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8/13/2019 Det Slides

1/17

Axiomatic approach to the determinant function

Recall the formula for determinants ofk k matrices for k= 1, 2, 3:

det[a] =a, det a1 b1

a2 b2

=a1b2 a2b1 and

det

a1 b1 c1a2 b2 c2a3 b3 c3

= a1b2c3a1b3c2a2b1c3+a3b1c2+a2b3c1a3b2c1.

Our approach to determinants ofn nmatrices is via their properties(rather than via an explicit formula as above).

Let d be a function that associates a scalar d(A) R with everyn n matrix A with entries in R. We use the following notation.If the columns ofA areA1, A2, . . . , An, we writed(A) =d([A1, A2, . . . , An]) simply as d(A1, A2, . . . , An). Thus d canbe regarded as a function on Rn1 Rn1, the n-fold product of

n1 column vectors.1 / 17
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2/17

Axioms for the determinant function

(i) d is called multilinear if for each j = 1, 2, . . . , n ,scalars , andn 1 column vectors A1, . . . , Aj1, Aj+1, . . . , An, B , C , we have

d(A1, . . . , Aj1, B+C, Aj+1, . . . , An) =

d(A1, . . . , Aj1, B , Aj+1, . . . , An) + d(A1, . . . , Aj1, C , Aj+1, . . . , An).

(ii) d is called alternating ifd(A1, A2, . . . , An) = 0 whenever Ai =Ajfor some i=j, i, j = 1, . . . , n.

(iii) d is called normalized ifd(I) =d(e1, e2, . . . , en) = 1, where ej isthe jth basic column vector, j = 1, . . . , n.

(iv) A normalized, alternating and multillinear function d on the set of alln n matrices is called a determinant function of order n.

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3/17

Formula for the determinant of a 2 2 matrix

Suppose d is an alternating multilinear normalized function on 2 2matrices A= [A1, A2]. We show that

d(A1, A2) =d a1

a2

, b1

b2

= a1b2 a2b1.

Write the first column as A1=a1e1+a2e2 and the second column asA2=b1e1+b2e2. Using the axioms for a determinant function, we obtain

d(e1, e2) +d(e2, e1) =d(e1+e2, e1+e2) = 0, and

d(A1, A2) = d(a1e1+a2e2, b1e1+b2e2)

= d(a1e1+a2e2, b1e1) +d(a1e1+a2e2, b2e2)

= d(a1e1, b1e1) +d(a2e2, b1e1)

+d(a1e1, b2e2) +d(a2e2, b2e2)

= a2b1d(e2, e1) +a1b2d(e1, e2)

= (a1b2 a2b1)d(e1, e2) =a1b2 b1a2.

Exercise: Can you do this for a 3 3 matrix A=[A1, A2, A3]?3 / 17
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4/17

Properties of a determinant function

We show that there can be only one determinant function of order n.

Lemma: Suppose that d(A1, A2, . . . , An) is a multilinear alternatingfunction on the set of all n n matrices. Then(a) If some Aj = 0, then d(A1, A2, . . . , An) = 0.(b) d(A1, A2, . . . , Aj , Aj+1, . . . An) =d(A1, A2, . . . , Aj+1, Aj , . . . , An).

(c) d(A1, A2, . . . , Ai, . . . , Aj , . . . , An) =d(A1, A2, . . . , Aj , . . . , Ai, . . . , An).

Proof: (a) Let Aj = 0. For any column C, by the multilinearity,d(A1, A2, . . . , Aj , . . . , An) =d(A1, A2, . . . , C C , . . . , An)=d(A1, A2, . . . , C , . . . , An) d(A1, A2, . . . , C , . . . , An) = 0.

(b) Put Aj =B, Aj+1=C. By the alternating property,

0 = d(A1, A2, . . . , B+C, B+C, . . . , An)= d(A1, A2, . . . , B , B+C, . . . , An) +d(A1, A2, . . . , C , B+C, . . . , An)

= d(A1, A2, . . . , B , C , . . . , An) +d(A1, A2, . . . , C , B , . . . , An)

Hence d(A1, A2, . . . , B , C , . . . , An) =d(A1, A2, . . . , C , B , . . . , An).

(c) follows from (b) by 2(j i) 1 transpositions.4 / 17
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5/17

Uniqueness of the determinant function

Lemma: Suppose f is a multilinear alternating function on n nmatrices and f(e1, e2, . . . , en) = 0. Then f is identically zero.

Proof: Let A= [aij ] be an n n matrix with columns A1, . . . , An. Forj = 1, . . . , n, write Aj as

Aj =a1je1+a2je2+ +anjen.

Since f is multilinear, we obtain

f(A1, . . . , An) =h

ah(1)1ah(2)2 ah(n)n f(eh(1), eh(2), . . . , eh(n)),

where the sum is over all functions h: {1, 2, . . . , n} {1, 2, . . . , n}.

