Proofs about Matrix Rank I'm trying to prove the following two statements. I can prove them easily by considering the matrix as a representation of a linear map with a given basis, but I don't know a proof which uses just the properties of matrices.  
First, I want to prove that similar matrices have the same rank. This seems obvious because the rank is just the dimension of the image of the linear map, but these matrices represent the same map (just in a different basis).  
Next, I want to prove that $rank(AB)\le \min(rank(A),rank(B))$. Again, this seems relatively obvious if we just consider $AB$ as the composition of the two maps , but I can't see how to do it with matrices.

Matrix similarity: $\DeclareMathOperator{\rank}{rank}$
We say that two similar matrices $A,B$ are similar if $B = SAS^{-1}$ for some invertible matrix $S$.  In order to show that $\rank(A) = \rank(B)$, it suffices to show that $\rank(AS) = \rank(SA) = \rank(A)$ for any invertible matrix $S$.
To prove that $\rank (A) = \rank(SA)$: let $A$ have columns $A_1,\dots,A_n$.  We note that
$$
SA = S[A_1 \cdots A_n] = [SA_1 \cdots SA_n]
$$
Note that a subset of columns $\{A_{n_k}\}_{k=1}^r$ are linearly independent if and only if the columns $\{SA_{n_k}\}_{k=1}^r$ are as well.  The conclusion follows.
To prove $\rank(A) = \rank(AS)$, use the fact that $\rank(A) = \rank(A^T)$ to note that
$$
\rank(AS) = \rank((S^TA^T)^T) = \rank(S^TA^T) = \rank(A^T) = \rank(A)
$$
Thus, $\rank(SAS^{-1}) = \rank(SA) = \rank(A)$.
The inequality:
Let $B_i$ denote the columns of $B$.  We note that
$$
AB = [AB_1 \cdots AB_n]
$$
Now, if the columns $\{AB_{n_k}\}_{k=1}^r$ are linearly independent, then so are the columns $\{B_{n_k}\}_{k=1}^r$.  Thus, $\rank(AB) \leq \rank(B)$.
Now, note that by the above, 
$$
\rank(AB) = \rank(B^TA^T) \leq \rank(A^T) = \rank(A)
$$
So, $\rank(AB) \leq \rank(B)$ and $\rank(AB) \leq \rank(A)$.  We conclude that
$$
\rank(AB)\le \min(\rank(A),\rank(B))
$$
