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 Results concerning Generators and Bases of Vector Spaces, 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 $$\mathrm{\ker} \left({\phi}\right)$$.

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

The result follows.