If a and B of the same type have the same rank, then R (a) = R (b) = R (a,

If a and B of the same type have the same rank, then R (a) = R (b) = R (a,

no, it isn't.

It is proved that for any real matrix A, R (ATA) = R (AAT) = R (a)

If you know singular value decomposition, then the conclusion is obvious. If you don't know, do it like this: if R (a) = k, then you can use Gauss elimination method to eliminate a into a echelon matrix, that is, CA = u, where C is the product of row elementary transformation, and u has only the first k rows which are non-zero and linearly independent

Let a be m * n real matrix, and prove that R (a'a) = R (AA ') = R (a)
A 'is the transpose matrix of A

If X1 is a'ax = 0, then a'ax1 = 0, so x1'a'ax1 = 0, so (ax1) '(ax1) = 0, so there is ax1 = 0, that is, the solution of a'ax = 0 is AX = 0, so AX = 0 and a'ax = 0 have the same solution