$T:V\to W$ How these vector space and matrix form equivalent I was reading Artin Algebra. Thwre was thia proposition.
Prove that these two statements are equivalent
Vector spaces form: Let $T:V\to W$ be a linear transformation between finite dimensional vector spaces. There are bases $\mathbf{B}$ and $\mathbf{C}$ of $V$ and $W$, respectively, such that the matrix of T with respect to these bases has the form
$$A'=\begin{array}{|c|c|}
\hline
I_r & 0\\
\hline
0&0\\
\hline
\end{array}$$
where $I_{r}$ is the $r \times r$ identity matrix and  $r $ is the rank of $T$.
Matrix form: Given an $m \times n$ matrix $A$, there are invertible matrices $Q$ and $P$ such that $A^{\prime}=Q^{-1} A P$ has the form shown above.
How these two are equivalent. It is written there but too short. Please help me to understand
I understood that $Q$ is product of row operation matrices to get to the row reduced echelon form but why the final form will be this
 A: In the first statement we can write explicitly what it means for $A'$ to represent $T$. Write $B = \{v_1, \dots, v_n\}$. We have $$A' = J_C^{-1}TJ_B,$$
where $J_B \colon \mathbb{F}^n \to V$ is the isomorphism that maps $B$-coordinates to vectors in $V$ defined by $J_B e_j = v_j$ on the standard basis vectors $e_j$, and $J_C$ is defined in the same way. Thus
$$T = J_C A' J_B^{-1}$$
The second statement is a special case of the first statement since an $m \times n$ matrix $A$ is identified with a linear transformation $A \colon \mathbb{F}^n \to \mathbb{F}^m$, so we can apply the first statement with $T = A$.
The first statement does follow from the second statement. Pick bases $B$ of $V$, $C$ of $W$. Then
$$T = J_C A J_B^{-1},$$
where $A$ is the matrix representation of $T$ with respect to bases $B$ and $C$. Then apply the second statement to $A$ to get
$$T = J_C Q A' P^{-1} J_B^{-1} = (J_C Q)A'(J_B P)^{-1}.$$
The desired bases are $\{J_BPe_1,\dots, J_BPe_n\}$ for $V$ and $\{J_C Qe_1, \dots, J_C Qe_m\}$ for $W$.
