MATRICES AND VECTOR SPACES
The matricesA,BandCare given byA=(
2 − 1
31
)
, B=
(
10
0 − 2
)
, C=
(
− 21
− 11
)
.
Find the matrixD=A+2B−C.D=
(
2 − 1
31
)
+2
(
10
0 − 2
)
−
(
− 21
− 11
)
=
(
2+2× 1 −(−2) −1+2× 0 − 1
3+2× 0 −(−1) 1+2×(−2)− 1
)
=
(
6 − 2
4 − 4
)
.
From the above considerations we see that the set of all, in general complex,M×Nmatrices (with fixedMandN) forms a linear vector space of dimension
MN. One basis for the space is the set ofM×NmatricesE(p,q)with the property
thatEij(p,q)=1ifi=pandj=qwhilstEij(p,q)= 0 for all other values ofiand
j, i.e. each matrix has only one non-zero entry, which equals unity. Here the pair
(p, q) is simply a label that picks out a particular one of the matricesE(p,q),the
total number of which isMN.
8.4.2 Multiplication of matricesLet us consider again the ‘transformation’ of one vector into another,y=Ax,
which, from (8.24), may be described in terms of components with respect to a
particular basis as
yi=∑Nj=1Aijxj fori=1, 2 ,...,M. (8.32)Writing this in matrix form asy=Axwe have
y 1
y 2
..
.
yM
=
A 11 A 12 ... A 1 N
A 21 A 22 ... A 2 N
..
...
...
...
.
AM 1 AM 2 ... AMN
x 1
x 2..
.xN
(8.33)where we have highlighted with boxes the components used to calculate the
elementy 2 : using (8.32) fori=2,
y 2 =A 21 x 1 +A 22 x 2 +···+A 2 NxN.All the other componentsyiare calculated similarly.
If instead we operate withAon a basis vectorejhaving all components zero