Since f is alternating, we obtain

f(A1, . . . , An) =h

ah(1)1ah(2)2 ah(n)n f(eh(1), eh(2), . . . , eh(n)),

where the sum is now over all bijections h: {1, 2, . . . , n}{1, 2, . . . , n}.5 / 17
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6/17

Uniqueness of the determinant function

By using part (c) of the lemma above, we see that

f(A1, . . . , An) =h

ah(1)1ah(2)2 ah(n)n f(e1, e2, . . . , en),

where the sum is over all bijections h: {1, 2, . . . , n} {1, 2, . . . , n}. Since

f(e1, . . . , en) = 0, we obtain f(A) = 0.Theorem: Let fbe an alternating multilinear function of order n, andlet d be a determinant function of order n. Then for every n n matrixA= [A1, A2, . . . , An], we have

f(A1, A2, . . . , An) =d(A1, A2, . . . , An)f(e1, e2, . . . , en).

In particular, iffis also a determinant function, thenf(A1, A2, . . . , An) =d(A1, A2, . . . , An).

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7/17

Proof of the uniqueness of a determinant function

Proof: Consider the function g given by

g(A1, A2, . . . , An) =f(A1, A2, . . . , An) d(A1, A2, . . . , An)f(e1, e2, . . . , en).

Since f, d are alternating and multilinear, so is g. Since

g(e1, e2, . . . , en) = 0,

the result follows from the previous lemma.

Notation: We shall denote the determinant ofAby det A.

Setting det[a] =a gives the existence for n= 1.

Assume that we have proved the existence of the determinant function detof order n 1. The determinant function of order n can be defined interms of the determinant function of order n 1 as follows.

Let Aij denote the (n 1) (n 1) matrix obtained from an nnmatrix

A by deleting the ith row and jth column ofA for i,j =1, . . . , n.7 / 17
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8/17

Existence of the determinant function

TheoremThe function f on nn matrices A= [aij ] given by

f(A) :=a11det A11 a12det A12+ + (1)

n+1

a1ndet A1nis multilinear, alternating, and normalized, and hence it is the determinantfunction of order n. (Expansion by the first row)

Proof: Fix j {1, . . . , n 1}. Suppose the columns Aj and Aj+1 ofA

are equal. Then A1i has two equal columns except when i=j, j+ 1.By induction det(A1i) = 0 ifi=j and i=j+ 1. Thus

f(A) = (1)j+1a1jdet(A1j) + (1)j+2a1(j+1)det(A1(j+1)).

Since Aj =Aj+1, we have a1j =a1j+1 and A1j =A1j+1. Thus f(A) = 0.Therefore the function f is alternating.

Let A= I= [e1, e2, . . . , en]. Then a12= = a1n = 0, so that

f(A) = 1 det(A11) = det(e1, e2, . . . , en1) = 1.

The multilinear property of the function f canbesimilarlyproved.8 / 17
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9/17

Determinants of triangular and elementary matrices

In a similar manner, we have the expansion by row k= 2, . . . n:det A=

n

j=1

(1)k+jakjdet Akj forA= [aij ].

Theorem: (i) Let Ube an upper triangular or a lower triangular matrix.Then det U is equal to the product of the diagonal entries ofU.

(ii) Let E:=I+ Eij be an elementary matrix of the type I, where Rand i=j. Then det E= 1.

(iii) Let E:=I+Eij+Eji Eii Ejj be an elementary matrix of typeII, where i=j. Then det E=1.

(iv) Let E:=I+ ( 1)Eii be an elementary matrix of type III, where0= R. Then det E=.

Proof: (i) Let U= [uij ] be upper triangular. Arguing as in the lemmaused for proving the uniqueness of the determinant function, we obtain

det U=

h uh(1)1uh(2)2 uh(n)n,

where the sum is over all bijections h:{1, 2, . . . , n} {1, 2, . . . , n}.

9 / 17
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10/17

Multiplicity of det

Since U is upper triangular the only choice of the bijection h yeilding a

nonzero term is the identity bijection (and this gives a plus sign).

(ii) follows from part (i).

(iii) E is obtained from the identity matrix by exchanging columns i and j.The result follows since the determinant function is alternating.

(iv) follows form part (i).

Theorem: (Multiplicativity ofdet) For any nn matrices Aand B,

det(AB) = det A det B.

Proof: Since AB =A[B1, . . . , Bn] = [AB1, . . . , A Bn], we need prove

det(AB1, AB2. . . , ABn) = det Adet B.

Keep A fixed, and for an arbitrary nn matrix B, define

f(B1

, B2

, . . . , Bn

) = det(AB1

, AB2

, . . . , A Bn

).

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11/17

det(AB) = detA detB

The function f is alternating, since f(B1, . . . , Bi, . . . , Bi, . . . , Bn) =det(AB1, . . . , A Bi, . . . , A Bi, . . . , A Bn) = 0.

To show that the function f is also multilinear, let C and D be n1column vectors, and let , R. Then

f(B1, . . . , C + D, . . . , Bn) = det(AB1, . . . , A(C+D), . . . , A Bn)

= det(AB1, . . . , A C + AD, . . . , ABn)= det(AB1, . . . , A C , . . . , A Bn)

+ det(AB1, . . . , A D , . . . , A Bn)

= f(B1, . . . , C , . . . Bn) +f(B1, . . . , D , . . . , Bn

Since f is alternating and multilinear, an earlier theorem givesf(B1, B2, . . . , Bn) = det(B1, . . . , Bn)f(e1, e2, . . . , en).

