[proofplan] We reduce $A$ to its canonical form $\begin{pmatrix} I_r & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{pmatrix}$ using the [Canonical Form for Linear Maps](/theorems/388) and observe that this form has row rank $= $ column rank $= r$. We then show that left-multiplication by an invertible matrix preserves row rank and right-multiplication by an invertible matrix preserves column rank, so the original matrix $A$ inherits equality of row and column rank from its canonical form. [/proofplan] [step:Observe that the canonical form has equal row and column rank] By the [Canonical Form for Linear Maps](/theorems/388), there exist invertible matrices $P \in \mathrm{GL}_m(\mathbb{F})$ and $Q \in \mathrm{GL}_n(\mathbb{F})$ such that $Q^{-1}AP = \begin{pmatrix} I_r & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{pmatrix}$, where $r = \mathrm{rank}(A)$. The canonical form has exactly $r$ non-zero rows (the first $r$ standard basis vectors, extended with zeros) and exactly $r$ non-zero columns. Both the row space and the column space have dimension $r$. [/step] [step:Show right-multiplication by an invertible matrix preserves column rank] [claim:Right Mult Preserves Column Rank] If $P \in \mathrm{GL}_m(\mathbb{F})$, then the column space of $AP$ equals the column space of $A$. [/claim] [proof] The columns of $AP$ are $\{Ap_1, \dots, Ap_m\}$ where $p_i$ denotes the $i$th column of $P$. Since $P$ is invertible, $\{p_1, \dots, p_m\}$ is a basis for $\mathbb{F}^m$. Every vector $x \in \mathbb{F}^m$ can be written as $x = \sum_{i=1}^m c_i p_i$, so \begin{align*} Ax = \sum_{i=1}^m c_i Ap_i, \end{align*} showing that the column space of $A$ (spanned by $\{Ae_1, \dots, Ae_m\}$) is contained in the span of $\{Ap_1, \dots, Ap_m\}$. Conversely, each $Ap_i = A(Pe_i)$ is a linear combination of columns of $A$. Hence the column spaces of $A$ and $AP$ coincide. [/proof] [/step] [step:Show left-multiplication by an invertible matrix preserves row rank] By a symmetric argument applied to rows (which are columns of the transpose), left-multiplication by an invertible matrix $Q^{-1}$ preserves the row space: the rows of $Q^{-1}A$ span the same subspace of $\mathbb{F}^m$ as the rows of $A$, because $Q^{-1}$ is invertible and acts by forming linear combinations of rows. [/step] [step:Conclude that row rank equals column rank for $A$] Since $Q^{-1}AP$ and $A$ have the same column rank (by the claim in the second step, applied to $Q^{-1}A$ and $(Q^{-1}A)P$, noting that $Q^{-1}$ preserves column rank by acting on rows) and the same row rank (by the third step), and the canonical form $Q^{-1}AP$ has row rank $=$ column rank $= r$, both the row rank and column rank of $A$ equal $r$. [guided] Let us trace the argument carefully. Start with $A$ and form $Q^{-1}AP$. First, right-multiplication: the column space of $AP$ equals the column space of $A$ (by the claim), so $\dim(\text{col space of } AP) = \dim(\text{col space of } A)$. Next, left-multiplication: the row space of $Q^{-1}(AP)$ equals the row space of $AP$ (by the symmetric argument for rows), and the row space of $AP$ equals the row space of $A$ (since multiplying on the right by $P$ does not change the row space -- each row of $AP$ is a linear combination of rows of $A$ weighted by entries of $P$, and invertibility of $P$ ensures the spans coincide). Therefore $A$ and $Q^{-1}AP$ have the same row rank and the same column rank. Since the canonical form $Q^{-1}AP = \begin{pmatrix} I_r & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{pmatrix}$ has row rank $r$ and column rank $r$, the row rank and column rank of $A$ are both $r$. [/guided] [/step]