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

Last updates: 14 September 2013


Let m and n be positive integers. Let R be a commutative ring.

Let Mm×n (R) be the set of m×n matrices with entries in R.


( 2 3 1 2 ) + ( 5 2 1 6 ) = ( 7 5 2 8 ) , ( 2 3 1 2 ) ( 5 2 1 6 ) = ( 13 22 7 14 ) and ( 2 3 1 2 ) t = ( 2 1 3 2 ) .

HW: Show that if a,b,c are n×n matrices then (ab)c =a(bc), a(b+c) = ab+ac and (b+c)a =ba+ca.

HW: Give an example of n×n matrices a and b such that abba.

Let S be a set.

Let m and n be positive integers and let R be a ring.
(a)   The set
Mm×n (R)= {m×n matrices with entries in R} ,
with operation addition is an abelian group with zero element 0 Mm×n (R) given by
0ij =0 for i=1,2,,m, and j=1,2,,n.
(b)   The set Mm×n (R) with operations addition and scalar multiplication is an R-module.
(c)   The set Mn(R) = Mn×n (R) with operations addition and product is a ring with identity 1 Mn(R) given by
1ij = δij for i=1,2,,n, and j=1,2,,n.


To show: A+B=B+A
To show: (A+B)ij= (B+A)ij
(A+B)ij = Aij+Bij = Bij+Aij, since𝔽 has commutative addition = (B+A)ij.
To show: (A+B)+C=A+(B+C)
To show: ((A+B)+C)ij= (A+(B+C))ij
(A+B)+C= A+(B+C) = (A+B)ij+Cij =(Aij+Bij) +Cij = Aij+ (Bij+Cij), since+ is associative in𝔽, = Aij+ (B+C)ij= (A+(B+C))ij.
(c) Define the zero matrix 0 by 0ij=0.
To show: (ca) 0+A=A
(cb) A+0=A
To show: (0+A)ij=Aij
(0+A)ij = 0ij+Aij =0+Aij= Aij.
To show: (A+0)ij=Aij
(A+0)ij = Aij+0ij =Aij+0= Aij.
To show: (AB)C=A(BC)
To show: ((AB)C)ij= (A(BC))ij
((AB)C)ij = k=1n (AB)ik Ckj = k=1n =1m (AiBk) Ckj = =1m k=1n (AiBk) Ckj = =1m k=1n Ai (BkCkj) = =1m Ai (BC)j.
To show: (ea) (A+B)·C=AC+BC
(eb) C(A+B)=CA+CB
To show: ((A+B)C)ij =(AC+BC)ij
((A+B)C)ij = k=1n (A+B)ik Ckj= k=1n (AikBik) Ckj = k=1n ( AikCkj+ BikCkj ) ,by the distributive property in𝔽 = ( k=1n Aik Ckj ) + ( k=1n Bik Ckj ) = (AC)ij+ (BC)ij = (AC+BC)ij.
To show: (C(A+B))ij= (CA+CB)ij
(C(A+B))ij = k=1m Cik (A+B)kj =k=1m Cik (Akj+Bkj) = k=1m ( CikAkj+ CikBkj ) ,by the distributive property in𝔽, = ( k=1m CikAkj ) + ( k=1m CikBkj ) = (CA)ij+ (CB)ij = (CA+CB)ij.
(f) Define the identity matrix 1 by 1ij= { 1, ifi=j, 0, ifij.
To show: (fa) 1·A=A
(fb) A·1=A
To show: (1·A)ij= Aij
(1·A)ij = k=1n 1ik Akj =1ii Aij+ ki 1ik Akj = 1·Aij+0 =Aij.
To show: (A·1)ij =Aij
(A·1)ij = k=1m Aik1kj =Aij1jj +kj Aik 1kj = Aij·1+0 =Aij.
(g) Define the matrix -A by (-A)ij= -Aij.
To show: (ga) A+(-A)=0
(gb) (-A)+A=0
To show: (A+(-A))ij =0ij
(A+(-A))ij = Aij+ (-A)ij = Aij+ (-Aij) = 0=0ij
To show: ((-A)+A)ij =0ij
((-A)+A)ij= (-A)ij+ Aij=-Aij +Aij=0=0ij.

HW: Show that Mn×1 (R) Rn as R-modules.


(a)   The set of triangular matrices and the set of diagonal matrices are subrings of Mn(R).
(b)   The ideals of Mn(R) are Mn(I), where I is an ideal of R.

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) .

Notes and References

A matrix is just a table of numbers, and hence matrices appear everywhere. These notes are taken from notes of Arun Ram from 1999. A nice solid reference is [HP].


[HP] W.V.D. Hodge and D. Pedoe, Methods of algebraic geometry. Vol. I. Reprint of the 1947 original. Cambridge Mathematical Library. Cambridge University Press, Cambridge, 1994. viii+440 pp. ISBN: 0-521-46900-7, 14-01 (01A75) Methods of algebraic geometry I, Cambrdige University Press, MR1288305.

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