Equivalence of matrices I can find very little on these matrices, from what I understand they're not actually special matrices, they just have the same dimensions of their images (ranks), thus can be thought of changing the basis. Which is great! Unfortunately I must try and find out (for homework) I confess if two matrices are the "equivalent" however, I have gotten "no" for both, a friend got "no" then "yes"
Rather than write them out I just want to confirm something, I am more thrown off by getting both being "no" and contradicting a friend.
If I get them into reduced row echelon form, which I know to be unique, and the forms differ does that mean that they are not equiv? 
Will column operations affect this form (I don't think they would, a simple proof or a link would be nice) 
Now for the embarrassing question, which is why I am asking, it seems the two matrices have the same rank (looking at their reduced forms) is there a link I've been too daft to spot? Or for some reason being ignorant of?

Definition of equivalent:
Theorem 11.5. Let A and B be m × n matrices over K. Then the following condi-
tions on A and B are equivalent.
(i) A and B are equivalent.
(ii) A and B represent the same linear map with respect to different bases.
(iii) A and B have the same rank.
(iv) B can be obtained from A by application of elementary row and column opera-
tions.
 A: No. Example: $A=\begin{pmatrix}1&0\\0&0\end{pmatrix}$ and $B=\begin{pmatrix}0&1\\0&0\end{pmatrix}$. Both of them are already in RREFs and they are different. However, they are equivalent because they have the same rank (condition iii). For this notion of matrix equivalence, $A$ and $B$ are equivalent if they can be brought into the same form that is both row reduced and column reduced (condition iv). If you interchange the two columns of $B$ (this is an elementary column operation), you get $A$, and $A$ is in both RREF and CREF. So condition (iv) is satisfied.
In general, condition (iv) means that by left- and right-multiplying some elementary matrices $E_i$s and $F_j$s, we have $(E_rE_{r-1}\cdots E_1)A(F_1F_2\cdots F_s)=\begin{pmatrix}I_k\\&0_{(n-k)\times(n-k)}\end{pmatrix}$ for some $k$ (which is the rank of $A$), and similarly for $B$ (by using perhaps a different number of different elementary matrices, but for the same $k$). So, two matrices $A$ and $B$ of the same size are equivalent iff $A=PBQ$ for some invertible matrices $P$ and $Q$.
Finally, let $\mathcal{A}=\left\{e_1=(1,0)^T,\,e_2=(0,1)^T\right\}$ denotes the canonical basis of $\mathbb{R}^2$. Define a linear map $f:\mathbb{R}^2\to\mathbb{R}^2$ by $f(e_1)=e_1$ and $f(e_2)=0$. Then $A$ is the matrix of $f$ w.r.t. the canonical basis. Now let $\mathcal{B}$ be the ordered basis $\{e_2,e_1\}$. Then $B$ is the matrix of $f$ w.r.t the bases $\mathcal{A}$ (for the domain of $f$) and $\mathcal{B}$ (for the range of $f$). This illustrates that condition (ii) is also satisfied.
