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 $\rho \left({\phi}\right)$ and $\nu \left({\phi}\right)$ be the rank and nullity respectively of $\phi$.

Then the image of $\phi$ is finite-dimensional, and $\rho \left({\phi}\right) + \nu \left({\phi}\right) = n$

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

$\dim \left({\operatorname {Im} \left({\phi}\right)}\right) + \dim \left({\ker \left({\phi}\right)}\right) = \dim \left({G}\right)$

Proof

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

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

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

$\exists r \in \N_n: \left\{{a_k: r + 1 \le k \le n}\right\}$ is a basis of $\ker \left({\phi}\right)$.

As a consequence, $\nu \left({\phi}\right) = n - r$ and by Unique Linear Transformation Between Vector Spaces, $\rho \left({\phi}\right) = r$.

The result follows.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 28$: Theorem $28.5$: Corollary