Rank Plus Nullity Theorem

Theorem

Let $G$ be an $n$-dimensional vector space.

Let $H$ be a vector space.

Let $\phi: G \to H$ be a linear transformation.

Let $\map \rho \phi$ and $\map \nu \phi$ be the rank and nullity respectively of $\phi$.

Then the image of $\phi$ is finite-dimensional, and:

$\map \rho \phi + \map \nu \phi = n$

By definition of rank and nullity, it can be seen that this is equivalent to the alternative way of stating this result:

$\map \dim {\Img \phi} + \map \dim {\map \ker \phi} = \map \dim G$

Proof

If $\phi = 0$ then the assertion is clear.

Let $\phi$ be a non-zero linear transformation.

By Dimension of Proper Subspace is Less Than its Superspace and Generator of Vector Space Contains Basis, there is an ordered basis $\sequence {a_n}$ of $G$ such that:

$\exists r \in \N_n: \set {a_k: r + 1 \le k \le n}$ is a basis of $\map \ker \phi$

As a consequence:

$\map \nu \phi = n - r$

and by Unique Linear Transformation Between Vector Spaces:

$\map \rho \phi = r$

The result follows.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.5$: Corollary