Column Rank of Matrix equals Row Rank/Proof using Rank Factorization
Theorem
Let $\mathbf A$ be a matrix.
The column rank of $\mathbf A$ is equal to the row rank of $\mathbf A$.
Proof
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Let $\map c {\mathbf A}$ denote the column rank of $\mathbf A$.
Let $\map r {\mathbf A}$ denote the row rank of $\mathbf A$.
Let $\mathbf A$ be an $m \times n$ matrix whose column rank is $r$. Therefore, the dimension of the column space of $\mathbf A$ is $r$. Let $\mathbf c_1, \ldots, \mathbf c_r$ be any basis for the column space of $\mathbf A$. Place these vectors as column vectors to form the $m\times r$ matrix $\mathbf C = \begin {bmatrix} \mathbf c_1 & \mathbf c_2 & \cdots & \mathbf c_r \end{bmatrix}$. Therefore, every column vector of $\mathbf A$ is a linear combination of the columns of $\mathbf C$. To be precise, if $\mathbf A = \begin {bmatrix} \mathbf a_1 & \mathbf a_2 & \cdots & \mathbf a_n \end {bmatrix}$ is an $m \times n$ matrix with $\mathbf a_j$ as the $j$-th column, then
- $\mathbf a_j = f_{1j} \mathbf c_1 + f_{2j} \mathbf c_2 + \cdots + f_{rj} \mathbf c_r\quad \mbox{for}\quad j=1,\ldots,n$
where $f_{ij}$ are the scalar coefficients of $\mathbf a_j$ in terms of the basis $\mathbf c_1, \mathbf c_2, \ldots, \mathbf c_r$. This implies that $\mathbf A = \mathbf C \mathbf F$, where $f_{ij}$ is the $\tuple {i, j}$th element of $\mathbf F$.
Using transposition of products, we note that $\mathbf A^\intercal = \mathbf F^\intercal \mathbf C^\intercal$. From the definition of matrix multiplication, this means that each column of $\mathbf A^\intercal$ is a linear combination of the columns of $\mathbf F^\intercal$. Therefore, the column space of $\mathbf A^\intercal$ is a subspace of the column space of $\mathbf F^\intercal$ and, hence, $\map c {\mathbf A^\intercal} \leq \map c {\mathbf F^\intercal}$. Now, $\mathbf F^\intercal$ is $n \times r$, so there are $r$ columns in $\mathbf F^\intercal$ and, hence, $\map c {\mathbf A^\intercal} \leq r = \map c {\mathbf A}$. This proves that $\map c {\mathbf A^\intercal} \leq \map c {\mathbf A}$. Now apply the result to the matrix $\mathbf A^\intercal$ to obtain the reverse inequality: since $\paren {\mathbf A^\intercal}^\intercal = \mathbf A$, we can write $\map c {\mathbf A} = \map c {\paren {\mathbf A^\intercal}^\intercal} \leq \map c {\mathbf A^\intercal}$. This proves $\map c {\mathbf A} \leq \map c {\mathbf A^\intercal}$. We have, therefore, proved $\map c {\mathbf A^\intercal} \leq \map c {\mathbf A}$ and the reverse inequality $\map c {\mathbf A} \leq \map c {\mathbf A^\intercal}$. Therefore, $\map c {\mathbf A} = \map c {\mathbf A^\intercal}$.
$\blacksquare$