# Rank is Dimension of Subspace

## Theorem

Let $K$ be a field.

Let $\mathbf A$ be an $m \times n$ matrix over $K$.

Then the rank of $\mathbf A$ is the dimension of the subspace of $K^n$ generated by the rows of $\mathbf A$.

## Proof

Let $u: K^n \to K^m$ be the linear transformation such that $\mathbf A$ is the matrix of $u$ relative to the standard ordered bases of $K^n$ and $K^m$.

Let $\map \rho {\mathbf A}$ be the rank of $\mathbf A$.

Let $\mathbf A^\intercal$ be the transpose of $\mathbf A$.

Similar notations on $u$ denote the rank and transpose of $u$.

We have:

$\map \rho {\mathbf A} = \map \rho u$

and:

$\map \rho {\mathbf A^\intercal} = \map \rho {u^\intercal}$

but from Rank and Nullity of Transpose:

$\map \rho {u^\intercal} = \map \rho u$

$\blacksquare$