Equivalent Matrices have Equal Rank

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

Let $\map \phi {\mathbf A}$ denote the rank of $\mathbf A$.

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

Then:
 * $\mathbf A \equiv \mathbf B$


 * $\map \phi {\mathbf A} = \map \phi {\mathbf B}$
 * $\map \phi {\mathbf A} = \map \phi {\mathbf B}$

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 $\sequence {a_n}$ of $S$ and $\sequence {b_m}$ of $T$.

Let $\psi: K^m \to T$ be the isomorphism defined as:
 * $\ds \map \psi {\sequence {\lambda_m} } = \sum_{k \mathop = 1}^m \lambda_k b_k$

Then $\psi$ takes the $j$th column of $\mathbf A$ into $\map u {a_j}$.

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

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

Now let $\map \LL {K^n, K^m}$ be the set of all linear transformations from $K^n$ to $K^m$.

Let $u, v \in \map \LL {K^n, K^m}$ 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 $r = \map \phi {\mathbf A}$.

By Linear Transformation from Ordered Basis less Kernel, there exist ordered bases $\sequence {a_n}, \sequence {a'_n}$ of $K^n$ such that:
 * $\sequence {\map u {a_r} }$ and $\sequence {\map v {a'_r} }$ are ordered bases of $\map u {K^n}$ and $\map v {K^n}$ respectively

and such that:
 * $\set {a_k: k \in \closedint {r + 1} n}$ and $\set {a'_k: k \in \closedint {r + 1} n}$ are respectively bases of the kernels of $u$ and $v$.

Thus, by Results concerning Generators and Bases of Vector Spaces there exist ordered bases $\sequence {b_m}$ and $\sequence {b'_m}$ of $K^m$ such that $\forall k \in \closedint 1 r$:

Let $z$ be the automorphism of $K^n$ which satisfies $\forall k \in \closedint 1 n: \map z {a'_k} = a_k$.

Let $w$ be the automorphism of $K^m$ which satisfies $\forall k \in \closedint 1 m: \map w {b'_k} = b_k$.

Then:
 * $\map {\paren {w^{-1} \circ u \circ z} } {a'_k} = \begin{cases}

\map {w^{-1} } {b_k} = b'_k = \map v {a'_k} & : k \in \closedint 1 r \\ 0 = \map v {a'_k} & : k \in \closedint {r + 1} n \end{cases}$

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

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

Then $\mathbf P$ and $\mathbf Q$ are invertible and:
 * $\mathbf Q^{-1} \mathbf A \mathbf P = \mathbf B$

and thus:
 * $\mathbf A \equiv \mathbf B$