triangular matrix U
satisfy (U)ij = 0 whenever i > j.
I The elements of a lower triangular matrix L
satisfy (L)ij = 0 whenever i < j.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 82 of 87

Products of Upper Triangular Matrices
Theorem
The product W = UV of any two upper triangular matrices U, V
is upper triangular,
with diagonal elements wii = uii vii (i = 1, . . . , n) equal
to the product of the corresponding diagonal elements of U, V.
Proof.
Given any two upper triangular n × n matrices U and V,
one has uik vkj = 0 unless both i ≤ k and k ≤ j.
So the elements (wij )n×n of their product W = UV satisfy
(P
j
k=i uik vkj if i ≤ j
wij =
0 if i > j

Hence W = UV is upper triangular.

Finally, when j = i the above sum collapses to just one term,
and wii = uii vii for i = 1, . . . , n.
University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 83 of 87
Triangular Matrices: Exercises

Exercise
Prove that the transpose:
1. U> of any upper triangular matrix U is lower triangular;
2. L> of any lower triangular matrix L is upper triangular.

Exercise
Consider the matrix Er +αq
that represents the elementary row operation
of adding a multiple of α times row q to row r , with r 6= q.
Under what conditions is Er +αq
(i) upper triangular? (ii) lower triangular?

Hint: Apply the row operation to the identity matrix I.

Answer: (i) iff q < r ; (ii) iff q > r .

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 84 of 87

Products of Lower Triangular Matrices

Theorem
The product of any two lower triangular matrices
is lower triangular.

Proof.
Given any two lower triangular matrices L, M,
taking transposes shows that (LM)> = M> L> = U,
where the product U is upper triangular,
as the product of upper triangular matrices.
Hence LM = U> is lower triangular,
as the transpose of an upper triangular matrix.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 85 of 87

Determinants of Triangular Matrices
Theorem
The determinant of any n × n upper triangular matrix U
equals the product of all the elements on its principal diagonal.

Proof.
Recall the expansion formula |U| = π∈Π sgn(π) ni=1 uiπ(i)
P Q
where Π denotes the set of permutations on {1, 2, . . . , n}.
Because U is upper triangular, one has uiπ(i) = 0 unless i ≤ π(i).
So ni=1 uiπ(i) = 0 unless i ≤ π(i) for all i = 1, 2, . . . , n.
Q

But the identity ι is the only permutation π ∈ Π

that satisfies i ≤ π(i) for all i ∈ Nn .
Because sgn(ι) = +1, the expansion reduces to the single term
Yn Yn
|U| = sgn(ι) uiι(i) = uii
i=1 i=1

This is the product of the n diagonal elements, as claimed.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 86 of 87
Invertible Triangular Matrices

Qn
Similarly |L| = i=1 `ii for any lower triangular matrix L.
Evidently:
Corollary
A triangular matrix (upper or lower) is invertible
if and only if no element on its principal diagonal is 0.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 87 of 87