Traces and Determinants

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last update: 13 August 2013


Let a be an n×n matrix with entries aij.

Let 𝔽 be a field and let n >0.
(a)   Up to constant multiples, tr: Mn(𝔽) 𝔽 is the unique function such that
(a1)   If c𝔽 and a,b Mn(𝔽) then
tr(a+b) =tr(a) +tr(b) and tr(ca)= ctr(a).
(a2)   If a,b Mn(𝔽) then
tr(ab) = tr(ba) .
(b)   Identify Mn(𝔽) with the 𝔽-module 𝔽n × × 𝔽n ntimes ,
Mn(𝔽) 𝔽n × × 𝔽n ntimes a (a1| a2|| an) ,      where ai are the columns of a.
The function det: Mn(𝔽)𝔽 is the unique function such that
(b1)   (columnwise linear) If i {1,2,,n} and c𝔽 then
det (a1| | ai+bi || an) = det (a1| | ai|| an) + det (a1| | bi|| an)
det (a1| | cai|| an) = c det (a1| | ai|| an) ,
(b2) If i {1,2,, n-1} then
det (a1| | ai| ai+1| | an) = - det (a1| | ai+1| ai| | an) ,
(b3)   if a,b Mn(𝔽) then
det(ab) = det(a) det(b) .

(Laplace expansion) det ( a11a12a1n a21a22a2n an1an2ann ) =ipidet ( ai1j1 ai1jp aipj1 aipjp ) ( aip+1jp+1 aip+1jn ainjp+1 ainjn ) where j1,,jn is a fixed permutation of 1,2,,n and the sum is over all possible divisions of 1,2,,n into two sets i1<<ipand ip+1<<in, and pi=+1 or -1 according as i1,..,in and j1,,jn are like or unlike derangements of 1,2,,m.

If A and B are n×n matrices then det(AB)= det(A)det(B).


To show: det(AB)= det(A)det(B)
det(AB) = wSn det(w) (AB)1,w(1) (AB)n,w(n) = wSndet(w) k1,,kn A1,k1 Bk1,w(1) An,kn Bkn,w(n) = k1,,kn A1,k1 An,kn wSndet(w) Bk1,w(1) Bkn,w(n) = k=(k1,,kn)Sn det(k) A1,k1 An,kn wSn det(k-1) det(w) Bk1,w(1) Bkn,w(n) = k=(k1,,kn)Sn det(k) A1,k1 An,kn wSn det(wk-1) Bk1,wk-1(k1) Bkn,wk-1(kn) = kSn det(k) A1,k1 An,kn wk-1Sn det(wk-1) B1,wk-1(1) Bn,wk-1(n) = det(A) det(B)

a) Let B be the matrix obtained by switching two rows of A. Then det(B)=-det(A).
b) Let B be the matrix obtained by adding a multiple of a row of A to another row of A. Then det(B)=det(A).
c) Let B be the matrix obtained by multiplying a row of A by a constant cR. Then det(B)=cdet(A).

HW: Show that if two rows of A are the same then det(A)=0.

Inverses and Cramer's rule"

i=1n aikAhi =δhkdetA.

If det(a) is a unit in R then a-1=det (a)-1 (Aij) . SHOULD THERE BE A TRANSPOSE HERE?

Cramer’s rule for AX=B.

Put Thms VII-X of Hodge and Padoe as exercises.


If A is a p×qλ-matrix of rank r and E1(λ),,Er(λ) are its invariant factors then there exist MGLp(K[λ]) and NGLq(K[λ]) such that MAN= ( E1(λ) E2(λ)0 Er(λ) 00 0 ) .

Two p×qλ-matrices A and B are equivalent if and only if they have the same invariant factors and if and only if they have the same elementary divisors.

Note that these proofs work for any Euclidean domain.

Cayley-Hamilton Theorem.

Note that the proof of Theorem II §10 Hodge and Padoe!
Theorem III (Hodge and Padoe) gives minimal polynomial.

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