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 $\rho \left({\mathbf A}\right)$ be the rank of $\mathbf A$.

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

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

We have $\rho \left({\mathbf A}\right) = \rho \left({u}\right)$ and $\rho \left({\mathbf A^t}\right) = \rho \left({u^t}\right)$, but $\rho \left({u^t}\right) = \rho \left({u}\right)$ from Rank and Nullity of Transpose.

Sources

• Seth Warner: Modern Algebra (1965): $\S 29$: Theorem $29.7$