Rank is Dimension of Subspace
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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:
- $\map \rho {u^\intercal} = \map \rho u$
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices: Theorem $29.7$