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 $$\mathbf A \equiv \mathbf B$$.