Page 140 - A Course in Linear Algebra with Applications

124 Chapter Five: Basis and Dimension

Hence the transactions correspond to column vectors

h

such that ti-\ \-t n = 0. Now vectors of this form are easily
n
seen to constitute a subspace T of the vector space R ; this
is called the transaction space. Evidently T is just the null
space of the matrix

(\ 1 1 ••• 1
0 0 0 ••• 0
A =
\ 0 0 0 0,

Now A is already in reduced row echelon form, so we can read
off at once the general solution of the linear system AX = 0 :

/ - c 2 - c 3 Cn\

c 2
X = C3

\ J

with arbitrary real scalars c 2 , C3,..., c n. Now we can find a
basis of the null space in the usual way. For i = ,..., n define
2
Ti to be the n-column vector with first entry —1, zth entry 1,
and all other entries zero. Then

X = c 2T 2 + C3T3 + • • • + c nT n

and {T2, 3,..., T n} is a basis of the transaction space T. Thus
T
dim(T) = n - 1. Observe that Ti corresponds to a simple
transaction, in which there is a flow of funds amounting to