Equivalent Matrices have Equal Rank

Theorem
Let $\mathbf A$ and $\mathbf B$ be $m \times n$ matrices over a field $K$.

Let $\phi \left({\mathbf A}\right)$ be the rank of $\mathbf A$.

Let $\mathbf A \equiv \mathbf B$ denote that $\mathbf A$ and $\mathbf B$ are equivalent.

Then $\mathbf A \equiv \mathbf B$ iff $\phi \left({\mathbf A}\right) = \phi \left({\mathbf B}\right)$.

Proof
Let $\mathbf A$ and $\mathbf B$ be $m \times n$ matrices over a field $K$ such that $\mathbf A \equiv \mathbf B$.

Let $S$ and $T$ be vector spaces of dimensions $n$ and $m$ over $K$.

Let $\mathbf A$ be the matrix of a linear transformation $u: S \to T$ relative to the ordered bases $\left \langle {a_n} \right \rangle$ of $S$ and $\left \langle {b_m} \right \rangle$ of $T$.

Let $\psi: K^m \to T$ be the isomorphism defined as $\psi \left({\left \langle {\lambda_m} \right \rangle}\right) = \sum_{k=1}^m \lambda_k b_k$.

Then $\psi$ takes the $j$th column of $\mathbf A$ into $u \left({a_j}\right)$.

Hence it takes the subspace of $K^m$ generated by the columns of $\mathbf A$ onto the codomain of $u$.

Thus $\rho \left({\mathbf A}\right) = \rho \left({u}\right)$, and equivalent matrices over a field have the same rank.


 * Now let $\mathcal L \left({K^n, K^m}\right)$ be the set of all linear transformations from $K^n$ to $K^m$.

Let $u, v \in \mathcal L \left({K^n, K^m}\right)$ such that $\mathbf A$ and $\mathbf B$ are respectively the matrices of $u$ and $v$ relative to the standard ordered bases of $K^n$ and $K^m$.

Let $r = \phi \left({\mathbf A}\right)$.

By Linear Transformation from Ordered Basis less Kernel, there exist ordered bases $\left \langle {a_n} \right \rangle, \left \langle {a'_n} \right \rangle$ of $K^n$ such that:
 * $\left \langle {u \left({a_r}\right)} \right \rangle$ and $\left \langle {v \left({a'_r}\right)} \right \rangle$ are ordered bases of $u \left({K^n}\right)$ and $v \left({K^n}\right)$ respectively, and such that:
 * $\left\{{a_k: k \in \left[{r+1 \, . \, . \, n}\right]}\right\}$ and $\left\{{a'_k: k \in \left[{r+1 \, . \, . \, n}\right]}\right\}$ are respectively bases of the kernels of $u$ and $v$.

Thus, by Results concerning Generators and Bases of Vector Spaces there exist ordered bases $\left \langle {b_m} \right \rangle$ and $\left \langle {b'_m} \right \rangle$ of $K^m$ such that $\forall k \in \left[{1 \,. \, . \, r}\right]$:

Let $z$ be the automorphism of $K^n$ which satisfies $\forall k \in \left[{1 \,. \, . \, n}\right]: z \left({a'_k}\right) = a_k$.

Let $w$ be the automorphism of $K^m$ which satisfies $\forall k \in \left[{1 \,. \, . \, m}\right]: w \left({b'_k}\right) = b_k$.

Then $\left({w^{-1} \circ u \circ z}\right) \left({a'_k}\right) = \begin{cases} w^{-1} \left({b_k}\right) = v \left({a_k}\right) & : k \in \left[{1 \,. \, . \, r}\right] \\ 0 = v \left({a_k}\right) & : k \in \left[{r + 1 \,. \, . \, n}\right] \end{cases}$.

So $w^{-1} \circ u \circ z = v$.

Now let $\mathbf P$ be the matrix of $z$ relative to the standard ordered basis of $K^n$, and let $\mathbf Q$ be the matrix of $w$ relative to the standard ordered basis of $K^m$.

Then $\mathbf P$ and $\mathbf Q$ are invertible and $\mathbf Q^{-1} \mathbf A \mathbf P = \mathbf B$, and thus $\mathbf A \equiv \mathbf B$.