$A$ is (nxn) full-rank matrix. $B$ is (nxm) matrix with rank k. I want to prove analytically that $$rk(AB) = k$$
I am not sure if my answer is correct or not, but here it is:
As vector space spanned by $B$’s columns is $k$-dimensional, we can express any $k$-dimensional vector (e.g. $s$) as a linear combination of columns of B:
$$s=Bx$$ with $s$ -> (nx1); $B$ -> (nxm); $x$ -> (mx1)
$x$ here is a (mx1) vector of coefficients of the linear combination. By notation $s$ is (nx1) vector, but it is only k-dimensional (only k entries are non-zero), as this vector is the linear combination of $B$’s columns, from which only $k$ columns are linearly independent from each other.
As matrix $A$ is (nxn) and full-rank, it hast n linearly independent columns. Then if we multiply $A$ by vector $s$, then $As$ will be a linear combination of only $k$ (out of n) linearly independent columns of $A$, as (n-k) entries of vector $s$ are zero.
Now again any k-dimensional vector can be represented by $As$.
$$As = A(Bx) = (AB)x$$
By that we prove that vector spaces generated by columns of $AB$ and by columns of $B$ arethe same. Subsequently, their dimensions are the same and that is the definition of rank, so
$$rk(B) = k = rk(AB)$$
Any help would be greatly appreciated.
This statement doesn't make any sense. The columns of $B$ are in $R^n$. The columns span a $k$-dimensional subspace of $\mathbb{R^n}$. It doesn't make any sense to represent a "k-dimensional vector" in terms of vectors in $R^n$.