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.
 A: One purely matrix-based definition of rank is decomposition rank: the rank of an $n\times m$-matrix is the smallest integer $r$ such that the matrix can be decomposed as product of an $n\times r$ and a $r\times m$ matrix.
It is now obvious that the rank of $AB$ cannot be larger than the rank of $A$, or than the rank of $B$ (a decomposition of $A$ or of $B$ immediately leads to a decomposition of $AB$). It is also clear that if $P,Q$ are invertible, then the rank of $Q^{-1}AP$ cannot be larger than that of $A$, nor can it be smaller, since the rank of $A=QQ^{-1}APP^{-1}$ would then be larger than that of $Q^{-1}AP$; ergo they are equal.
By the way, the relation between $A$ and $Q^{-1}AP$ is (unfortunately) called matrix equivalence. It is defined for rectangular matrices while similarity is only defined for square matrices; for square matrices similarity is much more restrictive than equivalence. So the result above is more general than what you mentioned.
A: If $n$  vectors $v_1,..v_n$ are linearly independent(dependent) then for non singular matrix $P$ the vectors $Pv_1,...Pv_n$ also  will be independent(dependent). From this fact and from the definition of the rank as a number of linearly independent columns (rows) we immediately can conclude that similar matrix have the same rank.  The second fact follows from  the dimension left and right kernels. 
A: 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
$$
\rank(AB)\le \min(\rank(A),\rank(B))
$$