But f(e1, e2, . . . , en) = det(Ae1, . . . , A en) = det(A1, . . . , An) = det A.Thus det(AB1, AB2, . . . , A Bn) = det(B1, . . . , Bn)det A= det A det B,

as desired.11 / 17
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12/17

Determinant and invertibility

Lemma: (i) Let Abe a square matrix. Then A is invertible if and only ifdet A= 0, and in that case,

det A1 = 1det A

.

(ii) Suppose A and B are square matrices such that AB =I. Then A andB are invertible and B =A1.

Proof: (i) Let A be invertible. Then(det A)(det A1) = det AA1 = det I= 1, so that det A= 0.

Suppose A is not invertible. Then there is a nonzero column vector x suchthat Ax= 0. Hence a linear combination of the column vectors of Ahaving at least one nonzero coefficient is equal to the zero column vector.

Thus some column ofAis a linear combination of other columns ofA. Itnow follows from the multilinearity and alternating properties of thedetetminant function that det A= 0.

(ii) det A det B = det(AB) = det(I) = 1. So det A= 0 and A is

invertible. Now B = (A1

A)B =A1

(AB) =A1

is alsoinvertible.12 / 17
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13/17

Determinant of the transpose of a matrix

Theorem: For any n n matrix A,

det At = det A.

Proof: IfAis not invertible, then At is also not invertible, anddet A= 0 = det At. Let now Abe invertible. Then there are elementarymatrices E1, . . . , E k such that

A= E1 Ek, and so det A= det E1 det Ek.Now the transpose of an elementary matrix is also an elementary matrix ofthe same type, and has the same determinant. Hencedet At = det(Etk E

t1) = det E

tk det E

t1= det Ek det E1= det A.

Corollary: The determinant is a multilinear, alternating and normalizedfunction of the rows of a square matrix, and so for A= [aij ], we have thefollowing expansion by column k, where k= 1, . . . , n:

det A=n

i=1(1)k+iaikdet Aik.

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14/17

Computation of determinant by Gauss-Jordan Method

Let A be an n n matrix. Reduce A to its row canonical form (rcf)

U by

elementary row operations (E.R.O.s) of type I (denoted by Ri+Rj withi=j), of type II (denoted by Ri Rj), and of type III (denoted byiRi with 0=i), that is, by premultiplying A by elementary matrices oftypes I, II and III.

To obtainU from A, ifr row interchanges Ri Rj have been used andif1, . . . , p are the nonzero scalars used for the row operations iRi,then detU= (1)r1 pdet A.This follows from the result about the determinants of elementary matricesand from the multiplicativity of the determinant function proved earlier.

Ifu11, . . . , unn are the diagonal entries ofU, then detU=u11 unn.Hence

det A= (1)r(1 p)1u11 unn.

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15/17

The cofactor matrix

Definition: Let A= [aij ] be an n n matrix. For i, j = 1, . . . , n,the cofactor of the entry aij , denoted by cofaij , is defined by

cofaij = (1)i+j det Aij .

Ifn= 1, we define cofa11= 1.

The cofactor matrix ofA, denoted by cofA, is the matrix

cofA = [cofaij ].Expansion ofdet A by row k and by column k can be written as

ni=1

aikcofaik = det A=

nj=1

akjcofakj for each k= 1, . . . , n .

Theorem: For any square matrix A,

A(cofA)t = (det A)I= (cofA)tA.

In particular, ifdet A is nonzero, then A1 = 1

det A

(cofA)t.

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16/17

Matrix inverse using its cofactor matrix

Proof: The (i, j)th entry ofC:= (cofA)tA is cij =n

k=1akjcofaki.

Ifi=j, then cij is the expansion ofdet Aby the jth column ofA.Let i=j. Consider the matrix B = [bij ] obtained by replacing the ithcolumn ofA by the jth column ofA. Then B has a repeated column.Expanding det B by its ith column, cij =

nk=1bkicofbki = det B = 0.

The other equation A(cofA)t = (det A)I is proved similarly.

Cramers Rule: Consider the system

a11 a12 a1n

a21 a22 a2n...

... ...

...an1 an2 ann

x1

x2...

xn

= b1

b2...

bn

ofn linear equations in the n unknowns x

1, x

2, . . . , x

n.

16 / 17

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17/17

Cramers Rule

Suppose the coefficient matrix A= [aij ] is invertible. Let Cj be the

matrix obtained from A by replacing its jth column by the right sideb= [b1, b2, . . . , bn]

t. Then the unique solution of the system Ax=b isgiven by

xj =det Cj

det A forj = 1, . . . n .

Proof: Let A1, . . . , An be the columns ofA. Ifx= [x1, . . . , xn]t satisfies

Ax= b, then b=x1A1+x2A2+ +xnAn.

Since the determinant function is multilinear and alternating, we obtaindet Cj = det[A1, . . . , Aj1, b , Aj+1, . . . , An] =xjdet A.

Hence xj =det Cj

det A forj = 1, 2, . . . , n .

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