In the vector space V with the base vectors e1, ··· , en, the vectors
cover a partial space T, the dimension of which, as we have seen in the preceding Section, equals the rank r of the matrix A = (aik). Form now the linear combinations
of the a1,··· , am, with the coefficient matrix B = (bli). Obviously, all the vectors b1,··· , bs belong to the vector space T and cover there a certain partial space T', which can also coincide with T. At the most, the dimension r' of T' can be as large as the dimension r of T:
r' 
r.
(24.2)
By (24.1), the vectors b1,··· , bs can be expressed with the aid of e1,··· , en:
The (s,n) matrix C = (cik) is then the product
C = BA.
The dimension r' of T' can be computed directly, following the results of Section 23:
r' = Rank C.
Hence (24.2) yields
Rank BA
Rank A. (24.3)
B can here be any matrix for which the product BA exists. Changing in (24.3) to the transposed matrix, one finds from (23.1)
Rank A'B'
Rank A'. (24.3)
Since you can view any matrix as a transposed one, it has been proved with full generality that the Rank of a matrix can at the most become smaller as you multiply it from the left or the right by any matrix. Thus, according to (24.3):
Rank PA Rank
A,
and moreover,
according to (24.4), Rank PAQ Rank PA,
whence follows the Rank equation
Rank PAQ Rank
A.
You obtain again the inequalities, which correspond to (24.3) and (24.4) for multiplication from the left or the right, by setting Q = E or P = E.
In particular, if the matrices P and Q multiplying A are regular, the above inequality can become an equation. First of all,
Rank PA Rank
A. (24.5)
Since, if P is regular, there exists its inverse matrix P-1 and you find from P-1PA = A that
Rank A
= Rank P-1PA = Rank P-1(PA)
Rank PA. (24.6)
Equations (24.5) and (24.6) yield Rank PA = Rank A. In the same manner, you prove that Rank AQ = Rank A if Q is a regular matrix. These two statements yield the Rank equation
Rank PAQ = Rank A.
The following example shows that the Rank can become less in the case of non-regular matrices:
.