Let $V$ and $W$ be vector spaces over the same field. If $\phi\colon V\to W$ is a linear mapping, then $$ \dim V = \dim(\ker\phi) + \dim(\im\phi). $$ In other words, the dimension of $V$ is equal to the sum of the rank and nullity of $\phi$

Note that if $U$ is a subspace of $V$ then this (applied to the canonical mapping $V\to V/U$ says that $$ \dim V = \dim U + \dim(V/U), $$ that is, $$ \dim V = \dim U + \codim U, $$ where $\codim$ denotes codimension.

An alternative way of stating the rank-nullity theorem is by saying that if $$ 0\to U\to V\to W\to 0 $$ is a short exact sequence of vector spaces, then $$ \dim(V) = \dim(U) + \dim(W). $$ In fact, if $$ 0\to V_1\to\cdots\to V_n\to 0 $$ is an exact sequence of vector spaces, then $$ \sum_{i=1}^{\floor{n/2}}V_{2i}=\sum_{i=1}^{\ceiling{n/2}}V_{2i-1}, $$ that is, the sum of the dimensions of even-numbered terms is the same as the sum of the dimensions of the odd-numbered terms.