Rank–nullity theorem for an m-by-n matrix

**Rank-Nullity Theorem:** If $A$ is an $m \times n$ matrix, then:

$$\text{rank}(A) + \text{nullity}(A) = n$$

**Proof:** The null space is a subspace of $\mathbb{R}^n$. Let $\{v_1, \ldots, v_k\}$ be a basis for the null space (so nullity $= k$). Extend this to a basis $\{v_1, \ldots, v_k, v_{k+1}, \ldots, v_n\}$ of $\mathbb{R}^n$. Then $\{Av_{k+1}, \ldots, Av_n\}$ is a basis for the column space, so rank $= n - k$.Copy