If a matrix has full rank, then its transpose also has full rank.
The product of two invertible matrices is invertible.
I don't think that quite answers the question! The OP didn't say they are invertible, just of "full rank". A square matrix is invertible if and and only if it has full rank but I imagine this problem is for non-square matrices.
If the matrix A is, say, m by n, then we can think of it as representing a linear transformation from a vector space of dimension n to a vector space of dimension m. If A has full rank, then m> n and A maps the vector space of dimension n one-to-one onto a subspace of dimension n.
Assuming that these are real matrices.
In general . To prove this we first note that in general so it suffices to prove that . But, for this (by first isomorphism theorem [the rank-nullity theorem if that previous theorem is unfamiliar to you]) it is sufficent to prove that . Evidently and the reverse inclusion follows by noticing that if then and thus and thus .