F. Xavier Malcata - Mathematics for Enzyme Reaction Kinetics and Reactor Performance

Consider now ( m × n ) matrix A, ( n × p ) matrix B, and ( p × m ) matrix C; ( m × p ) matrix ABexists, and its product by ( p × m ) matrix Cwill eventually lead to ( m × m ) matrix ABC– so there will be a true main diagonal of ABCfor it being square, and its trace can accordingly be calculated. Recall the associative property of multiplication of matrices, i.e.

(4.89)

as per Eq. (4.57); assuming matrices Aand Bare defined by their generic elements a i,jand b j,k, as per Eq. (4.1), and

(4.90)

en lieu of Eq. (4.46), respectively, one may multiply Aby Bto get

(4.91)

A similar reasoning may be applied to multiplication of matrix ABby matrix C, with generic element c k,l, viz.

(4.92)

ordered multiplication of Eqs. (4.91)and (4.92)leads indeed to

(4.93)

where the associative property of multiplication of scalars was taken on board. The trace will pick up the sum of only the elements in the main diagonal, i.e. those abiding to l = i , according to

(4.94)

where advantage was implicitly taken of the commutativity of addition of scalars; however, the order of factors in each term and corresponding summations is arbitrary – because both addition and multiplication of scalars are commutative, while the limits of the said summations are independent of each other. Consequently, Eq. (4.94)may be rewritten as

(4.95)

where the definition of trace of a matrix was recalled once more; on the other hand,

(4.96)

as per Eqs. (4.90)- (4.92)– following convenient relabeling of subscript (dummy) lto i, since C≡ [ c k,i] ≡ [ c k,l] plays the role of generic element, with j holding no relationship to i (or l , for that matter). Hence, the product of BCby Alooks like

(4.97)

where A≡ [ a i,j] ≡ [ a i,l] for absence of constraints encompassing j and l ; Eq. (4.2)was again followed, coupled with the distributive property of multiplication of scalars. The trace of ( BC) Aabides to

(4.98)

based on its definition; the right‐hand side of Eq. (4.98)is identical to the right‐hand side of Eq. (4.95), so one readily concludes that

(4.99)

– whereas combination with Eq. (4.57)leads finally to

(4.100)

Remember that BCis an ( n × m ) matrix and Ais an ( m × n ) matrix, so ( BC) A = BCA is an ( n × n ) matrix – and thus distinct from the ( m × m ) matrix ( AB) C = ABC as outlined above, due to the product of matrices not being commutative; nevertheless, the traces of BCAand ABCare the same. A similar derivation would prove that

(4.101)

where ( p × n ) matrix CAis compatible with ( n × p ) matrix B, thus yielding a ( p × p ) matrix CABthat possesses a trace for being square – and equal to that of BCA, despite CABbeing distinct from ( n × n ) matrix BCA. Note that the next similar move of swapping the first factor (i.e. C) to the last position without modifying the sequence of the other two (i.e. AB) would transform tr { CAB} to tr { ABC} again – so tr { CAB} = tr { ABC} would close the cycle with Eqs. (4.100)and (4.101).

A particular situation covered by Eq. (4.100)pertains to an ( n × m ) matrix Aand an ( m × n ) matrix C, together with I mplaying the role of matrix B; the products ABCand BCAin Eq. (4.100)look like

(4.102)

which degenerates to

(4.103)

at the expense of Eqs. (4.61)and (4.64). Therefore, the trace of the product of two matrices remains unchanged when the said matrices switch positions (should that be compatible with multiplication).

Recall again matrix A, as defined by Eqs. (4.1)and (4.2); if generic element a i,j, initially located in the i th row and j th column, were swapped with element a j,i, initially located in the j th row and i th column, then a transpose matrix would result – given by