Given this proposition
Let $f: U → V$ be a linear function. Then there exists basis of $U$ and $V$ such that the matrix associated with f is of the form $$ \pmatrix{I_r & 0\\0 & 0} $$ where $I_r$ is identity matrix and r is the range of $f$.
Show that
Let $I^{m, n}_r$ be a $m\times n$ matrix whose first $r$ lines are the canonical base of $K^r$ and whose remaining lines are zero, i.e., $$ I^{m, n}_r = \pmatrix{I_r & 0\\0 & 0} $$ Prove that, if $A, B \in M_{m × n}(K)$ then $A$ is equivalent to $I^{m, n}_r \iff rank(A)=r$. Prove that $A$ is equivalent to $B$ iff their ranks are equal.
My attempt:
$\Rightarrow$ Let A be a matrix associated to a linear transformation $T:U\rightarrow V$, then, by the proposition, there exists basis of $U$ and $V$ such that $$ A =\pmatrix{I_r & 0\\0 & 0} $$ where $I_r$ is identity matrix and $rank(T)=r \implies rank(A)=r$.
$\Leftarrow$ We know that $rank(A)=r \implies \exists$ a linear transformation $T:U\rightarrow V$ s.t. $rank(T)=r$, so by the proposition, there exists basis of $U$ and $V$ such that $$ A =\pmatrix{I_r & 0\\0 & 0} $$ So clearly A is equivalent to $I^{m, n}_r$
Now, to prove that $A$ is equivalent to $B$ iff their ranks are equal, just make $B = I^{m, n}_r$
Is it right?
Hint: The key to the second part is to observe that if $A$ is equivalent to $C$ and $C$ is equivalent to $B$, then $A$ is equivalent to $B$.