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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
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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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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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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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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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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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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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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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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}.
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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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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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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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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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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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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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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.
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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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