Row rank equals column rank

Statement

Let $F$ be a field and let $A$ be an $m \times n$ matrix over $F$. Then the dimension of the row space of $A$ equals the dimension of the column space of $A$. Equivalently, $\rank A^T = \rank A$, where $A^T$ is the transpose.

Remarks

One proof uses row reduction: row operations keep the row space and the dimension of the column space. In reduced row echelon form, both dimensions equal the number of pivots. Another proof uses the annihilator: the dual map has the transposed matrix and the same rank. Over $\C$, also $\rank A^* = \rank A$.

Explanations

L1 human

The number of independent rows of a matrix always equals the number of independent columns. This is surprising, because rows and columns live in different spaces. For a $3 \times 5$ matrix, it shows at once that the rank is at most $3$.