Tell me if this makes sense.
Let Tij mean the ijth element of matrix T.
Let Ti mean the ith row of matrix T. Ti = aug(Ti1, Ti2, ..., Tin).
A combo rule means "to get the new row i, take old row i, and add q times row j to it." In other words, given that F is the matrix obtained by applying a combo rule to C,
Fi = aug(Ci1, Ci2, ..., Cin) + q * aug(Cj1, Cj2, ..., Cjn)
= aug(Ci1, Ci2, ..., Cin) + aug(q*Cj1, q*Cj2, ..., q*Cjn)
= aug(Ci1 + q*Cj1, Ci2 + q*Cj2, ... Cin + q*Cjn).
Now we will discuss matrix multiplication, specifically multiplication against I. In matrix multiplication, given AB = R, n
Rij = Σ Aik * Bkj.
k=1
This is particularly easy to evaluate when A or B is I, since only Iii is non-zero (and, in fact, equals 1): n
Rij = Σ Iik * Bkj = 1*Bij.
k=1
Ri = aug(Ri1, Ri2, ..., Rin) = aug(Bi1, Bi2, ..., Bin) = Bi.
This is nearly as trivial to do when we've applied a combo rule to I to obtain E. The only thing that changes by letting Ei = Ii + q*ij is that Eij is changed to q. Therefore, n
Rij = Σ Eik * Ckj = 1*Cij + q*Cjj.
k=1
Therefore, Ri = aug(Ci1 + q*Cj1, Ci2 + q*Cj2, ..., Cin + q*Cjn), and all other rows are unchanged.
Since all other rows are unchanged in both this type of multiplication and a combo operation, it is sufficient to show that the changed rows are equal.
(Ready for some of my math teacher's odd little notation?) Trying to prove Fi = Ri, where R = EC.
LHS = Fi
= aug(Ci1 + q*Cj1, Ci2 + q*Cj2, ..., Cin + q*Cjn)
= Ri
= RHS
So the changed rows are equal, and thus F = EC.
Spencer |
