Does the rank-nullity theorem hold for infinite dimensional $V$?

Solution 1:

The rank formula also holds in infinite dimensions, whether you use cardinal arithmetic for the dimensions, or just say $\infty + n = \infty$, and $\infty + \infty = \infty$ (but one should use cardinal arithmetic). The proof is basically the same as in the finite-dimensional case, you choose a basis $\mathcal{B}_1$ of $\ker T$, a basis $\mathcal{B}_2$ of $\operatorname{im} T$, let $\mathcal{B}_3$ consist of preimages of the elements of $\mathcal{B}_2$ (choose one preimage per element), then $\mathcal{B}_1 \cup \mathcal{B}_3$ is a basis of $V$. In the infinite-dimensional case, some form of the axiom of choice is required, while the finite-dimensional case can be proved without that.